I have a problem that is giving me a headache. I'm fairly new to PHP and MySQL interacting, and coding in general (about 3 months since I first dived into it), so I'm still learning the ropes, so to speak.
The thing is, I have a page that receives data through $_POST from another. All is fine for well. The page receives the data, which is an ID number, and echoes the item's characteristics available in the database corresponding to that ID.
This part of the code works just fine.
$sql = "SELECT nome, preco, `status` FROM produtos WHERE id = $_POST[id]";
$result = mysqli_query($conn, $sql);
if (mysqli_num_rows($result) > 0){
while($row = mysqli_fetch_assoc($result)){
echo "Nome: "."<span style='color: red;'>".$row["nome"].
"</span>".
"<br>"."Preço: "."<span style='color: red;'>".
"R$".$row["preco"]."</span>"."<br><br>".$row["status"];
}
}
What I want, is to turn the values echoed into variables, so that I can set them as default value in another form and send that one to another page. Apparently, $row['nome'], for example, isn't available for re-use outside of the instance above.
<form method="post" action="?p=venda">
<input type="text" name="nome" value="<?php echo $row["nome"]; ?>">
<input type="text" name="preco" value="<?php echo $row["preco"]; ?>">
<input type="submit" name="change" value="Efetuar">
</form>
I know this code is prone to SQL injection, but I'm not looking into it right now. This will be a sort of offline program to help me with organizing some of my stuff, so, for now, I don't have to worry about security (not that I'll keep ignoring these issues).
Thanks in advance.
Assign $row to a variable and treat it as an array() outside the while loop.
$sql = "SELECT nome, preco, `status` FROM produtos WHERE id = $_POST[id]";
$result = mysqli_query($conn, $sql);
if (mysqli_num_rows($result) > 0){
while($row = mysqli_fetch_assoc($result)){
$array = $row; // assign $row a variable.
echo "Nome: "."<span style='color: red;'>".$row["nome"].
"</span>".
"<br>"."Preço: "."<span style='color: red;'>".
"R$".$row["preco"]."</span>"."<br><br>".$row["status"];
}
}
// example
echo($array['nome']);
OK I'll tell you this, its actually unnecessary to sanitize every post. Its only necessary if your authenticated users are web users and there is a blank that they type code in.
In closed production environments, where your users are, you can safe guard the environment in several ways, but this is off topic here.
anyways, you have a DB chart that has a an Id column, the item name, and I am guessing stock or production status.
your display, I don't use spans, but I'll show you how I do it with tables.
So lets make your query for your colums: nome, preco, status
Here are the two methods, the first one is called looped result method which is mostly used for more than one row in the db table:
<?php
//load it in a variable and trim the outside spaces away if you are entering this from a blank form
$id=trim($_POST[id]);
///this should look familiar to you
$sql = "SELECT nome, preco, status FROM produtos WHERE id ='".$id."'";
///but I use the shorthand method here to get my results.
$result = $conn->query($sql);
///Now we loop and shift colum information into variables
/// in a incremental loop, the colums are numbered starting with 0
echo "<table align=center bgcolor=fff2e2 border=1>\n";
while ($data = $result->fetch_row()) {
////I print my table header, and start the data row, if I want it as several ids I will reset here too (which I will do here if you want to play with this)
echo '<tr><th>Nome</th><th>Preco</th><th>Status</th></tr>';
echo '<tr>';
for ($m=0; $m<$result->field_count; $m++) {
if ($m==0){
$nome='';
$nome=$data[$m];
echo '<td bgcolor="#ff0000">'.$nome.'</td>';
} else if ($m==1){
$preco='';
$preco=$data[$m];
echo '<td bgcolor="#ff0000">'.$preco.'</td>';
}else if ($m==2){
$status='';
$status=$data[$m];
echo '<td bgcolor="#ffffff">'.$status.'</td>';
}
}
//////now if I was building a query chart with submit to another I would put my form and my hidden inputs here before i close the table row with /tr, and my submit button would be my first cell value
echo "</tr>";
}
echo "</table>";
You could do the regular colum /row method since it is one ID, which would look like this I used the span format here so you can get the idea of how html is typically expressed in php:
$id=trim($_POST[id]);
$sql = "SELECT nome, preco, status FROM produtos WHERE id ='".$id."'";
$result = mysqli_query($conn, $sql);
$row = mysqli_fetch_assoc($result);
$nome=stripslashes($row['nome']);
$preco=stripslashes($row['preco']);
$status=stripslashes($row['status']);
echo 'Nome: <span style="color: red;">'.$nome.'</span><br>Preço: <span style="color: red;">'.$preco.'</span><br><br>'.$status;
///notice my quote usage above
here is my table example if I was going to list the whole db table in a scrollable list, and submit it to a file called detail.php
<?php
$sql = "SELECT nome, preco, status FROM produtos";
$result = $conn->query($sql);
echo "<table align=center bgcolor=e3fab5 ><td>";
echo '<div style="width: 500px; height: 450px; overflow: auto; border 5px dashed black; background color: #ccc;">';
echo "<table align=center bgcolor=fff2e2 border=1>\n";
while ($data = $result->fetch_row()) {
echo '<tr><th>Nome</th><th>Preco</th><th>Status</th></tr>';
echo '<tr>';
for ($m=0; $m<$result->field_count; $m++) {
if ($m==0){
$nome='';
$nome=$data[$m];
echo '<td bgcolor="#ff0000"><form action="detail.php" method="post"><input type="submit" name="id" value="'.$nome.'"></td>';
} else if ($m==1){
$preco='';
$preco=$data[$m];
echo '<td bgcolor="#ff0000">'.$preco.'<input type="hidden" name="preco" value="'.$preco.'"></td>';
}else if ($m==2){
$status='';
$status=$data[$m];
echo '<td bgcolor="#ffffff">'.$status.'<input type="hidden" name="status" value="'.$status.'"></td>';
}
}
echo "</form></tr>";
}
echo "</table>";
echo "</div></table>";
?>
Related
So I have been working on this code for a while. I believe I am really close. My if statement that is inside my while loop isn't showing any data in the area it's suppose to show. I know mysql is old and deprecated. I am going to change it once I figure this out.
$result = mysql_query("SELECT * FROM inventoryTable",$db);
$result2 = mysql_query("SELECT * FROM users WHERE username='$username' and sub = 'yes'",$db);
echo "<TABLE style=\"background-color: #FFFFFF; border: 10px solid A4A4A4;\">";
echo"<TR><TD>"."<B>Title</B>"."</td>";
echo"<TD>"."<B>Authors First Name</B>"."</td>";
echo"<TD>"."<B>Authors Last Name</B>"."</td>";
echo"<TD>"."<B>ISBN</B>"."</td>";
echo"<TD>"."<B>Publisher</B>"."</td>";
echo"<TD>"."<B>Course Number</B>"."</td>";
echo"<TD>"."<B>Source</B>"."</td></TR>";
while ($myrow = mysql_fetch_array($result))
{
echo "<TR><TD>".$myrow["title"]."</td>";
echo "<TD>".$myrow["authorsFirst"]."</td>";
echo"<TD>".$myrow["authorsLast"]."</td>";
echo "<TD>".$myrow["ISBN"]."</td>";
echo "<TD>".$myrow["publisher"]."</td>";
echo "<TD>".$myrow["courseNum"]."</td>";
while ($subResults = mysql_fetch_row($result2))
{
If($subResults == 'yes' )
{
echo "<td>".$myrow["source"]."</td>";
} else {
echo "<TD>"."Please subscribe to View"."</td>";
}
echo "</TABLE>";
}
}
?>
This is the part of my code that isn't showing any results.
while
while ($myrow = mysql_fetch_array($result))
{
echo "<TR><TD>".$myrow["title"]."</td>";
echo "<TD>".$myrow["authorsFirst"]."</td>";
echo"<TD>".$myrow["authorsLast"]."</td>";
echo "<TD>".$myrow["ISBN"]."</td>";
echo "<TD>".$myrow["publisher"]."</td>";
echo "<TD>".$myrow["courseNum"]."</td>";
while ($subResults = mysql_fetch_row($result2))
{
If($subResults == 'yes' )
{
echo "<td>".$myrow["source"]."</td>";
} else {
echo "<TD>"."Please subscribe to View"."</td>";
}
echo "</TABLE>";
}
}
I want my session user to be able to see the source from my inventory table if they have a yes in the sub field. If they do not have a yes in the sub field, they will see please subscribe to view. Am i doing the mysql_fetch incorrectly or is there a problem because I have 2 while loops going on at once.
you need to have "==" to compare two values, otherwise you assign the second value to the first variable:
...If($username == $subResults)...
or to use a strict comparison of type and content, use "==="
If($username === $subResults)
also I am thinking the code should be
...If($subResults ==="yes"){echo"....///desired content";}else{echo"...//alternate content";}...
and you are missing the echo statement and closing </td> in the code
"<td>".$myrow["source"];
should be
echo"<td>".$myrow["source"]."</td>";
in fact - aren't you missing the closing td's in all of the cells?
I am currently running into an issue, where I have this form consisting of checkboxes. I get the values of user preferences for the checkboxes from a database. Everything works great, and does what is supposed to do, however after I change and check some boxes and then hit the submit button, it will still show the old values to the form again. If I click again in the page again it will show the new values.
The code is shown below with comments.
<form action="myprofile.php" method="post">
<?php $usr_cats=array();
$qry_usrcat="SELECT category_id_fk
FROM user_categories
WHERE user_id_fk='".$_SESSION['user_id']."';";
$result = mysqli_query($conn,$qry_usrcat);
while($row = mysqli_fetch_array($result)){
$usr_cats[] = $row[0]; // getting user categories from db stored in array
}
$query_allcats="SELECT category_id,category_name, portal_name
FROM categories
INNER JOIN portals on categories.portal_id=portals.portal_id
ORDER BY category_id;"; // select all category queries
$result = mysqli_query($conn,$query_allcats);
while($row = mysqli_fetch_array($result)){
echo $row['portal_name'] . "<input "; //print categories
if(in_array($row['category_id'], $usr_cats)){ // if in array from db, check the checkbox
echo "checked ";
}
echo "type='checkbox' name='categories[]' value='";
echo $row['category_id']."'> ". $row['category_name']."</br>\n\t\t\t\t\t\t";
}
?>
<input type="submit" name="submit" value="Submit"/>
<?php
$qry_del_usrcats="DELETE FROM user_categories
WHERE user_id_fk='".$_SESSION['user_id']."';"; //delete all query
if(isset($_POST['submit'])){
if(!empty($_POST['categories'])){
$cats= $_POST['categories'];
$result = mysqli_query($conn,$qry_del_usrcats); //delete all
for ($x = 0; $x < count($cats); $x++) {
$qry_add_usrcats="INSERT INTO `user_categories` (`user_id_fk`, `category_id_fk`)
VALUES ('".$_SESSION['user_id']."', '".$cats[$x]."');";
$result = mysqli_query($conn,$qry_add_usrcats);
}
echo "success";
}
elseif(empty($_POST['categories'])){ //if nothing is selected delete all
$result = mysqli_query($conn,$qry_del_usrcats);
}
unset($usr_cats);
unset($cats);
}
?>
I am not sure what is causing to do that. Something is causing not to update the form after the submission. However, as i said everything works great meaning after i submit the values are stored and saved in the DB, but not shown/updated on the form. Let me know if you need any clarifications.
Thank you
Your procedural logic is backwards and you're doing a bunch of INSERT queries you don't need. As #sean said, change the order.
<?php
if(isset($_POST['submit'])){
if(isset($_POST['categories'])){
$cats= $_POST['categories'];
// don't do an INSERT for each category, build the values and do only one INSERT query with multiple values
$values = '';
for($x = 0; $x < count($cats); $x++) {
// add each value...
$values .= "('".$_SESSION['user_id']."', '".$cats[$x]."'),";
}
// trim the trailing apostrophe and add the values to the query
$qry_add_usrcats="INSERT INTO `user_categories` (`user_id_fk`, `category_id_fk`) VALUES ". rtrim($values,',');
$result = mysqli_query($conn,$qry_add_usrcats);
echo "success";
}
elseif(!isset($_POST['categories'])){ //if nothing is selected delete all
// you may want to put this query first, so if something is checked you delete all, so the db is clean and ready for the new data.
// and if nothing is checked, you're still deleting....
$qry_del_usrcats="DELETE FROM user_categories WHERE user_id_fk='".$_SESSION['user_id']."';"; //delete all query
$result = mysqli_query($conn,$qry_del_usrcats);
}
unset($usr_cats);
unset($cats);
}
?>
<form action="myprofile.php" method="post">
<?php $usr_cats=array();
$qry_usrcat="SELECT category_id_fk FROM user_categories WHERE user_id_fk='".$_SESSION['user_id']."';";
$result = mysqli_query($conn,$qry_usrcat);
while($row = mysqli_fetch_array($result)){
$usr_cats[] = $row[0]; // getting user categories from db stored in array
}
$query_allcats="SELECT category_id,category_name, portal_name FROM categories INNER JOIN portals on categories.portal_id=portals.portal_id ORDER BY category_id;"; // select all category queries
$result = mysqli_query($conn,$query_allcats);
while($row = mysqli_fetch_array($result)){
echo $row['portal_name'] . "<input "; //print categories
if(in_array($row['category_id'], $usr_cats)){ // if in array from db, check the checkbox
echo "checked ";
}
echo "type='checkbox' name='categories[]' value='";
echo $row['category_id']."'> ". $row['category_name']."</br>\n\t\t\t\t\t\t";
}
?>
<input type="submit" name="submit" value="Submit"/>
Typically this occurs due to the order of your queries within the script.
If you want to show your updated results after submission, you should make your update or insert queries to be conditional, and have the script call itself. The order of your scripts is fine, but you just need to do the following:
Take this query:
$qry_del_usrcats="DELETE FROM user_categories
WHERE user_id_fk='".$_SESSION['user_id']."';"
and put it inside the if statement so it looks like this:
if (isset($_POST['submit'] {
$qry_del_usrcats="DELETE FROM user_categories
WHERE user_id_fk='".$_SESSION['user_id']."';"
$result = mysqli_query($conn,$qry_del_usrcats);
[along with the other updates you have]
}
Also, you will need to move this entire conditional above the form itself; typically any updates, inserts, or deletes should appear year the top of the form, and then call the selects afterward (outside of the conditional)
Struggling with a one page form that i want to first populate a form from a mysql query, then enable a user to update some of the values in each of several table rows from text input fields in the form.
The code's intent is to update the field by referencing the row ID.
But for some reason I'm only able to update the last row in the form (the last row from the array). I've included some troubleshooting code to see what the ID variable is and it always comes up as the last iteration of the array. I think I'm either overwriting the ID variable in the while loop or the for loop.
I have tried moving the POST/update to a 2nd file but get the same results. Any thoughts/suggestions would be greatly appreciated
Here is the code minus the db connection:
$saveClicked = $_POST["saveClicked"];
{ // SAVE button was clicked
if (isset($saveClicked)) {
unset($saveClicked);
$ID = $_POST["ID"];
$win = $_POST["Winner"];
$winScr = $_POST["WinnerScore"];
$losScr = $_POST["LoserScore"];
$tschedule_SQLupdate = "UPDATE tschedule SET ";
$tschedule_SQLupdate .= "Winner = '".$win."', ";
$tschedule_SQLupdate .= "WinnerScore = '".$winScr."', ";
$tschedule_SQLupdate .= "LoserScore = '".$losScr."' ";
$tschedule_SQLupdate .= "WHERE ID = '".$ID."' ";
if (mysql_query($tschedule_SQLupdate)) {
echo '<p> the number of mysql affected rows is '.mysql_affected_rows().'</p>';
echo 'this is the value for post id '.$ID;
} else {
echo '<span style="color:red; ">FAILED to update the game.</span><br /><br />';
echo mysql_error();
}
}
// END: SAVE button was clicked ie. if (isset($saveClicked))
}
{ // Get the details of all associated schedule records
// and store in array: gameArray with key >$indx
$indx = 0;
$tschedule_SQLselect = "SELECT * ";
$tschedule_SQLselect .= "FROM ";
$tschedule_SQLselect .= "tschedule ";
$tschedule_SQLselect .= "WHERE week = 1 ";
$tschedule_SQLselect_Query = mysql_query($tschedule_SQLselect);
while ($row = mysql_fetch_array($tschedule_SQLselect_Query, MYSQL_ASSOC)) {
$gameArray[$indx]['ID'] = $row['ID'];
$gameArray[$indx]['Date'] = $row['Date'];
$gameArray[$indx]['week'] = $row['week'];
$gameArray[$indx]['Favorite'] = $row['Favorite'];
$gameArray[$indx]['Line'] = $row['Line'];
$gameArray[$indx]['Underdog'] = $row['Underdog'];
$gameArray[$indx]['OU'] = $row['OU'];
$gameArray[$indx]['Winner'] = $row['Winner'];
$gameArray[$indx]['WinnerScore'] = $row['WinnerScore'];
$gameArray[$indx]['LoserScore'] = $row['LoserScore'];
$indx++;
}
$numGames = sizeof($gameArray);
mysql_free_result($tschedule_SQLselect_Query);
}
{ // Output
echo '<form name ="postGame" action="'.$thisScriptName.'" method="post">';
echo '<table border="1">';
echo '<tr>
<th>ID</th>
<th class="date">Date</th>
<th class="num">Week</th>
<th>Favorite</th>
<th class="num">Line</th>
<th>Underdog</th>
<th class="num">OU</th>
<th>Winner</th>
<th>WScore</th>
<th>LScore</th>
<th>Save</th>
</tr> ';
for ($indx = 0; $indx < $numGames; $indx++) {
$thisID = $gameArray[$indx]['ID'];
$saveLink = '<input type = "submit" value = "Save" />';
$fld_ID = '<input type="text" name="ID" value="'.$thisID.'"/>';
$fld_saveClicked = '<input type="hidden" name="saveClicked" value="1"/>';
echo $fld_ID;
echo $fld_saveClicked;
echo '<tr>
<td>'.$gameArray[$indx]['ID'].'</td>
<td>'.$gameArray[$indx]['Date'].'</td>
<td>'.$gameArray[$indx]['week'].'</td>
<td>'.$gameArray[$indx]['Favorite'].'</td>
<td>'.$gameArray[$indx]['Line'].'</td>
<td>'.$gameArray[$indx]['Underdog'].'</td>
<td>'.$gameArray[$indx]['OU'].'</td>
<td><input type="text" size =5 name="Winner">'.$gameArray[$indx]['Winner'].'</td>
<td><input type="number" size=5 name="WinnerScore">'.$gameArray[$indx]['WinnerScore'].'</td>
<td><input type="number" size=5 name="LoserScore">'.$gameArray[$indx]['LoserScore'].'</td>
<td>'.$saveLink.'</td>
</tr> ';
}
echo '</table>';
echo '</form>';
echo' View Schedule';
}
You're using the same names for each field in each row, so when you post the form, only the last is accessible. Use array notation for the fields like this:
<input type="text" size =5 name="Winner[]">
^^
This will give you an array for $_POST['Winner'] instead of a single value. Do the same for the other <input> elements.
Also, the code that processes the form after it's submitted only processes one value. You'll need to modify that to loop through these arrays.
Warnings:
don't use mysql_*() for new code - it's depracated. Switch to mysqli_*() or PDO now.
Your code is susceptible to SQL injection. Escape your input variables with mysql_real_escape_string() (or the mysqli equivalent) or better, switch to prepared statements.
After some more research I think I understand the two answers already shared much better. However I have chosen a different path and have resolved my issue -I wrapped the form tags directly around each row:
echo '<form name ="postGame'.$indx.'" action="'.$thisScriptName.'" method="POST">';
$fld_saveClicked = '<input type="hidden" name="saveClicked" value="1"/>';
echo $fld_saveClicked;
$fld_ID = '<input type="text" name="ID" value="'.$thisID.'"/>';
echo $fld_ID;
echo '<tr>
<td>'.$gameArray[$indx]['ID'].'</td>
<td>'.$gameArray[$indx]['Date'].'</td>
<td>'.$gameArray[$indx]['week'].'</td>
<td>'.$gameArray[$indx]['Favorite'].'</td>
<td>'.$gameArray[$indx]['Line'].'</td>
<td>'.$gameArray[$indx]['Underdog'].'</td>
<td>'.$gameArray[$indx]['OU'].'</td>
<td><input type="text" size=5 name="Winner" id="Winner">'.$gameArray[$indx]['Winner'].'</td>
<td><input type="number" size=5 name="WinnerScore" id="WinnerScore">'.$gameArray[$indx]['WinnerScore'].'</td>
<td><input type="number" size=5 name="LoserScore" id="LoserScore">'.$gameArray[$indx]['LoserScore'].'</td>
<td><button type="submit" />Save</button></td>
</tr></form>';
}
One of the key trouble shooting steps was to use var_dump to validate that the $_POST actually contained data. I understand there are several ways this could be done including the responses shared by Hobo and Syed, as well as using javascript, but was really glad I could accomplish my goal with just php.
Your For Loop is storing the last value of the array in your form.
$fld_ID = '<input type="text" name="ID" value="'.$thisID.'"/>';
Store the ID value as an array in HTML form and when a form is posted get all the ID values and update using your same mysql update query.
Your winner and loser score are also returning the last array values.
I creating a simple site with PHP where the users can submit blogs and other users (who are logged in) can post comments on them. I have made a link called "comments" below each blog that when clicked will show / hide all the comments relevant to the specific blog (also if the user is logged in, it will show a form field in which they can submit new comments). So basically each blog will have multiple comments. I have done two different codes for this but they both have the same problem that each comment appears twice (everything else works fine). Could anyone point out why?
mysql_select_db ("ooze");
$result = mysql_query ("select * from blog") or die(mysql_error());
$i = 1;
while($row = mysql_fetch_array($result))
{
echo "<h1>$row[title]</h1>";
echo "<p class ='second'>$row[blog_content]</p> ";
echo "<p class='meta'>Posted by .... • $row[date] • Comments<div id='something$i' style='display: none;'>";
$i++;
$a = $row["ID"];
$result2 = mysql_query ("select * from blog, blogcomment where $a=blogID") or die(mysql_error());
while($sub = mysql_fetch_array($result2))
{
echo "<p class='third' >$sub[commentdate] • $sub[username]</p><p>said:</p> <p>$sub[comment]</p>";
}
if ( isset ($_SESSION["gatekeeper"]))
{
echo '<form method="post" name="result_'.$row["ID"].'" action="postcomment.php"><input name="ID" type = "hidden" value = "'.$row["ID"].'" /><input name="comment" id="comment" type="text" style="margin-left:20px;"/><input type="submit" value="Add comment" /></form>';
}
else
{
echo '<p class="third">Signup to post a comment</p>';
}
echo "</div>";
}
mysql_close($conn);
//second version of inner loop://
if ( isset ($_SESSION["gatekeeper"]))
{
while($sub = mysql_fetch_array($result2))
{
echo "<p class='third' >$sub[commentdate] • $sub[username] said:</p> <p>$sub[comment]</p>";
}
echo '<form method="post" name="result_'.$row["ID"].'" action="postcomment.php"><input name="ID" type = "hidden" value = "'.$row["ID"].'" /><input name="comment" id="comment" type="text" style="margin-left:20px;"/><input type="submit" value="Add comment" /></form>';
}
else
{
while($sub = mysql_fetch_array($result2))
{
echo "<p class='third' >$sub[commentdate] • $sub[username] said:</p> <p>$sub[comment]</p>";
}
echo '<p class="third">Signup to post a comment</p>';
}
echo "</div>";
}
mysql_close($conn);
Your problem lies in this query from the first example.
$result2 = mysql_query ("select * from blog, blogcomment where $a=blogID")
You have already queried the blog table so there is no need to query it again. Simply changing this to
$result2 = mysql_query ("select * from blogcomment where $a=blogID")
should solve the problem.
However there are many things you need to think about.
Why are you re-inventing the wheel? There are plenty of good blog applications out there. You'd be better off using one of them.
It's not recommended to use the mysql_ family of functions any more. Go away and learn mysqli_ or better still PDO.
You should learn about separation of concerns. At the very least you should make sure your data access/business logic is separate from your display logic. MVC is very common in PHP.
You should also learn about JOINs. Even in this simple inline script you have a query within a loop which is not very efficient. You can combine your queries into one (as you've tried with the inner query). The difference is the one query should be outside your main loop.
I'm trying to display the new dynamic list by clicking dynamic list. Why do i call them dynamic list? Because the data is from database.
My idea is generating a list of companies, when i click one company, a list of all sites in the company is displayed; And then when i click the one site of one company, a list of all employees in the site is displayed.
Now i have met a problem. When i click any item in list of companies, a list of sites in the last item of company list shows. And when i click any item in the list of sites, a list of employees of last item in sites is showed.
Do you know why?
Here is the code and result image:
<script language="JavaScript">
function toggle(id,id2,id3) {
var state = document.getElementById(id).style.display;
if (state == 'block') {
document.getElementById(id).style.display = 'none';
if (id2 != undefined)document.getElementById(id2).style.display = 'none';
if (id3 != undefined)document.getElementById(id3).style.display = 'none';
} else {
document.getElementById(id).style.display = 'block';
}
}
</script>
<style type="text/css">
#main{
position:relative;
top:20px;
left:20px;
width:200px;
background: lightblue;
}
#hidden {
position:relative;
top:5px;
left:280px;
width:200px;
background: lightgrey;
display: none;
}
#hidden2 {
position:relative;
top:-12px;
left:580px;
width:200px;
background: lightgreen;
display: none;
}
#hidden3 {
position:relative;
top:100px;
left:20px;
width:200px;
background: lightpink;
display: none;
}
</style>
<?php
error_reporting(E_ALL ^ E_NOTICE);
include("./conn/connect.php");
$query = "SELECT * FROM entreprise ORDER BY id";
$result = mysql_query($query) or die("result failed: ".mysql_error());
?>
<div id="main" >
<?php
echo "<ul>";
while($row = mysql_fetch_assoc($result)){
echo "<li onclick=\"toggle('hidden','hidden2','hidden3');\">".$row['name'].'<li>';
$query2 = "SELECT * FROM site WHERE eid = '".$row['id']."'";
//$query2 = "SELECT * FROM site WHERE eid = ".$row['id'];
//$result2 = mysql_query($query2) or die("query2 result error".mysql_error());
$result2 = mysql_query($query2) or die("query2 result error".mysql_error());
}
echo "</ul>";
?>
</div>
<div id="hidden" >
<?php
echo "<ul>";
while($row2 = mysql_fetch_assoc($result2)){
echo "<li onclick=\"toggle('hidden2','hidden3')\">".$row2['name'].'< >';
$query3 = "SELECT * FROM salarie WHERE siteid =".$row2['id'];
//echo $query3;
$result3 = mysql_query($query3) or die("query3 result error".mysql_error());
}
echo "</ul>";
?>
</div>
<div id="hidden2" >
<?php
echo "<ul>";
while($row3 = mysql_fetch_assoc($result3)){
echo "<li onclick=\"toggle('hidden3')\">".$row3['prenom'].'< >';
$query4 = "SELECT * FROM salarie WHERE id =".$row3['id'];
$result4 = mysql_query($query4) or die("query4 result error".mysql_error());
}
echo "</ul>";
?>
</div>
<div id="hidden3">
<?php
echo "<table>";
while($row4 = mysql_fetch_assoc($result4)){
echo "<tr><td>".$row4['prenom'].'</td>';
echo "<td>".$row4['nom'].'</td></tr>';
}
echo "</table>";
?>
</div>
Result image:
Pretty simple: Your PHP code is executed ONCE when you access the site.
So for example the result of this block
while($row = mysql_fetch_assoc($result)){
echo "<li onclick=\"toggle('hidden','hidden2','hidden3');\">".$row['name'].'<li>';
$query2 = "SELECT * FROM site WHERE eid = '".$row['id']."'";
//$query2 = "SELECT * FROM site WHERE eid = ".$row['id'];
//$result2 = mysql_query($query2) or die("query2 result error".mysql_error());
$result2 = mysql_query($query2) or die("query2 result error".mysql_error());
}
is that $result2 holds all the sites of the last company in your list. This is then used in the next loop to generate the corresponding list of sites. Look at the source of the generated HTML file.
PHP is a server side language, the code is executed at the server and it is not re-executed by your Javascript functions (i.e. not executed in the browser).
What you are after is dynamically loading the data from your server with AJAX and pass into the generated HTML.
Edit:
You could also do it without Ajax: Rewrite your PHP like this:
$sitequeries = array()
while($row = mysql_fetch_assoc($result)){
echo "<li onclick=\"showSites('sites_$row['id']');\">".$row['name'].'<li>';
$query = "SELECT * FROM site WHERE eid = '".$row['id']."'";
$sitequeries[$row['id']] = mysql_query($query2 or die("query2 result error".mysql_error());
}
and
<?php
foreach($sitequeries as $id => $query) {
echo "<ul class='sites' id='sites_$id'>";
while($row2 = mysql_fetch_assoc($query)){
//...
}
echo "</ul>";
}
?>
This is not a working example but should give you the right idea. You have to adjust your JS accordingly to show only the corresponding lists, e.g.:
function showSites(id) {
// Hide all lists with class 'site' here and only display the list with id 'id' e.g. 'sites_5'
}
But note that this is not a good solution if you have a lot of companies, site, employes, etc. as the generation of the HTML may take a while. Then Ajax is a better choice.
Your toggle() Function needs 3 Parameters
You set on some places just 2 parameters
echo "<li onclick=\"toggle('hidden2','hidden3')\">".$row2['name'].'< >';
Shoud be
echo "<li onclick=\"toggle('hidden1','hidden2','hidden3')\">".$row2['name'].'< >';
I would do some things about your code:
Split the data acquiring stuff from the rest. At the beginning, get the data from the required tables and keep it in PHP variables. Then, do something with them using a JS framework... something according to your requirements
The problem with your approach is that you are NEVER telling anyone which row's ID should be sent... hence, it sends the id from the row selected by default, which happens to be the last one parsed by HTML parser on the browser. It means, the last one...
Your PHP code does not match your goal. $result2 will always be the last ID found in $result1, and so on.
If you need to generate result2 based on what the user selects in result1, then you need to either create rows for every possible selection then use javascript to show or hide, or utilize Ajax calls (much better).
Might wanna look into JQuery instead of doing it the way you currently are. But streetparade is correct.
Your approach to this task is a bit wrong, I think.
What do you do in the first loop? You setting $result2 variable and you want to access it in the next loop. And in next loop $result2 is set to the last record of first loop.
Have you heard about AJAX? jQuery may be?