I am having trouble making my drop down sticky (to make it so that after the form is submitted the first time, the selection chosen will be pre-selected in the form on the following page). I removed some code I deemed irrelevant. I tried making the value $_GET['continent'] but that didn't work. Does anyone have any thoughts? See function createpulldown
<!DOCTYPE html>
<head>
<title>Homework 14</title>
</head>
<body>
<?php
if (isset($_GET['submitted']))
handleform($_GET['country']);
displayform("country");
?>
</body>
function displayform($menuname) {
echo "<fieldset><legend>Select a continent and I will show you information from the CIA about it.</legend>
<form method = 'get'>";
createpulldown($menuname);
echo "<input type='submit' name='submitted' value='Search'>
</form>
</fieldset>";
}
function createpulldown($menuname) {
echo "<select name='$menuname'>";
$dbc = connectToDB();
$query = "SELECT Continent FROM countries GROUP BY Continent";
$result = performQuery($dbc, $query);
while ($row=mysqli_fetch_array($result, MYSQLI_ASSOC)){
$continent = $row['Continent'];
if (isset($_GET[$menuname]))
echo "<option name='continent' value = $continent selected>$continent</option>\n";
else
echo "<option name='continent' value = $continent>$continent</option>\n";
}
echo "</select>";
disconnectFromDB($dbc, $result);
}
?>
You want to check if the value retrieved from $_GET is equal to one of the option values.
Try this for your while statement:
while ($row=mysqli_fetch_array($result, MYSQLI_ASSOC)){
$continent = $row['Continent'];
if ($_GET[$menuname] == $continent)
echo "<option name='continent' value='$continent' selected>$continent</option>\n";
else
echo "<option name='continent' value='$continent'>$continent</option>\n";
}
I also fixed a syntax error. You need to wrap your option values in single quotes.
while ($row=mysqli_fetch_array($result, MYSQLI_ASSOC)){
$continent = $row['Continent'];
if (isset($_GET[$menuname]))
echo "<option name='continent' value = $continent selected>$continent</option>\n";
else
echo "<option name='continent' value = $continent>$continent</option>\n";
}
You are checking here, if a GET parameter named $menuname was passed to your script – but this is the same for all options you create, either the parameter is there o not.
What you want to do instead, is compare the value of the parameter to the current $continent value you are writing – if they are equal, select the current option, if not, don’t.
Related
I have a nested drop down choice where, basically, the second select input should update its content depending on first selected option.
I get it working ok when it open the page but did not exactly understood how to update the variable once the page is loaded. As a matter of fact if the last option of the SQL table in the first option match the name of one or more record in the second input those may correctly be selected but - changing the option - the second input box doesn’t update.
I tried some java but hadn’t succeeded, also thought about AJAX. Can any one help me with that?
My code is following, the $namev variable is the one, I guess, should need some correct action to be handled toward second input box:
<head>
…..
</head>
<body>
…..
<form>
…….
<div class="divFLOATleft">
<label>voce</label>
<?php
$result = $conn->query("select id_voce,id_gruppo_voce,gruppo_voce,voce FROM voci");
echo "<select name='voce'><option value='' disabled selected>scegli voce</option>";
while ($row = $result->fetch_assoc()) {
unset($idv, $namev);
$idv = $row['id_voce'];
$namev = $row['voce'];
$id_gv = $row['id_gruppo_voce'];
$id_v = $row['gruppo_voce'];
echo '<option value="'.$idv.'">'.$namev.'</option>';
}
echo "</select>";
echo"</div>";
echo"<div class='divFLOATleft'>";
echo"<label>categoria</label>";
$result = $conn->query("select id_voce,id_gruppo_voce,gruppo_voce,voce FROM voci WHERE gruppo_voce LIKE '$namev'");
echo "<select name='voce'><option value='' disabled selected>scegli categoria</option>";
while ($row = $result->fetch_assoc()) {
unset($id, $name);
$idc = $row['id_voce'];
$namec = $row['voce'];
$id_gc = $row['id_gruppo_voce'];
$id_c = $row['gruppo_voce'];
echo '<option value="'.$idc.'">'.$namec.'</option>';
}
echo "</select>";
?>
</div>
</form>
…..
</body>
This is a dynamic dropdown in PHP/mySQL.
I want to store the name in the database server but the tag outputs the integer value.
If i change the code from <option value="<?php echo $row["id"]; ?>"> to <option value="<?php echo $row["name"]; ?>"> It shows my_sqli_fetch_array expects parameter 1 error.
My objective being to store the corresponding $row["name"] that is being displayed on the dropdown instead of $row["id"].
<?php
$link = mysqli_connect("localhost","root", "");
mysqli_select_db($link,"loginsystem");
?>
<form name="form1" action="" method="post">
<table>
<tr>
<td>Select Assembly Line</td>
<td><select id ="assemblylinedd" onChange="change_assemblyline()">
<option>Select</option>
<?php
$i=1;
$res=mysqli_query($link, "SELECT * FROM assemblyline");
$count=mysqli_num_rows($res);
if ($count >0){
while($row=mysqli_fetch_array($res))
{
?>
<option value="<?php echo $row["id"]; ?>"><?php echo $row["name"]; ?></option>
}
<?php $i++;} }else{
echo "No record Found !";
} ?>
</select></td>
</tr>
Scripting code :
<script type="text/javascript">
function change_assemblyline()
{
var xmlhttp = new XMLHttpRequest();
xmlhttp.open("GET","ajax.php?assemblyline="+document.getElementById("assemblylinedd").value, false);
xmlhttp.send(null);
alert(xmlhttp.responseText);
document.getElementById("devices").innerHTML=xmlhttp.responseText;
}
This is my ajax.php
$link = mysqli_connect("localhost","root", "");
mysqli_select_db($link,"loginsystem");
$assemblyline = isset($_GET['assemblyline']) ? $_GET['assemblyline'] : '';
$devices = isset($_GET['devices']) ? $_GET['devices'] : '';
if($assemblyline!="")
{
$res=mysqli_query($link, "SELECT * FROM devices WHERE devices_id=$assemblyline");
echo "<select id='devicesdd' onchange='change_devices()'>";
while($row=mysqli_fetch_array($res))
{
echo "<option value='$row[id]'>";echo $row["name"]; echo "</option>";
}
echo "</select>";
}
Please do ignore onchange_devices() as it follows the same for next consecutive dropdown.
Though, its your requirement to save device name in DB, it is advised to save numeric id.
Reason: Name may change, but, id will persist.
If say your device id:name is 99 : iPhone 6 and you save in DB: iPhone 6, later the name gets changed to iPhone6.
In this scenario if you search records with name iPhone6, clearly, your above record will not show.
If you save numeric id, it will show irrespective of name change.
Coming back to your question:
I cannot write code here. But a pseudo code logic will help (hope so):
Take a hidden field device_name.
On change of drop down, with jQuery, assign value to hidden field.
$("#assemblylinedd option:selected").text();
Now, after submit, you will get device_name in hidden field.
$devices = isset($_GET['device_name']) ? $_GET['device_name'] : '';
Save this to DB.
$link = mysqli_connect("localhost","root", "");
mysqli_select_db($link, "loginsystem");
$assemblyline = isset($_GET['assemblyline']) ? $_GET['assemblyline'] : '';
$devices = isset($_GET['devices']) ? $_GET['devices'] : '';
if(!empty(trim($assemblyline)))
{
$res = mysqli_query($link, "SELECT * FROM devices WHERE devices_id = '$assemblyline'");
echo "<select id='devicesdd' onchange='change_devices()'>";
while($row = mysqli_fetch_array($res))
{
echo "<option value='" . $row["id"] . "'>" . $row["name"] . "</option>";
}
echo "</select>";
}
I've added a proper empty check instead of your != "", which didn't previously prevent a single space from being passed.
I've quoted your query value, I would definitely use prepared statements instead of passing values directly.
I've quoted your $row[id].
I've concatenated your string correctly.
Note: It would be preferable to return a JSON array object with the IDs and the names instead of outputting HTML via the AJAX, it would make your code-base much cleaner and adaptable in the future.
Reading Material
empty
trim
http://i.stack.imgur.com/Gy3o0.png
That is what the site looks like now. What I want to do is when you click on the approve registration on the table, it will extract the value of the id no and the name of that particular record. I thought i was on the right track. I knew how to get the id no. But it doesn't get the value of the name at the same time.
This is how the code looks like:
while($row = mysql_fetch_array($mayor))
{
$id = $row['identification_no'];
$name = $row['lastname'].", ".$row['firstname'];
echo "<tr>";
echo "<td><form method=post action=approvedmayor.php><input type='radio' name=id value='$id'></td>";
}
approvedmayor.php
$query = mysql_query("insert into tbcandidates VALUES ($id, '$name', 'mayor')");
if ($query)
{
echo "You appproved ";
echo $name;
}
else
echo "error";
you can try like this...
<?php
while($row = mysql_fetch_array($mayor))
{
$id = $row['identification_no'];
$name = $row['lastname'].", ".$row['firstname'];
echo "<tr><td><a href='approvedmayor.php?id=$id&name=$name'>Approve</a></td></tr>";
}
?>
in this type don't use the form, and Approve button... try this alone
Actually it is bad practice to insert that kind of data directly from POST data.
If you have all these candidates already stored in the database, you should run a SELECT query in your approvedmayor.php first, and if the candidate still exists, insert it's data to another table.
$query = mysql_query('SELECT * FROM `candidates` WHERE `id` = '.$id.' LIMIT 1');
if(mysql_num_rows($query)) {
$candidate = mysql_fetch_assoc($query);
$insertQuery = mysql_query("insert into tbcandidates VALUES ($candidate['id'], $candidate['name'], $candidate['mayor'])");
if ($insertQuery) {
echo "You appproved ";
echo $name;
} else echo "error";
} else echo 'This candidate is no longer available';
I understand your question,
But thats not the best way go ahead, Before we move let us understand some little elements functions
Radio Button : Its an input type element, that allows the user to choose only one [ 1 ] of option given list.
Check Boxes : Its an input type element, that allows the user to select n number of options or selections from give list.
Fore info - http://www.w3schools.com/html/html_forms.asp
Now comming to your question..
You need to modify your code to check boxes as below
<input type='checkbox' name=id[] value='$id'>
Notice : elements name is in Array mode.. ie whenever a user one or more than one, the values are stored in array.
Once the values are stored in array, call it / use if however you want.
For your mentioned code
echo "<form method=post action=approvedmayor.php>';
while($row = mysql_fetch_array($mayor))
{
$id = $row['identification_no'];
$name = $row['lastname'].", ".$row['firstname'];
echo "<tr>";
echo "<td><input type='radio' name=id[] value='$id'></td>";
}
echo "</form>";
approvedmayor.php
$temp_app_arr = $_POST['id'];
foreach ($temp_app_arr as $pos => $val) {
$query = mysql_query("insert into tbcandidates VALUES ('$val', '$name', 'mayor')");
if ($query) {
echo "You appproved ";
echo $name;
} else {
echo "error";
}
}
And i believe this should gonna be good code / algorithm for your project.
Okay so I have a table called Countries and it looks like this:
---------------------------
|Country | Code |
---------------------------
|Afganastan | AF |
|ÅLAND ISLANDS| AX |
| etc. | etc. |
---------------------------
The thing that I want to do is create a dynamic menu in which the user chooses a country and that itself gets stored as a value that I can call after the user hits submit.
I did try something here but I'm not sure what its doing because I am still new to PHP and HTML to the point where I just type things in to see what would happen.
Anyways I am really stuck and I tried using google and the search feature in this site and nothing I found worked for me...
The code I tried is this:
<select>
<?php
$result = mysql_query('SELECT Country FROM Countries');
echo '<select name="country">';
while ($row = mysql_fetch_array($result))
{
echo '<option value="'.$row['id'].'">'.$row['name'].'</option>';
}
echo '</select>';
?>
</select>
The result is supposed to look like a dropdown menu with the list of countries from the database in it. But this doesn't work and just shows this in the drop down:
.$row['name']
Which is nothing close to what I want because that's not even a country. when I remove that part of the code, then there is no option for the user to choose, the menu is empty.
EDIT
My code so far that still doesn't work:
<select name = 'country'>
<?php
include ("account.php");
include ("connect.php");
$result = mysql_query('SELECT Code , Country FROM Countries');
while ($row = mysql_fetch_array($result))
{?>
<option value="<?php echo $row['Code']?>"><?php echo $row['Country']?></option>
<?php}
?>
</select>
The include ("account.php"); and include ("connect.php"); lines allow me to connect to my database.
you code should be something like this
$host = "localhost";
$user = "root";
$pass = "yourpassword";
$db = "databasename";
// This part sets up the connection to the
// database (so you don't need to reopen the connection
// again on the same page).
$ms = #mysql_connect($host, $user, $pass);
if ( !$ms )
{
echo "Error connecting to database.\n";
}
// Then you need to make sure the database you want
// is selected.
#mysql_select_db($db);
<form method = "POST" action = "abc.php">
<select name = 'country'>
<?php
$result = mysql_query('SELECT id , name FROM Countries');
while ($row = mysql_fetch_array($result))
{?>
<option value="<?php echo $row['id']?>"><?php echo $row['name']?></option>
<?php}
?>
<input type = "submit" value = "Submit">
</form>
Now in php use this
echo '<pre>';
print_r($_POST);
And you will see what user selected. Check your settings there might be some problem.
Your single and double quotes are messing you up:
echo '<option value="'.$row['id'].'">'.$row['name'].'</option>';
should be:
echo "<option value=\"" . $row['id'] . "\">" . $row['name'] . "</option>";
You can use a single quote around your script but when you jump out of it to do the $row['id'] and $row['name'] you are running into issues because it thinks you are jumping back into your quoted code... Either use my example above, starting/ending with double-quotes and escaping all double-quotes inside that need to display, or escape your single quotes in the $row[\'id\'] and $row[\'name\']
Thant should help you out.
Try this code
<?php
$result = mysql_query('SELECT * FROM Countries');
?>
<select name="country">
<?php
while ($row = mysql_fetch_array($result))
{
?>
<option value="<?php echo $row['id']; ?>"><?php echo $row['name']; ?></option>
<?php
}
?>
</select>
Firstly your table doesn't have a column id. Try changing your query like
SELECT Country, Code FROM Countries
Then the code and the html should be like this
<?php
$host = "localhost";
$user = "user"; //username
$pass = "pass"; //password
$db = "db"; //database
$con = #mysql_connect($host, $user, $pass);
if ( !$con )
{
echo "Error connecting to database.\n";
}
#mysql_select_db($db);
?>
<select name="country">
<option value="0" selected="selected">Choose..</option>
<?php
//echo '<select name = \'country\'>';
$result = mysql_query('SELECT Country, Code FROM Countries');
while ($row = mysql_fetch_array($result))
{
echo '<option value="'.$row['Code'].'">'.$row['Country'].'</option>';
}
?>
</select>
<select>
<?php
$result = mysql_query('SELECT Country FROM Countries');
echo '<select name="country">';
$row = mysql_fetch_array($result)
for ($i=0; $i<count($row ); $i++)
{
echo '<option value="'.$row[$i]['id'].'">'.$row[$i]['name'].'</option>';
}
echo '</select>';
?>
next page use print_r($_POST);
or var_dump($_REQUEST);
If you are using mysql_fetch_array you can use either the field names or their selected index to read them from the fetched row. You can also use either the sprintf or printf functions to merge content into a string to help keep the HTML fragment clean of the quotes needed to merge in values otherwise.
$result = mysql_query('SELECT Country, Code FROM Countries');
while ($row = mysql_fetch_array($result)) {
printf('<option value="%1$s">%2$s</option>',
$row['Code'], $row['Country']);
}
Your SQL statement selected only 'Country' from the 'Countries' table; as a result, it's impossible for you to use $row['id'] and $row['name'].
Use this instead:
echo '<select name="country">';
while ($row = mysql_fetch_array($result))
{
echo '<option value="'.$row['Code'].'">'.$row['Country'].'</option>';
}
echo '</select>';
?>
That should solve your problem.
I figured out the problem I was having. The first being that there was an extra select tag in the page and the second that the file was saved as a html page instead of a php file. Thank you to everyone that helped me figure this out!
Try my code, I'm using this and it really works... just change the values...
<?php
include ('connect.php');
$sql = "SELECT * FROM casestatusfile";
$result = mysql_query($sql);
echo "<select name = 'txtCaseStatus'/>";
echo "<option value = ''>--- Select ---</option>";
$casestatus = $_POST['txtCaseStatus'];
$selected = 'selected = "selected" ';
while ($row = mysql_fetch_array($result)) {
echo "<option " .($row['CASESTATUSCODE'] == $casestatus? $selected:''). "value='". $row['CASESTATUSCODE'] ."'>" . $row['CASESTATUS'] ."</option>";
}
echo "</select>";
?>
I'm sure that is working because that is the one that i'm using....
I'm currently using php to populate a form with selections from a database. The user chooses options in a select style form and submits this, which updates a summary of the selections below the form before a second submit button is used to complete the interaction.
My issue is that every time a user uses the first submit, the selections that were there previously do not stick. They have to go through the whole form again.
Is there anyway to keep these selections present without resorting to php if statements? There are a ton of options so it would be a pain to use php for each one. Also, form is being submitted via POST.
Sample from form:
<?php
// GRAB DATA
$result = mysql_query("SELECT * FROM special2 WHERE cat = 'COLOR' ORDER BY cat")
or die(mysql_error());
echo "<div id='color'><select id='color' name='product_color'>";
while($row = mysql_fetch_array( $result )) {
$name= $row["name"];
$cat= $row["cat"];
$price= $row["price"];
echo "<option value='";echo $name;echo"'>";echo $name;echo" ($$price)</option>";}
echo "</select>";
echo "<input type='hidden' name='amount_color' value='";echo $price;echo"'></div>";
?>
I tried using this js snippet to repopulate the selections, but it does not seem to work properly...
<script type="text/javascript">document.getElementById('color').value = "<?php echo $_GET['proudct_cpu'];?>";</script>
This does not seem to work. Any suggestions other than php if statements?
Thanks!
edit: This is basically the form set up I'm using, though I've shortened it significantly because the actual implementation is quite long.
// Make a MySQL Connection
<?php mysql_connect("localhost", "kp_dbl", "mastermaster") or die(mysql_error());
mysql_select_db("kp_db") or die(mysql_error());
?>
<br />
<form action="build22.php" method="post">
<input type="hidden" name="data" value="1" />
<br />
<br />
<?php
// GRAB DATA
$result = mysql_query("SELECT * FROM special2 WHERE cat = 'color' ORDER BY cat")
or die(mysql_error());
echo "<div id='color'><select id='color' name='product_color'>";
while($row = mysql_fetch_array( $result )) {
$name= $row["name"];
$cat= $row["cat"];
$price= $row["price"];
echo "<option value='";echo $name;echo"'>";echo $name;echo" ($$price)</option>";}
echo "</select>";
echo "<input type='hidden' name='amount_color' value='";echo $price;echo"'></div>";
?>
<input type="submit" value="Update Configuration">
</form>
The selections from the form above get echoed after submission to provide the user with an update as such:
<div id="config" style="background-color:#FFF; font-size:12px; line-height:22px;">
<h1>Current Configuration:</h1>
<?php echo "<strong>Color:</strong>    ";echo $_POST['product_color']; ?>
</div>
I assume you're storing the user's selections in a separate table. If that's the case, you'll need to add some logic to determine if you should display the form values or what's already been stored.
<?php
// form was not submitted and a config id was passed to the page
if (true === empty($_POST) && true === isset($_GET['config_id']))
{
// make sure to properly sanitize the user-input!
$rs = mysql_query("select * from saved_configuration where config_id={$_GET['config_id']}"); // make sure to properly sanitize the user-input!
$_POST = mysql_fetch_array($rs,MYSQL_ASSOC); // assuming a single row for simplicity. Storing in _POST for easy display later
}
?>
<div id="config" style="background-color:#FFF; font-size:12px; line-height:22px;">
<h1>Current Configuration:</h1>
<?php echo "<strong>Color:</strong>    ";echo $_POST['product_color']; ?>
</div>
So after storing the user's selections in the database, you can redirect them to the page with the new config_id in the URL to load the saved values. If you're not storing the selected values in a table, you can do something similar with cookies/sessions.
echo the variables into the value tag of the form elements. If you post all your code I'm sure I can help you.
UPDATE
ah, so they are dropdown lists that you need to remember what was selected? Apologies, I read your post in a rush yesterday and thought it was a form with text inputs.
I just did a similar thing myself but without trying your code let me see if I can help.
Basically what you need to do is set one value in the dropdown to selected="selected"
When I had to do this I had my dropdown values in an array like so:
$options = array( "stack", "overflow", "some", "random", "words");
// then you will take your GET variable:
$key = array_search($_GET['variablename'], $options);
// so this is saying find the index in the array of the value I just told you
// then you can set the value of the dropdown to this index of the array:
$selectedoption = $options[$key];
This is where it might be confusing as my code is different so if you want to use it you will probably need to restructure a bit
I have a doSelect function to which I pass the following parameters:
// what we are passing is: name of select, size, the array of values to use and the
// value we want to use as the default selected value
doSelect("select_name", 1, $options, $selectedoption, "");
// these are the two functions I have:
// this one just processes each value in the array as a select option which is either
// the selected value or just a 'normal' select value
FUNCTION doOptions($options, $selected)
{
foreach ($options as $option)
{
if ($option == $selected)
echo ("<option title=\"$title\" id=\"$value\" selected>$option</option>\n");
else
echo ("<option title=\"$title\" id=\"$value\">$option</option>\n");
}
}
// this is the function that controls everything - it takes your parameters and calls
// the above function
FUNCTION doSelect($name, $size, $options, $selected, $extra)
{
echo("<select class=\"\" id=\"$name\" name=\"$name\" size=\"$size\" $extra>\n");
doOptions($options, $selected);
echo("</select>\n");
}
I know that's a lot of new code that's been threw at you but if you can get your select values from the db into the array then everything else should fall nicely into place.
The only thing I would add, is at the start where we call doSelect, I would put that in an if statement because you don't want to set something as selected which hasn't been set:
if (isset($_GET['variable']))
{
$key = array_search($_GET['variablename'], $options);
$selectedoption = $options[$key];
doSelect("select_name", 1, $options, $selectedoption, "");
}
else
{
doSelect("select_name", 1, $options, "", "");
}
I hope that helps!