if ($_SERVER['REQUEST_METHOD'] != 'POST'){
echo '<form method="post" action="">
<select name="class_name">
<option value="class_id">name</option>
</select>
<input type="submit" value="Proceed">
</form>';
}
else{
$_SESSION['var_name'] = $_POST['class_name'];
$sql = "SELECT (...) students.class_id = " . $_SESSION['var_name'];
$que = mysqli_query($conn, $sql);
echo '<table border 1>
<tr><th>Last</th><th>Name</th><th>Add</th></tr>';
while ($row = mysqli_fetch_array($que)){ //<-HERE I GOT A ERROR
echo '<tr><td>' . $row['user_last'] . '</td><td>' . $row['user_name'] . '</td>
<td> <form method="post" action="">
<input type="hidden" name="usr_id" value = "' . $row['user_id'] . '"/>
<input type="submit" name="button" value="+" />
</form>
</td>
</tr>';
}
echo '</table>';
if (isset($_POST['button'])){
//DOESEN'T MATTER
}
}
Ok, so my problem is that after clicking on my button from the second i got the error mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given on line I've commented above. I know that after clicking the value of _SESSION var is lost. My question is how to store this thing. I've tried session management but nothing.
I forgot to add, that I'm opening a session in included file.
Use Sessions, and make sure you have a
session_start();
on each page.
Related
I have a list of items in one file (fruit.php):
<form action="delete_items.php" method="post">
<input type="radio" value="apple" name="fruit">Apple<br>
<input type="radio" value="pear" name="fruit">Pear<br>
<input type="radio" value="banana" name="fruit">Banana<br>
<input type="submit" value="delete" name="deleteButton"><br>
</form>
In the delete_items.php, I have a variable that stores the selection:
<?php
$selection = $_POST["fruit"];
echo "are you sure you want to delete it?";
echo '<form action="delete_confirmation.php" method="post"><br>';
echo '<input type="submit" name="deleteYes" value="Yes"><br>';
echo '<input type="submit" name="deleteNo" value="No"><br>';
echo '</form>';
?>
Than in the confirmation file (delete_confirmation.php):
<?php
include 'delete_items.php';
if($_POST["deleteYes"]){
$query = 'DELETE FROM databaseName WHERE fruitName="' . $selection . '"';
$result = mysqli_query($connection, $query);
if(!$result){
die("Delete after check query failed!");
} else {
echo "Delete after check query success!";
}
} elseif($_POST["deleteNo"]){
echo "The course was not deleted!";
}
?>
But the varibale $selection in the delete_confirmation.php file is always null even though i included the delete_items.php file (when i echo it, nothing is shown). Is there a way to pass the selection variable from delete_items.php to the delete_confirmation.php file?
I'm in the middle of making a simple inventory system for keeping track of equipment going in and out of our doors. The inventory is stored in MYSQL, with a table looking like this: id name storage used location_storage location
This is all fun and games when I create a simple form with PHP, so it stays dynamic with the content from the server. I can update all values with no problem.
But for the sake of simplicity I'm looking into having a drop down menu, with a button, that creates/shows input fields in a form. The reason being is that I will have many rows in my table in the upcoming time. As the forms earlier have been made from server information, I will also need the scripts to be dynamic. Right now I'm stuck thinking about what I should do.
As of now, my code for the bits look like this:
"Static" PHP form:
<form action="<?php $_PHP_SELF ?>" method="POST">
<?php
//conn stuff
$sql = "SELECT id, name, storage, used, location FROM inventory";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo ' ' . $row["id"]. ' ' . $row["name"]. '<input type="text" name="newamount[' . $row["id"]. ']" />';
echo '<br>';
}
} else {
echo "0 results";
}
$conn->close();
?>
<input type="submit" name="checkout" value="Check out"/>
<input type="submit" name="checkin" value="Check in"/>
</form>
Check in PHP (check out is identical except for change of + and minus):
if(isset($_POST['sjekkinn'])){
//conn stuff
mysql_select_db( 'experimental' );
$newamount = $_POST['newamount'];
foreach($newamount as $key => $value){
$sql = "UPDATE inventory ". "SET storage = (storage + $value), used = (used - $value)". "WHERE ID = $key " ;
if (empty($value)) continue;
$retval = mysql_query( $sql, $conn );}
if(! $retval ) {
die('Could not update data: ' . mysql_error());
}
echo "Updated data successfully!<br>";
header("Refresh:1");
mysql_close($conn);
}
Those are all working great, but I would like to use something like the code below for a neater setup... Anyone got any advice?
Drop down list generated from MYSQL:
<form action="#" method="post">
<select name="selectinventory">
<?php
//conn stuff
$sql = "SELECT * FROM inventory";
$result = $conn->query($sql);
while ($row = $result->fetch_array()){
echo '<option value="' . $row["id"]. '">' . $row['name'] . '</option>';
}
?>
</select>
<input type="submit" name="submit" value="Add line">
</form>
Okay, so I found my own solution.
I ended up using my table-populated drop down list as I showed you in the last code box of the question. What came to my mind was that I could simply use the jQuery show/hide function.
What I did was to make a script that told (in "readable") div 'x' to show and move when option 'x' was selected and button clicked. I will show you my complete code in the end.
That way I could have my input fields each created inside a div from my table (the same that populated the drop down list), so I could manage them easier (probably possible to do it easier - but if it ain't broken, don't fix it). These were created outside the form. When I then click the previously mentioned button, the div will move inside the form. The next one I select will end up UNDER the previous one, instead of just showing up in table-order.
Feel free to shout any questions!
Here's the complete code:
<form action="<?php $_PHP_SELF ?>" method="post">
<select name="selectinventory" id="selectinventory">
<option selected="selected">Choose one</option>
<?php
//conn stuff
$sql = "SELECT id, name FROM inventory";
$result = $conn->query($sql);
$sql = "SELECT * FROM inventory";
$result = $conn->query($sql);
while ($row = $result->fetch_array()){
echo '<option value="' . $row["id"]. '">' . $row['name'] . '</option>';
}?>
</select>
<div id="btn">Add line</div>
</form>
<script type="text/javascript">
$(document).ready(function(){
$("#btn").click(function(){
$("#" + $("#selectinventory").val()).show();
$("#" + $("#selectinventory").val()).appendTo($(".selecteditems"));
});
});
</script>
<?php
//conn stuff
$sql = "SELECT id, name FROM inventory";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo '<div style="display:none" id="' . $row["id"]. '"> ' . $row["id"]. ' ' . $row["name"]. '<input type="text" name="newamount[' . $row["id"]. ']" /><br></div>
';
}
} else {
echo "0 results";
}
$conn->close();
?>
<form action="<?php $_PHP_SELF ?>" method="POST" name="isthisthearrayname">
<div class="selecteditems">
</div>
<input type="submit" name="checkout" value="Check out"/>
<input type="submit" name="checkin" value="Check in"/>
</form>
I'm in need of a bit help. I'm trying to find out how to associate a specific query (deletion of a record) with not the id of a record, but the record with which another query (selection of a record) is echoed out.
This line of code totally works when the id is specified, but again I need it for the record that gets called, where the id can skip numbers if I delete a record.
$querytwo = "DELETE FROM `paginas` WHERE id = 5";
I've got a table in my phpmyadmin database with columns 'id', 'pagetitle', 'toevoeging' (addition in Dutch) , 'message'. First one is an INT, rest are varchars/text.
This may be a stupid question, I'm sorry for that. I'm still new to PHP, and to programming in general.
Here is the code. I've commented on lines code to clarify. Thanks you!.
<?php
if (isset($_SESSION['email'])) //if the admin is active, forms can be written out.
{
echo '</nav>
<br><br> <div class="inlogscript">
<form action="verstuurd.php" method="post">
<input type="text" placeholder="Titel" method="POST" name="pagetitle" /><br><br>
<input type="text" placeholder="Toevoeging" method="POST" name="toevoeging" /><br><br>
<textarea class="pure-input-1-2" placeholder="Wat is er nieuws?" name="message"></textarea><br>
<input type="submit" value="Bevestigen" />
</form></div>';
}
?>
<div class="mainContent">
<?php
include_once("config.php"); //this is the database connection
$query = "SELECT * FROM paginas "; //selects from the table called paginas
$result = mysqli_query($mysqli, $query);
while($row = mysqli_fetch_assoc($result))
{
$pagetitle = $row['pagetitle'];
$toevoeging = $row['toevoeging'];
$message = $row['message'];
echo '<article class="topcontent">' . '<div class="mct">' . '<h2>' . "$pagetitle" .'</h2>' . '</div>' . "<br>" .
'<p class="post-info">'. "$toevoeging" . '</p>' . '<p class="post-text">' . '<br>'. "$message" . '</p>' .'</article>' . '<div class="deleteknop">' . '<form method="post">
<input name="delete" type="submit" value="Delete Now!">
</form>' . '</div>' ;
} //This long echo will call variables $pagetitle, $toevoeging and &message along with divs so they automatically CSS styled,
//along with a Delete button per echo that has the 3 variables
$querytwo = "DELETE FROM `paginas` WHERE id = 5";
if (isset($_POST['delete'])) //Deletes the query if 'delete' button is clicked
{
$resulttwo = $mysqli->query($querytwo);
}
?>
</div>
</div>
Also here is the Insert INTO query of the records. Thanks again!
$sql = "INSERT INTO paginas (pagetitle,toevoeging, message)
VALUES ('$_POST[pagetitle]','$_POST[toevoeging]','$_POST[message]')";
//the insertion into the table of the database
if ($MySQLi_CON->query($sql) === TRUE) {
echo "";
} else {
echo "Error: ". $sql . "" . $MySQLi_CON->error;
}
This won't be sufficient but, to begin with your echo :
echo '<article class="topcontent">
<div class="mct">
<h2>' . $pagetitle .'</h2>
</div><br>
<p class="post-info">'. $toevoeging . '</p>
<p class="post-text"><br>'.$message.'</p>
</article>
<div class="deleteknop">
<form method="post">';
// you ll want to use $_POST["id"] array to delete :
echo '<input type="hidden" name="id" value="'.$row['id'].'">
<input name="delete" type="submit" value="Delete Now!">
</form>
</div>' ;
What is wrong in this code? Here I am trying to show the data fetched from a mySQL database when someone clicks on the "Show" button (Which is a submit button).
<?php
if(isset($_POST['submit'])){
$mysqli = new mysqli("localhost", "rms", "sarangi", "rms");
if($mysqli === false){
die("Connection Error.");
}
else{
echo'<table border="1"><tr><th> Name </th><th> Username </th></tr>';
$sql = "SELECT * FROM users";
if($result = $mysqli->query($sql)){
if($result->num_rows > 0){
while($data = $result->fetch_array()){
echo'<tr><td>' . $data['firstname'] . " " . $data['lastname'] . '</td><td>' . $data['username'] . '</td></tr>';
}
}
}
echo'</table>';
}
$mysqli->close();
}
else{
?>
<form action="index.php" method="POST">
<input type="submit" id="submit" value="Show" />
</form>
<?php
}
?>
You have to name your input field
<input type="submit" id="submit" value="Show" />
should be
<input name="submit" type="submit" id="submit" value="Show" />
change your code to this
<form action="" method="POST">
<input type="submit" id="submit" name="submit" value="Show" />
</form>
I have changed form action to "" since its going to submit to the same page. or you can leave it as index.php if this page is index.php. You need to put the name attribute of input submit as submit
edit - I solved my "add friend" button issue, now I'm trying to get the userid from the loop below. I want to be able to get the userid of the name that the user looks up (the name that gets submitted to findUsers function, $friend). So basically I want to be able to use result['userid'] and be able to submit that into a database.
I commented in the code where I'm having trouble getting the value for the userid to set.
<input type="hidden" name="userId" value="' . $result['userid'] . '" />
Is there a certain way to use hidden inputs, or is the value just not being set correctly?
<?php
include_once 'config.php';
class Friends{
function addFriend($userId) {
return $userId; //this is supposed to return the value of the user's id selected in the loop below via the if statements towards the bottom.
}
function findUsers($friend){
$search = mysql_query("SELECT * from users where username='$friend'");
if (mysql_num_rows($search) > 0){
// $this->addFriend($friend);
$userLocation = mysql_query("select * from userinfo where username='$friend'");
$locationResult = mysql_fetch_array($userLocation);
$locationResultArray = $locationResult['userlocation'];
$locationExplode = explode("~","$locationResultArray");
if (mysql_num_rows($search)) {
// Table column names
echo '<table><tr><td>Username</td><td>Location</td></tr>';
while($result = mysql_fetch_array($search)) {
echo '<tr>
<td>'. $result['username'] . '</td>
<td>' . $locationExplode[0] . ', ' . $locationExplode[1] . '</td>
<td>
<form method="post" name="friendRequest" action="">
<input type="hidden" name="userId" value="' . $result['userid'] . '" />
<input type="submit" name="addFriend" value="Add Friend" />
</form>
</td></tr>';
}
}
}
}
}
$friends = new Friends();
if (isset($_POST['userId'], $_POST['addFriend'])) {
echo "friend button pressed"; //this message is displayed
if ($friends->addFriend($_POST['userId'])) {
echo "userID set"; //this message is displayed
echo $_POST['userID']; //this is not displayed
} else {
// some error code here
}
}
// Edit this to test here
// $friends->findUsers('<username>');
?>
That way to add friend is incorrect way, because when you click the "Add friend" button, that will send a $_POST['addFriend'] and then in the loop the check are going to add all users as friend.
The correct code is here:
<?php
function addFriend($userId){
// check is 'userId' exist, if not, then return 0;
}
if (isset($_POST['userId'], $_POST['addFriend'])) {
if (addFriend($_POST['userId'])) {
// some display code here
} else {
// some error code here
}
}
while($result = mysql_fetch_array($search)) {
?>
<tr><td>
<form method="post" name="friendRequest" action="">
<input type="hidden" name="userId" value="<?php echo $result['userid']; ?>" />
<input type="submit" name="addFriend" value="Add Friend" />
</form>
</td></tr>
<?php } ?>
EDIT1:
You can't use the code above into a function. I fixed a lot of bug that I can see in your code, but still look strange.
I don't get what you want to do with your code, but I made this:
<?php
function addFriend($userId) {
return 1; //using 1 for testing purposes
}
function findUsers($friend) {
$search = mysql_query('SELECT `userid`, `username`, `userlocation` FROM `users` JOIN `userinfo` ON `users`.`username` = `userinfo`.`username` WHERE `user`.`username` = ' . $friend);
if (mysql_num_rows($search)) {
// Table column names
echo '<table><tr><td>Username</td><td>Location</td></tr>';
while($result = mysql_fetch_array($search)) {
$locationExplode = explode('~', $result['userlocation']);
echo '<tr>
<td>'. $result['username'] . '</td>
<td>' . $locationExplode[0] . ', ' . $locationExplode[1] . '</td>
<td>
<form method="post" name="friendRequest" action="">
<input type="hidden" name="userId" value="' . $result['userid'] . '" />
<input type="submit" name="addFriend" value="Add Friend" />
</form>
</td></tr>';
}
}
}
if (isset($_POST['userId'], $_POST['addFriend'])) {
if (addFriend($_POST['userId'])) {
echo "test"; //I'm simply trying to get the input to work, can't get it to post. Just using this for a test.
} else {
// some error code here
}
}
// Edit this to test here
// findUsers('<username>');
?>
EDIT2:
Well, you just need to put my functions code into the class and then use the other code outside the class, like this:
<?php
include_once 'config.php';
class Friends{
function addFriend($userId) {
return 1; //using 1 for testing purposes
}
function findUsers($friend) {
$search = mysql_query('SELECT `userid`, `username`, `userlocation` FROM `users` JOIN `userinfo` ON `users`.`username` = `userinfo`.`username` WHERE `user`.`username` = ' . $friend);
if (mysql_num_rows($search)) {
// Table column names
echo '<table><tr><td>Username</td><td>Location</td></tr>';
while($result = mysql_fetch_array($search)) {
$locationExplode = explode('~', $result['userlocation']);
echo '<tr>
<td>'. $result['username'] . '</td>
<td>' . $locationExplode[0] . ', ' . $locationExplode[1] . '</td>
<td>
<form method="post" name="friendRequest" action="">
<input type="hidden" name="userId" value="' . $result['userid'] . '" />
<input type="submit" name="addFriend" value="Add Friend" />
</form>
</td></tr>';
}
}
}
}
$friends = new Friends();
if (isset($_POST['userId'], $_POST['addFriend'])) {
if ($friends->addFriend($_POST['userId'])) {
echo "test";
} else {
// some error code here
}
}
// Edit this to test here
// $friends->findUsers('<username>');
?>
EDIT3:
That's because the function addFriend is incorrect... You need to pass the user ID value as argument and then display it like this:
function addFriend($userId) {
return $userId; //this is supposed to return the value of the user's id selected in the loop below via the if statements towards the bottom.
}