PHP Dynamic HTML Question - php

I have this piece of PHP/HTML, wrapped in a POST form. How do I pass only the ID for the row where the delete button is clicked back to the server?
<table>
<tr>
<th>File Path</th>
<th>Expiration</th>
</tr>
<?php
$result = mysql_query ($list_query);
$row_count = mysql_numrows ($result);
for ($i = 0; $i < $row_count; $i++) {
$id = mysql_result ($result, $i, "id");
$path = mysql_result ($result, $i, "path");
$expiration = mysql_result ($result, $i, "expires");
?>
<tr>
<td width="60%">
<?php echo $path; ?>
</td>
<td>
<?php echo $expiration; ?>
</td>
<td>
<input type="submit" value="Delete Expiration" />
</td>
</tr>
<?php
}
?>
</table>

Use a hidden field within the form.
<input type="hidden" name="id" value="<?php echo $id ?>">
You should also start a new form for each new row too.
for ($i = 0; $i < $row_count; $i++) {
$id = mysql_result ($result, $i, "id");
$path = mysql_result ($result, $i, "path");
$expiration = mysql_result ($result, $i, "expires");
?>
<tr>
<td width="60%">
<?php echo $path; ?>
</td>
<td>
<?php echo $expiration; ?>
</td>
<td>
<form method="POST" action="?">
<input type="hidden" name="id" value="<?php echo $id ?>">
<input type="submit" value="Delete Expiration" />
</form>
</td>
</tr>
<?php
}
?>

I agree with the solution Extrakun proposes, however, for completeness I would like to point you to the option of using JavaScript and DOM. You could use JQuery as explained in this question:
jquery + table row edit - String problem

You should set a hidden field containing the id somewhere in your row, and wrap the hidden field and the submit button in a form per row:
<form>
<input type="hidden" name="id" value="<?php echo $id; ?>" />
<input type="submit" value="Delete" />
</form>
After clicking, get the id with $_POST['id'].
Greetz,
XpertEase

Related

How to use form and its php processing inside a loop [duplicate]

I have outputted the results of a MySQL table to an HTML table. In the last column, I want to add a delete option which calls another form and deletes the user from the MySQL table. I can't seem to get it to work though.
This is my code for the results page:
<?php
$contacts = mysql_query("
SELECT * FROM contacts ORDER BY ID ASC") or die( mysql_error() );
// If results
if( mysql_num_rows( $contacts ) > 0 )
?>
<table id="contact-list">
<thead>
<tr>
<th>Name</th>
<th>Email</th>
<th>Telephone</th>
<th>Address</th>
<th>Delete</th>
</tr>
</thead>
<tbody>
<?php while( $contact = mysql_fetch_array( $contacts ) ) : ?>
<tr>
<td class="contact-name"><?php echo $contact['name']; ?></td>
<td class="contact-email"><?php echo $contact['email']; ?></td>
<td class="contact-telephone"><?php echo $contact['telephone']; ?></td>
<td class="contact-address"><?php echo $contact['address']; ?></td>
<td class="contact-delete"><form action='delete.php' method="post">
<input type="hidden" name="name" value="">
<input type="submit" name="submit" value="Delete">
</form></td>
</tr>
<?php endwhile; ?>
</tbody>
</table>
and, this is my delete.php script
<?php
//Define the query
$query = "DELETE FROM contacts WHERE name={$_POST['name']} LIMIT 1";
//sends the query to delete the entry
mysql_query ($query);
if (mysql_affected_rows() == 1) {
//if it updated
?>
<strong>Contact Has Been Deleted</strong><br /><br />
<?php
} else {
//if it failed
?>
<strong>Deletion Failed</strong><br /><br />
<?php
}
?>
I cannot figure out why this is not working.
You have to pass a variable in the delete link. You have to pass <?php echo $contact['name']; ?> (the name value) in a hidden field or pass this value in URL:
Replace
<td class="contact-delete">
<form action='delete.php' method="post">
<input type="hidden" name="name" value="">
<input type="submit" name="submit" value="Delete">
</form>
</td>
With
<td class="contact-delete">
<form action='delete.php?name="<?php echo $contact['name']; ?>"' method="post">
<input type="hidden" name="name" value="<?php echo $contact['name']; ?>">
<input type="submit" name="submit" value="Delete">
</form>
</td>
USe javascript
<input name="Submit2" type="button" class="button" onclick="javascript:location.href='delete.php?id=<?php echo $your_id;?>';" value="« Back" />
and in delet.php
$id=$_GET['id'];
and put $id in your sql statement.
You are missing to pass name in this line:
<input type="hidden" name="name" value="">
You need to have something (<?php echo $contact['name']; ?>) in the value attribute.
BTW, do not use deprecated mysql_* functions, use PDO or mysqli_* instead.
<input type="hidden" name="name" value="">
You are missing a value which wil be picked up by this line in your delete file.
$query = "DELETE FROM contacts WHERE name={$_POST['name']} LIMIT 1";
Right now it isn't receiving anything, which is why it will not work.
So add a value to it and it will work. Example:
<input type="hidden" name="name" value="<?php echo $contact['name']; ?>">
First, you should not write the code in that way; the code has no protection against SQL injection.
1. Try to use primary IDs instead of using a name (what happens if 2 people has the same name?).
So, you can create a hidden field to know which 'person' you are dealing with.
<input type="hidden" name="contact_id" value="<?php $contact['contact_id']; ?>">
2. Sanitize variables to avoid attacks:
<?php $contact_id = isset($_POST['contact_id'])?intval($_POST['contact_id']):0;
// proceed with the query
if($contact_id>0) { $query = "DELETE FROM contacts WHERE contact_id = '$contact_id'";
}
// redirect to the main table with header("location: main.php");
?>

How to update and delete specific value in html table without mysql

I have a form tag in which I have an input name and on click of a button, the value is displayed in table form. Now I want to update that value. How do I do it? This code does not have any database.
<form>
<input type="text" name="fname" value="" />
<input type="submit" value="submit" name="btn_submit" />
</form>
<br>
<br>
<table border="1">
<th>Name</th>
<th>Delete</th>
<th>Update</th>
<?php
session_start();
$na = array();
if (isset($_GET['btn_submit']))
{
if (isset($_SESSION['name']))
{
$na = $_SESSION['name'];
}
$na[] = $_GET['fname'];
$_SESSION['name'] = $na;
for ($i = 0; $i < count($na); $i++) {
?>
<tr>
<td><?php echo $na[$i]; ?></td>
<td>DELETE</td>
<?php
if(isset($_GET['del']))
{
}
?>
<td>UPDATE</td>
</tr>
<?php
}
}
?>

Insert data from while loop into a table with php

I'm creating a form using HTML and PHP. I have created a form which I want to submit and save that data in database.
I'm trying to submit a form with data that comes from a while loop. All input values are getting generated by while loop.
The code looks like this.
<table width="1348" border="0" class="table table-striped" >
<tr>
<td width="106"> </td>
<td width="332"><strong>Product Code</strong></td>
<td width="375"><strong>Product Name</strong></td>
<td width="211"><strong>QTY</strong></td>
</tr>
<?php
$i = 0;
$rowset = mysql_query("select * from product_detail where productID='".$data['productCode']."'");
while($stuff = mysql_fetch_array($rowset)){
?>
<tr>
<td><input type="text" name="code[<?php echo $i?>]" value="<?php enter code hereecho $stuff['code'];?>"/></td>
<td><input type="text" name="name[<?php echo $i?>]" value="<?php echo $stuff['name'];?>" size="50"/></td>
<td><input type="text" name="qty[<?php echo $i?>]" value="<?php echo $stuff['qty'];?>" size="10"/></td>
</tr>
<?php $i++; }?>
<tr id="last">
</table>
<input type="submit" name="save id="save" class="btn btn-primary btn-lg"/>
This is the code to add the data to database.
$code=$_POST['code'.$i];
$name=$_POST['name'.$i];
$qty=$_POST['qty'.$i];
$query = mysqli_query($con,"insert into stock(productCode, productName, qty) values ('".$code."', '".$name."','".$qty."')") or die(mysqli_error($con));
First, use prepared statement with bind_param as your script is totally exposed to sql injection.
Second, you can add input type hidden for the number of rows
<form action="" method="POST">
<table width="1348" border="0" class="table table-striped" >
<tr>
<td width="106"> </td>
<td width="332"><strong>Product Code</strong></td>
<td width="375"><strong>Product Name</strong></td>
<td width="211"><strong>QTY</strong></td>
</tr>
<?php
$data['productCode'] = "1"; // sample data
$stmt = $con->prepare("SELECT * FROM product_detail WHERE productID = ?");
$stmt->bind_param("i", $data['productCode']);
$stmt->execute();
$result = $stmt->get_result();
$i = 0;
while($stuff = $result->fetch_assoc()) {
?>
<tr>
<td></td>
<td><input type="text" name="code[<?php echo $i; ?>]" value="<?php echo $stuff['code'];?>"/></td>
<td><input type="text" name="name[<?php echo $i; ?>]" value="<?php echo $stuff['name']; ?>" size="50" /></td>
<td><input type="text" name="qty[<?php echo $i; ?>]" value="<?php echo $stuff['qty']; ?>" size="10" /></td>
</tr>
<?php
$i++;
}
?>
<input type="hidden" name="count" value="<?php echo $i; ?>" />
<tr id="last">
</table>
<input type="submit" name="save" id="save" class="btn btn-primary btn-lg"/>
</form>
post count with the form
<?php
if (isset($_POST['save'])) {
$count = $_POST['count'];
for ($i = 0; $i < $count; $i++) {
$code = $_POST['code'][$i]; // check empty and check if interger
$name = $_POST['name'][$i]; // check empty and strip tags
$qty = $_POST['qty'][$i]; // check empty and check if interger
$stmt = $con->prepare("INSERT INTO stock (productCode, productName, qty) VALUES (?, ?, ?)");
$stmt->bind_param("iss",$code,$name,$qty);
$stmt->execute();
}
}
?>
You may also want to check if post values are empty with other necessary validation before insert
Since the table is dynamically filled, you need to use an array as the name attribute
<table>
<tr>
<th>Name</th>
<th>Present</th>
<th>Excused</th>
<th>Unexcused</th>
<th>Ext</th>
</tr>
<?php
$query = "select * from TbCard";
$sql = mysqli_query($connect, $query);
$count = 0;
while ($data = mysqli_fetch_array($sql)) {
?>
<tr>
<td>
<input name="tableRow[<?php echo $count; ?>]['dataName']" id='name' type='text' value="<?php echo $data['Name'];?>" readonly style='border:none;width:350px'></input>
</td>
<td>
<input name="tableRow[<?php echo $count; ?>]['status']" type="radio" value="Present"> Present
</td>
<td>
<input name="tableRow[<?php echo $count; ?>]['status']" type="radio" value="Excused"> Excused
</td>
<td>
<input name="tableRow[<?php echo $count; ?>]['status']" type="radio" value="Unexcused"> Unexcused
</td>
</tr>;
<?php
$count++;
}
?>
</table>
The php would be something like this, assuming that the data has values in it:
$tableRow = $_POST['tableRow'];
foreach($tableRow as $row){
/* here insert data from post */
echo $row['dataName'].' '.$row['status'].'<br/>';
}
To see the content of the array, use print_r($tableRow)
in this case i use a name tableRow

How to add a delete button to a PHP form that will delete a row from a MySQL table

I have outputted the results of a MySQL table to an HTML table. In the last column, I want to add a delete option which calls another form and deletes the user from the MySQL table. I can't seem to get it to work though.
This is my code for the results page:
<?php
$contacts = mysql_query("
SELECT * FROM contacts ORDER BY ID ASC") or die( mysql_error() );
// If results
if( mysql_num_rows( $contacts ) > 0 )
?>
<table id="contact-list">
<thead>
<tr>
<th>Name</th>
<th>Email</th>
<th>Telephone</th>
<th>Address</th>
<th>Delete</th>
</tr>
</thead>
<tbody>
<?php while( $contact = mysql_fetch_array( $contacts ) ) : ?>
<tr>
<td class="contact-name"><?php echo $contact['name']; ?></td>
<td class="contact-email"><?php echo $contact['email']; ?></td>
<td class="contact-telephone"><?php echo $contact['telephone']; ?></td>
<td class="contact-address"><?php echo $contact['address']; ?></td>
<td class="contact-delete"><form action='delete.php' method="post">
<input type="hidden" name="name" value="">
<input type="submit" name="submit" value="Delete">
</form></td>
</tr>
<?php endwhile; ?>
</tbody>
</table>
and, this is my delete.php script
<?php
//Define the query
$query = "DELETE FROM contacts WHERE name={$_POST['name']} LIMIT 1";
//sends the query to delete the entry
mysql_query ($query);
if (mysql_affected_rows() == 1) {
//if it updated
?>
<strong>Contact Has Been Deleted</strong><br /><br />
<?php
} else {
//if it failed
?>
<strong>Deletion Failed</strong><br /><br />
<?php
}
?>
I cannot figure out why this is not working.
You have to pass a variable in the delete link. You have to pass <?php echo $contact['name']; ?> (the name value) in a hidden field or pass this value in URL:
Replace
<td class="contact-delete">
<form action='delete.php' method="post">
<input type="hidden" name="name" value="">
<input type="submit" name="submit" value="Delete">
</form>
</td>
With
<td class="contact-delete">
<form action='delete.php?name="<?php echo $contact['name']; ?>"' method="post">
<input type="hidden" name="name" value="<?php echo $contact['name']; ?>">
<input type="submit" name="submit" value="Delete">
</form>
</td>
USe javascript
<input name="Submit2" type="button" class="button" onclick="javascript:location.href='delete.php?id=<?php echo $your_id;?>';" value="« Back" />
and in delet.php
$id=$_GET['id'];
and put $id in your sql statement.
You are missing to pass name in this line:
<input type="hidden" name="name" value="">
You need to have something (<?php echo $contact['name']; ?>) in the value attribute.
BTW, do not use deprecated mysql_* functions, use PDO or mysqli_* instead.
<input type="hidden" name="name" value="">
You are missing a value which wil be picked up by this line in your delete file.
$query = "DELETE FROM contacts WHERE name={$_POST['name']} LIMIT 1";
Right now it isn't receiving anything, which is why it will not work.
So add a value to it and it will work. Example:
<input type="hidden" name="name" value="<?php echo $contact['name']; ?>">
First, you should not write the code in that way; the code has no protection against SQL injection.
1. Try to use primary IDs instead of using a name (what happens if 2 people has the same name?).
So, you can create a hidden field to know which 'person' you are dealing with.
<input type="hidden" name="contact_id" value="<?php $contact['contact_id']; ?>">
2. Sanitize variables to avoid attacks:
<?php $contact_id = isset($_POST['contact_id'])?intval($_POST['contact_id']):0;
// proceed with the query
if($contact_id>0) { $query = "DELETE FROM contacts WHERE contact_id = '$contact_id'";
}
// redirect to the main table with header("location: main.php");
?>

arrange checkbox values in two columns

I have a form with check box values i want to arrange the checkbox in two columns instead of
one single long column.
How can i split it to two columns ?
Here is the code :
<form id="form" name="form" method="post" action="">
<table width="502" border="0">
<tr>
<td>
Name :
<input type="textbox" name="rcv_group_name" value="">
<input type="hidden" name="add" value="add" />
</td>
</tr>
<tr>
<td align="left">
<strong>HEADING</strong>
<?php
$q = "select * from Config_RCV";
$r = mysqli_query ($dbc, $q);
if (mysqli_num_rows($r) > 0)
{
$k=0;
while ($row = mysqli_fetch_array($r,MYSQLI_NUM))
{
?>
<br> <input type="checkbox" name ="rcv_val[]" value ="<? echo $row[0];?>" /> <? echo $row[1];?>
<?
$k++;
}
}
?>
</td>
</tr>
<tr>
<td align="right"><input type="submit" name="Submit" id="Submit" value="Submit" style="background-color:#999" />
< /td>
<td height="26" align="left"> </td>
<td> </td>
</tr>
</table>
<form>
add another table and make 2 columns by splitting 2'nd result:
if (mysqli_num_rows($r) > 0) {
$k=0;
echo '<table><tr>';
while ($row = mysqli_fetch_array($r,MYSQLI_NUM)) {
?>
<td><input type="checkbox" name ="rcv_val[]" value ="<?php echo $row[0];?>" /> <?php echo $row[1];?> </td>
<?php
$k++;
if($k%2 == 0){
echo '</tr><tr>';
}
}
echo '</table>';
}
use your $k
if ($k%2==0){
echo "<br />";
}
instead of the lone now
Using tables like above will probably be better for looks
You could just add a second column to your table:
echo "<tr>";
$i = 0;
while ($row = mysqli_fetch_array($r,MYSQLI_NUM)) {
if ($i++%2==0) {
echo "</tr><tr>";
}
echo "<td>".$CELL_CONTENT_HERE."</td>";
}
echo "</tr>";

Categories