PHP Variable with Array

Question

i'm trying to make a "update user's power" page. It is something similar to those you can find in say, invisionfree forums.

I need it to generate a list of members with checkbox [done] Add an option for it [done]

What i don't know how to do is to update, to say, give all the selected users the selected power.

Then i went searching for something and found most of them uses array to do this, but i never found one that actually explains how it works.

The example i took (with my own modification) was this:

while($row = mysql_fetch_array($result))
{
echo    '<tr>'.$id[]=$rows['id'].'';
echo    '<td width="50px" align="center" class="TableFormCell"><input type="checkbox" name="option[]" /></td>';
echo    '<td width="170px" align="center" class="TableFormCell">'.$row['uname'].'</td>';
echo    '</tr>';
}

I'm not sure what exactly $id[]=$rows['id'] does

I know after the row, my option[] would become an array of option[1], option[2], option[3]

for what power should be given, i've got no problem with that but on how to update the database i'm got no clue...

for($i=0;$i<$count;$i++){
$sql1="UPDATE ninos SET power='$power' WHERE id='$option[$i]'";
$result1=mysql_query($sql1);
}

So Say i have 5 users, Aye, Bee, Cee, Dee, Eee with IDs of 1,2,3,4,5 how can i make it so that my script would run like

$sql1="UPDATE ninos SET power = '$power' Where id='1','2','3','4','5'";

Please help, thanks.

Update

for nuqqsa

Here's the selecting page

$result = mysql_query("SELECT * FROM ninos");
while($row = mysql_fetch_array($result))
  {
echo    '<tr>';
echo    '<td width="50px" align="center" class="TableFormCell"><input     type="checkbox" name="option[]" value="' . $row['id'] . '" /></td>';
echo    '<td width="170px" align="center" class="TableFormCell">'.$row['uname'].'</td>';
echo    '</tr>';
}
?>

Here's the updating page

<?php
include('../openconn.php');

$power = $_POST['power'];

$ids = array();
foreach($_POST['option'] as $id)
{
$ids[] = (int)$id;
}   

if(!empty($ids)) {
// if there's at least one selected user
$sql1 = "UPDATE ninos SET power = '$power' Where id IN (" . implode(',', $ids) . ")";
// execute the query (...)
}


include('../closeconn.php');
?>

The sample code is vulnerable to SQL injection, which is a very serious security risk. To fix this hole, switch from the outdated mysql extension to PDO and use prepared statements. If you need a PDO tutorial, try "Writing MySQL Scripts with PHP and PDO". The site you save may just be your own. — outis
– outis, Commented Apr 23, 2011 at 20:05

nuqqsa · Accepted Answer · 2010-05-11 11:12:59Z

1

First, this doesn't make much sense (the assignment doesn't output anything printable):

echo '<tr>'.$id[]=$rows['id'].'';

This does the same yet it's clearer (the first line, since in a loop, will go on storing all the ids in an array):

$id[] = $rows['id'];
echo '<tr>';

UPDATE: Anyways there's no need to store the ids in an array here, just use $row['id'] to print the user identifier in the option value (right now no value is set):

while($row = mysql_fetch_array($result))
{
echo    '<tr>';
echo    '<td width="50px" align="center" class="TableFormCell"><input type="checkbox" name="option[]" value="' . $row['id'] . '"/></td>';
echo    '<td width="170px" align="center" class="TableFormCell">'.$row['uname'].'</td>';
echo    '</tr>';
}

The form action script will receive a $_POST['option'] variable containing the selected ids. An example of how it can be used:

$ids = array();
// input filtering: convert all values to integers
foreach($_POST['option'] as $id)
{
    $ids[] = (int)$id;
}    
if(!empty($ids)) {
    // if there's at least one selected user
    $sql1 = "UPDATE ninos SET power = '$power' Where id IN (" . implode(',', $ids) . ")";
    // execute the query (...)
}

edited May 11, 2010 at 11:12

answered May 11, 2010 at 10:00

nuqqsa

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6 Comments

Crays Over a year ago

After submitting ,this is how i get the ID right? $id = $_POST['id']; then i have this as you stated $sql1="UPDATE ninos SET power='$power' WHERE id IN (".implode(',', $id).")"; but i've got this Warning: implode() [function.implode]: Invalid arguments passed

nuqqsa Over a year ago

The second parameter of implode must be an array. This code assumes that $id is an array containing the user identifiers, NOT a $_POST variable. The array is generated through the instruction $id[] = $rows['id']; within the loop that iterates the query results.

nuqqsa Over a year ago

Oh, I misunderstood your question, now I get it also includes the post part. Let me update the answer.

Crays Over a year ago

Thanks for the reply, but it doesn't seem to work. Do you have a working example i could have a look at :) ?

Crays Over a year ago

I've updated my post with my full script so you probably can have a better view. I've no idea what's wrong with it, it goes through everything but when i check my records, nothing is changed :(.

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