How can I send over multiple check box checks in POST to be deleted from a database?

Question

I've been trying think of a way to do this. I want it to where users can check off items, hit submit and it goes to the code on the next page and deletes all of the checked items from the database. Problem one is that in the post its only sending over the last checked item. Here is how I have it set up right now.

echo "<form name='fm1' METHOD ='POST' ACTION ='displaydelete.php' > ";

//Draws up the table headers echo ""; echo ""; echo "Fund Number "; echo "Hours "; echo "Percentage"; echo "Delete"; echo ""; //While there are query results data is pushed into table cells while ($row = mysql_fetch_array($queryResult2)) { $hours = $row['hours']; $percentage = $hours / 160 * 100;

   echo "<tr>";
    echo "<td>";
    echo $row['funnumber'];
    echo "</td>";
    echo "<td>";
    echo $hours;
    echo "</td>";
    echo "<td>";
    echo $percentage ."%";
    echo "</td>";
    echo "<td>";
   echo "<input type='checkbox' name='id' value='$row[id]'/>";
  echo "</td>";
    echo "</tr>";




}
    //End of tabel 
echo "</table>";

echo" "; echo "";

What I would like to do is push all of the items into a variable and maybe delete them that way. I'm not really sure how you would handle multiple deletes. I'm doing my delete like this for something else if this helps any.

 $query = "DELETE FROM users

WHERE ninenumber = '$ninenumber'"; $result = mysql_query($query) or die("Query Failed: " .mysql_error());

        mysql_close($conn);

Boy, everyone really jumps on these easy ones :D

Jonah
– Jonah

2010-12-01 17:16:15 +00:00
Commented Dec 1, 2010 at 17:16 — Jonah
– Jonah, Commented Dec 1, 2010 at 17:16

Brian H · Accepted Answer · 2010-12-01 17:10:59Z

3

In your form:

<input type='checkbox' name='id[]' value='$row[id]'/>

Then, in the file you post to:

if(is_array($_POST['id'])){
   foreach($_POST['id'] as $id){
   ...do something to $id;
   }
}

answered Dec 1, 2010 at 17:10

Brian H

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Comments

Jonah · Accepted Answer · 2010-12-01 17:20:40Z

2

Instead of this:

echo "<input type='checkbox' name='id' value='$row[id]'/>";

You need this:

echo "<input type='checkbox' name='id[]' value='$row[id]'/>";

Note the difference. I added [] after the input name. This tells the client and server that there are multiple inputs with that name. $_POST['id'] will be an array you can loop through on the next page.

foreach ($_POST['id'] as $checkbox) {
    // DELETE FROM users WHERE ninenumber = $checkbox
}

isset, is_array, and mysql_real_escape_string omitted for brevity.

edited Dec 1, 2010 at 17:20

answered Dec 1, 2010 at 17:11

Jonah

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1 Comment

zanlok Over a year ago

agreed that brevity is nice. but when it's this kind of question, they probably like seeing it all spelled out and working together. props on at least mentioning the right ingredients, though :)

zanlok · Accepted Answer · 2010-12-01 17:15:18Z

1

In the form-generating code, make the name in the html have" []" after it:

...
echo "<input type='checkbox' name='id[]' value='$row[id]'/>";
...

Then, in the form-reading code, your post'ed id will be an array.

$id_array = isset($_POST['id']) && is_array($_POST['id']) ? $_POST['id'] : array();
foreach( $id_array as $id ) {
   $query = "DELETE FROM users WHERE ninenumber = '" . mysql_real_escape_string($id) . "'";
   // execute the delete query
}

edited Dec 1, 2010 at 17:15

answered Dec 1, 2010 at 17:09

zanlok

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Comments

Ignacio Vazquez-Abrams · Accepted Answer · 2010-12-01 17:09:48Z

0

Putting [] after the name of a control will turn it into an array in the superglobal that you can then iterate over to get all the values from.

answered Dec 1, 2010 at 17:09

Ignacio Vazquez-Abrams

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Comments

Jason · Accepted Answer · 2010-12-01 17:10:13Z

0

You need to have the same name for all of your checkboxes, then all values are passed as array POST variable.

<input type='checkbox' name='id[]' value='$row[id]'/>

answered Dec 1, 2010 at 17:10

Jason

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Comments

Rostyslav Dzinko · Accepted Answer · 2012-08-18 09:13:38Z

0

Change

name=id

to

name=id[]

this will then give you an array.

edited Aug 18, 2012 at 9:13

Rostyslav Dzinko

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answered Dec 1, 2010 at 17:10

Shaun Hare

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