How to submit the multiple rows form data with a single submit in PHP

Question

Here is my code. I made a form with multiple rows which will be updated in a single submit. I searched in forums, but didn't find the exact answer.

<form method='POST' action='demo24.php'>
  <table  width="500px" height="500px">
    <tr><th>SETNAME</th><th>POST</th></tr>
    <?php
      $query = "SELECT name,setid FROM `set` LIMIT 0 ,10";
      $result = mysql_query($query) or die(mysql_error()."[".$query."]");
      while($row = mysql_fetch_array($result))
      {
        $output .='<tr><td><input type="text" value="'.$row['name'].' '.$row['setid'].'" name="name'.$row['setid'].'">'.$row['name'].'</td><td><input type="checkbox" value="'.$row['setid'].'" name="setid"></td></tr>';
      }  

      echo $output;
    ?>


    </td><td><input type='submit' value='submit' name='submit'/></td></tr>
  </table>        
</form>

metho= != method="" (typo)

user3769335
– user3769335

2014-06-28 09:11:37 +00:00
Commented Jun 28, 2014 at 9:11 — user3769335
– user3769335, Commented Jun 28, 2014 at 9:11

chwcz7F · Accepted Answer · 2014-06-28 09:19:43Z

1

Note that you can use something like name[] in multiple input fields to send an array to the server:

<input type="text" name="row[]" />
<input type="text" name="row[]" />

On the server side $_POST['row'] will look like this:

Array (
  [0] => first_input,
  [1] => second_input
)

You can do this for all fields (id, row, etc.) and then loop through $_POST['id'] etc. to get all entries back together. Be sure to validate enough, i.e. make sure that those fields are indeed arrays, that all are of the same size, etc.

The MySQL family of PHP is deprecated and support thereof will disappear. Please look into PDO or MySQLi to execute SQL code instead.

answered Jun 28, 2014 at 9:19

chwcz7F

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Comments

Suman Biswas · Accepted Answer · 2014-06-28 09:16:23Z

0

If your form method as 'post' then you can use this code :

if(isset($_POST['submit']))
{
   $query = "SELECT name,setid FROM `set` LIMIT 0 ,10";
   $result = mysql_query($query) or die(mysql_error()."[".$query."]");
   while($row = mysql_fetch_array($result))
   {
      $id = $row['setid'];
      $name = $_POST['name'.$id];
     //here will be you update code. set value will be $name and update where setid = $id
   }
}

answered Jun 28, 2014 at 9:16

Suman Biswas

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