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up vote 0 down vote favorite

I am trying to create a table showing multiple users data. The data in the table will then be able to be edited and updated. Below is an example of the way the form is laid out:

    echo "<form action=AdminUpdateLecInfo.php method=post>";
    while ($return = mysql_fetch_assoc($result))    {
    $phonenumber = "$return[PhoneNumber]";
    $number = str_pad($phonenumber, 11, "0", STR_PAD_LEFT);
    echo " <tr class='data'>
           <input type='hidden' name='id'".$return['ID']."'' value= '".$return['ID']."' />
           <td class = 'title'><input class = 'title' type='text' name='title'".$return['ID']."'' value= '".$return['Title']."' /></td>
    }
    echo "</table>
    <input class='submit' type='submit' value='Update Info' />
    </form>

Once the table is created the information is passed to the 'update.php' script. 

    $sql="UPDATE completeinfo SET   Title='".$_POST['title'][$return['ID']]."'      
                          WHERE ID = '".$_POST['id'][$return['ID']]."'";
                          header("Location:Home.html");

The problem I'm having is that I need to add the '".$return['ID']."' to the name of each input field so that not all users details are updated with the same values. I am unsure if I need to apply a foreach loop around this query so that it applies to each user and updates their details. Currently however the update query is not working presumably because the post method is not fetching the values from the form correctly. Please bare with me because as im sure you can tell PHP is extremely new to me. Thanks for the help.

share|improve this question
1  
Please, before you write any more SQL interfacing code, you must read up on proper SQL escaping to avoid severe SQL injection bugs. Also, mysql_query should not be used in new applications. It's a deprecated interface that's being removed from future versions of PHP. A modern replacement like PDO is not hard to learn. A guide like PHP The Right Way will help you avoid making mistakes like this. – tadman Feb 28 '14 at 19:19
up vote 0 down vote

Your problem is the name of your fields, use that :

$return_id = $return['ID'];

echo "
<tr class='data'>
    <td class = 'title'>
        <input type='hidden' name='id$return_id' value='$return_id' />
        <input class='title' type='text' name='title$return_id' value='$return_id' />
    </td>
</tr>";

Before doing your mysql update, do var_dump($_POST);, you'll be shown the content of the HTTP POST parameters so you can see what they are and how to use them in the query.

share|improve this answer
up vote 0 down vote

You have to use the names of your text fields as array so you can iterate your array

name='title'".$return['ID']."'
name='id'".$return['ID']."'
share|improve this answer

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