Only the first character shows up after assigning one var to another

[David Lee Davis]

 <?php $counter = 1 ; ?>
 <?php foreach( $data as $row ) { ?>
 <tr>
    <td> 
      <?php $name[$counter]=$row['name'];?>
      <?php echo( $row['name'] ); ?>			[B]<!-- I get Brooklyn -->[/B]  	
      <?php echo( $name[$counter] ); ?>		
        [B]<!-- I get only first letter B -->[/B]
    </td>

shouldn't they both be the same??

[Sudaraka]

This depends on what is in your $data array, if it's a array of strings what you mentioned above could happen.
Other wise if it's a array of arrays with 'name' element in each, it should work.

$data=array(
	array('name'=>'Bytes.com'),
	array('name'=>'PHP'),
	array('name'=>'Question'),
	array('name'=>'908173'),
);

[Samishii23]

First off, as a general coding practice, try to do a little amount of <?php ?> in your page as possible. This reduces performance drastically. 2nd you don't need to put () around anything you want to echo.

$counter = 1 ;
echo $name[$counter];

If your only getting the 1st character of the $name in your foreach iteration, then whatever is being returned through the $name variable is a string.

In PHP if you want a specific character of a string variable, you can access it as if it were an array.

$str = 'hello world';
echo $str[0]; // h
echo $str[6]; // w

[Markus Igeland]

I think that adding this inside at the end of your foreach-loop will fix your problem:

$counter++;

If it doesn't help, I haven't really understood your problem, and will reread it after you give me feedback. :) Good luck to you, mister!

[NetDynamic]

Well, its bad practice to load an array with a counting integer anyway, i suggest you use:

$name[]=$row['name'];

the increment is automatic. Sorry i didnt read the rest I just saw something to tip you about. :)