PHP foreach nested table

Question

<table style="width: 600px" class="slicedTable">
     <tr>
      <th>Spetsialist</th>
      <th>Tunnid</th>
     </tr>


<tr>            
       <?php foreach($specs as $specName => $spec): ?>
        <td><?php echo $specName?>
        <tr>
       <?php endforeach; ?>


<td>
        <?php foreach($tunnid as $tund): ?>
           <td><?php echo $tund?></td>

       </td>
       <?php endforeach; ?>
     </tr>
 </table>

I need this to print out a regular table where it goes like:
name 1 - value1
name2 - value2

Specs:

$specs = array();

            foreach($data as $row) {
                $workerName = $row['Worker']['name'];
                if (!isset($specs[$workerName])) {
                    $specs[$workerName] = array('procs' => array(), 'extras' => array());
                }

However I can't seem to accomplish this, help please.

Accidentally left it in, i tried two tables next to eachother from desperation. — Kenksk
– Kenksk, Commented Jun 19, 2013 at 7:32

Jaime Gris · Accepted Answer · 2013-06-19 08:13:39Z

1

Here is the code I fix for you:

Assuming the number of elements in $spec is same as or more than the number of elements in $tunnid. In coder's perspective, count($spec) >= count($tunnid).

<table style="width: 600px" class="slicedTable">
   <tr>
      <th>Spetsialist</th>
      <th>Tunnid</th>
   </tr>
   <?php 
      $i = 0;
      foreach($specs as $specName => $spec): 
    ?>
   <tr>            
       <td><?php echo $specName; ?></td>
       <td><?php echo isset($tunnid[$i]) ? $tunnid[$i] : '-'; ?></td>
    </tr>
    <?php 
         $i++;
      endforeach; 
    ?>
 </table>

Not really an elegant solution, I admit.

edited Jun 19, 2013 at 8:13

answered Jun 19, 2013 at 7:40

Jaime Gris

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4 Comments

Kenksk Over a year ago

but what if it's not? right now the table looks fine but the right values are next to eachother only when the number is the same.

Jaime Gris Over a year ago

If the number of elements in $tunnid is greater than number of elements in $specs, the remaining elements in $tunnid will not be shown.

Kenksk Over a year ago

okay, then there must be something wrong with my query, i'm not getting the right values next to the right names..

Jaime Gris Over a year ago

I updated the answer, now includes if count($spec) >= count($tunnid).

gkalpak · Accepted Answer · 2013-06-19 09:21:48Z

0

UPDATE:
Seems like your arrays might have different lengths, so I updated the code below to handle such a situation (it doesn't matter which array is longer).

You could do it like this:

<table style="width: 600px" class="slicedTable">
    <tr>
        <th>Spetsialist</th>
        <th>Tunnid</th>
    </tr>
<?php $len1 = count($specs); $len2 = count($tunnid);
      $specsKeys = array_keys($specs);
      for ($i = 0; $i < max($len1, $len2); $i++) {
          $name  = ($i < $len1) ? $specsKeys[$i] : ""; 
          $value = ($i < $len2) ? $tunnid[$i] : ""; ?>
    <tr>
        <td><?php echo($name); ?></td>
        <td><?php echo($value); ?></td>
    </tr>
<?php } ?>
</table>

See, also, this short demo (*updated*).

edited Jun 19, 2013 at 9:21

answered Jun 19, 2013 at 7:45

gkalpak

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2 Comments

gkalpak Over a year ago

If you take a look at the demo provided, you can see there is no error. The only reason I can think of to raise an error in your code would be if $specs and $tunnid don't have the same length or if you are using a variable named $i elsewhere in your code. In any case without the exact error (and code) I can't figure out what the problem in your code is.

gkalpak Over a year ago

@user2485439: Taking into account your comments in another answer it seems like your arrays might have different lengths. I know you have probably solved your problem by now, but in any case I updated my answer to be able to handle arrays of different lengths.

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