使用属性从多维数组中删除项目

Question

I am trying to display an array of data produced from an sql query. this works fine. However I now wish not to display 1 item where the id attribute = a value. code:

<?php foreach($sheet_list as $key => $sheet) : ?> 
        <?php if($sheet['id'] == $selected1){unset ($sheet_list[$key]);}?>
        <tr>
            <td><?=$sheet['title'];?></td>
            <td><?php echo date('d F Y', strtotime($sheet['startDate'])); ?></td>
            <td><?php echo date('d F Y', strtotime($sheet['endDate'])); ?></td>
            <td><input type="radio" name="selected2" id="selected2" value="<?=$sheet['id'];?>" <?php 
            if ($key == 0) echo ' checked ';
            ?>/></td>
        </tr>
        <?php endforeach; ?>

so this is producing a table which is fine, however it is in the foreach loop where i want NOT to display 1 array item and all its attributes where the "id" field is = to the variable $selected1 I have seen lots of questions posted here, however many for 1 or two dimensional arrays. Please help.

Answer 1

You're close, you should be using continue; instead of unset():

<?php if($sheet['id'] == $selected1){ continue; } ?>

continue^{PHP Docs} makes the loop skip to the next iteration, effectively skipping the current iteration, which is what you want to do.

The benefit here is that you never modify for you $sheet_list array, unlike with unset() where you'll remove that item from the array.