Question

我有以下代码：

<form id="insertStepsForm" method="POST" name="maxSteps" action="<?php $PHP_SELF; ?>" enctype="multipart/form-data">
  <input type="hidden" name="catId" value="<?php echo $category->cat_id; ?>" />
  <table id="insertStepsTable">
    <?php

    $amountOfSteps = $wpdb->get_results("SELECT `steps` FROM `metal_work_max_steps` WHERE `cat_id` = $category->cat_id");

    $stepsData = $wpdb->get_results("SELECT * FROM `metal_work_steps` WHERE `cat_id` = $category->cat_id");      


    foreach($amountOfSteps as $steps){

      $stepAmount = $steps->steps;

      for($i = 1 ; $i <= $stepAmount; $i++){

      ?>
        <tr>

          <td><label for=""><b>Step <?php echo $i; ?></b></label></td>
          <td><label>Name </label><input class="stepName" name="steps[<?php echo $i; ?>][name]" type="text" value="<?php foreach($stepsData as $stepData){ if($stepData->step_number == $i){ echo $stepData->step_name; } } ?>" /></td>
          <td><label for="">Type</label>


            <select name="steps[<?php echo $i; ?>][type]">

              <option <?php if($stepData->type == 'radio'){ echo "selected='selected'"; } ?> value="radio">Radio</option>
              <option <?php if($stepData->type == 'checkbox'){ echo "selected='selected'"; } ?> value="checkbox">Checkbox</option>
              <option <?php if($stepData->type == 'dropdown'){ echo "selected='selected'"; } ?> value="dropdown">Dropdown</option>

            </select>
          </td>
        </tr>
      <?php

      }

    }
    ?>
  </table><!-- #insert-steps -->
  <input type="submit" value="Update Steps" name="submitSteps" />
</form><!-- #insertStepsForm -->

输出以下代码：

    <select name="steps[1][type]">
        <option value="radio">Radio</option>
        <option value="checkbox" selected="selected">Checkbox</option>
        <option value="dropdown">Dropdown</option>
    </select>

现在所有的盒子都有相同的选择复选框，我明白这是因为我没有每个选择框的唯一标识符，但我不确定最佳实践方法来解决这个问题。？

由于

填充选中=＆＃34;选中＆＃34;来自db PHP的选项列表

0 个答案: