doule6314 2016-10-19 03:19
浏览 23

将每个值传递给php脚本

I am passing a table data value within a for each loop to the php script. The problem here is that when I try to echo $exchange;. I don't get any values of the specific product name. The product name being displayed in the form is item1.

<form action="exchangeItem" method="post">    
 <tbody>
                    <?php
                      $query = "SELECT * FROM product";
                      $data = $MySQLi_CON->query($query);
                      foreach ($data as $key ) {              
                    ?>
                    <tr>                            
                      <td class="id"><h3><input type="hidden" name="productName" ><?php echo $key['product_Name'];?></h3>
                      </td>
                    <input name = "id" type = "hidden" value="<?php echo $id; ?>">
                    <td id="exchange_button">
                          <input value="Exchange" name="exchange_submit" type="submit" class="btn btn-primary btn-md raised exchange_submit">
                      </td>
                    </tr>
                    <?php
                      }
                      ?>                     
    </tbody>
</form>

exchangeItem.php

<?php
    $exchange = $_POST['productName'];
    echo $exchange;
    exit;
?>
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5条回答 默认 最新

  • douyi6168 2016-10-19 03:23
    关注

    This

    <td class="id"><h3><input type="hidden" name="productName" ><?php echo $key['product_Name'];?></h3>
    

    Should be:

    <td class="id"><h3><input type="hidden" name="productName" value="<?php echo $key['product_Name'];?>" ></h3>
    
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