Question

问题：

使用jQuery检测在通过PHP发送表单后按下了哪个提交按钮。

情境：

我有三个按钮：

submit-update1
提交-UPDATE2
提交-UPDATE3

如果在PHP中设置了$ _POST ['submit-update1']，那么jQuery应该将一个类添加到id为“collapseOne”的div中。例如：

if (PHP POST submit button 1 is submitted)
{
    $('#collapseOne').addClass('in');
}

if (PHP POST submit button 2 is submitted)
{
    $('#collapseTwo').addClass('in');
}

if (PHP POST submit button 3 is submitted)
{
    $('#collapseThree').addClass('in');
}

DIV元素：

<div class="accordion" id="accordion2">
    <div class="accordion-group">
      <div class="accordion-heading">
        <a class="accordion-toggle" data-toggle="collapse" data-parent="#accordion2" href="#collapseOne">
          Title 1
        </a>
      </div>
      <div id="collapseOne" class="accordion-body collapse">
        <div class="accordion-inner">
            Text here
        </div>
      </div>
    </div>
    <div class="accordion-group">
      <div class="accordion-heading">
        <a class="accordion-toggle" data-toggle="collapse" data-parent="#accordion2" href="#collapseTwo">
          Title 2
        </a>
      </div>
      <div id="collapseTwo" class="accordion-body collapse">
        <div class="accordion-inner">
          Text here
        </div>
      </div>
    </div>
    <div class="accordion-group">
      <div class="accordion-heading">
        <a class="accordion-toggle" data-toggle="collapse" data-parent="#accordion2" href="#collapseThree">
          Title 3
        </a>
      </div>
      <div id="collapseThree" class="accordion-body collapse">
        <div class="accordion-inner">
          Text here
        </div>
      </div>
    </div>
</div>

提交按钮：

<button type="submit" id="submit-update1" name="submit-update1">Update 1</button>
<button type="submit" id="submit-update2" name="submit-update2">Update 2</button>
<button type="submit" id="submit-update3" name="submit-update3">Update 3</button>

欢迎任何建议，并提前致谢！

Answer 1

  <form method="post" id="many_buttons">
    <input type="hidden" id="submit_button" name="button_pressed" />

    <button type="submit" id="submit-update1" name="submit-update1">Update 1</button>
    <button type="submit" id="submit-update2" name="submit-update2">Update 2</button>
    <button type="submit" id="submit-update3" name="submit-update3">Update 3</button>
  </form>

  <script>
  $('#many_buttons button[type=submit]').click(function(){
    $('#submit_button').val($(this).attr('name'))
  })
  </script>

然后在服务器端：

  <?php
  if ($_SERVER['REQUEST_METHOD'] == 'POST') {
    $button = $_POST['button_pressed'];

    if ($button == 'submit-update1')
      $div = 'collapseOne';
    elseif ($button == 'submit-update2')
      $div = 'collapseTwo';
    elseif ($button == 'submit-update3')
      $div = 'collapseThree';
  }
  ?>

最后你打印这个javascript：

  <?php if (!empty($div)) : ?>
  <script>$('#<?php echo $div ?>').addClass('in')</script>
  <?php endif ?>

Answer 2

我假设您已经找到了帖子提交按钮，但是根据这些变量操作页面时遇到了问题。您可以使用动态脚本根据post变量设置一些变量，如下所示：

echo "<script type='text/javascript'>";
if (isset($_POST['submit-update1']) && $_POST['submit-update1'] == true) {
    echo "var submitUpdate1 = true;";
}
else {
    echo "var submitUpdate1 = false;";
}
///...do similar things for the rest of your post variables
echo "</script>";

然后在您的javascript中，您只需在文档准备就绪后访问上面定义的那些变量。

Answer 3

将此添加到您的html：

<input type="hidden" name="submit" value="" />

然后这是jQuery：

$('button[type="submit"]').on('click', function(e){
    $('form input[name="submit"]').val($(this).attr('name'));
    $('form').serialize();
});

如何检测PHP中通过POST发送的提交按钮

3 个答案: