Question

您好我正在尝试制作DOM事件。当我加载到类的顶部时，我打算初始化一个事件，它将改变我的表单的动作，然后提交表单。

我的代码：

<?php $id = 1 ?>

<html>
<head>
<style>
div.box_images{
    background-color:black;
    color:  white;
    margin: 50px 0 50px 0;
    padding: 10px;
} 

</style>    
</head>
<body>
<form name ="form_test" action="test_form2.php" method="post">

    <input hidden="hidden" name="id_image" value="<?php echo $id; ?>" >
    <div id = "<?php echo $id; ?>" class="box_images" onclick="action_form()">
        <h2>IMAGE_<?php echo $id; ?></h2>
        <img class="main_img" src="arquivo/fotos/images_<?php echo $id; ?>.jpg">
    </div>


</form> 
<script>
function action_form(){
    document.('form_test').action = "form_upload.php";
    document.form_test.submit();
}   
</script>
</body>
</html>

甚至需要帮助，我完全被困在这部分代码中。

提前致谢！

Answer 1

您的语法略有偏差。改变这个

document.('form_test').action = "form_upload.php";

到

document.form_test.action = "form_upload.php";

＆＃13;

<form name="form_test" action="test_form2.php" method="post">
  <div class="box_images" onclick="action_form()">
    <h2>IMAGE_</h2>
  </div>
</form>
<script>
  function action_form() {
    document.form_test.action = "form_upload.php";
    console.log(document.form_test.action);
  }
</script>

＆＃13;

div类的DOM事件触发器

1 个答案: