如何从Ajax回调提交表单 [英] How to submit a FORM from ajax callback

问题描述

在我了解ajax的过程中，我遇到了这个问题，我不知道如何首先通过在我的表中请求数据来解决这个问题

through my journey on understanding ajax i came up with this problem and i don't know how to solve it first im requesting for datas in my table

load_data();
function load_data(page)  
      {  
           $.ajax({  
                url:"data.php",  
                method:"POST",  
                data:{page:page},  
                success:function(data){  
                     $('#result').html(data);  
                },
                error:function(exception){alert('Exeption:'+exception);}                
           })  
      }

这只是要输出的表格

    <table>
        <thead>
            <tr>
              <th>name</th>
              <th>details</th>
              <th>price</th>
              <th></th>
            </tr>
        </thead>
        <tbody id = "result">

这是网络"标签中的输出

and this is the output from network tab

来自网络"标签的响应

响应中有一个按钮，我试图使用此jquery激活它以提交其表单，但这不起作用

there's a button from the response i tried to activate it using this jquery to submit its form but this doesn't work

  $(document).on('click', '#add_details', function(){  
       $(this).parents("form").submit();
  });

推荐答案

将ID添加到您需要通过单击按钮提交的表单中，例如

add id to the form you need to submit with button click like

<form id="myForm1">
//Other Code Here

并使用您的按钮，只需使用您添加的ID

and with your button just Use the id you added

$(document).on('click', '#add_details', function(){  
       $("#myForm1").submit();
});

这篇关于如何从Ajax回调提交表单的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！