jQuery Ajax不工作问题

Question

用于使用PHP更新数据库表项的jQuery。阿贾克斯代码不工作后，请帮助。

$("#dynamic-table").on("click", ".submit", function() { 
 
    var rowID = $(this).attr("id"); 
 
    var allottedValue = $(this).parent().find('input').val(); 
 
    alert('Row id = ' + rowID + ' Enrollment no = ' + allottedValue); 
 

 
    var dataString = 'allottedEnroll=' + allottedValue + '&rowid=' + rowID; 
 

 
    // After this line it is not working 
 
    $.ajax({ 
 
     type: "POST", 
 
     url: "request/allot_enrollmentNo_gov.php", 
 
     data: dataString, 
 
     success: function (html) { 
 
      $(this).parents(".success1").replaceWith(html); 
 
     } 
 
    }); 
 
    //$(this).parents(".success1").animate({backgroundColor: "#003"}, "slow").animate({opacity: "hide"}, "slow"); 
 
});

// HTML to Show Multiple Inputbox for multiple upload with link 
 

 
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script> 
 
<div id="dynamic-table"> 
 
    <div class="success1"> 
 
     <input name="enrollNo" type="text" value="" class="postEnroll"/> 
 
     <br/>&nbsp; 
 
     <a href="#" id="TakeFromDB" class="text-success submit">Allot Enrollment No</a> 
 
    </div> 
 
</div>

Answer 1

您想通过父母来代替，它必须是孩子。使用下面的代码

success: function (html) { 
    $("#dynamic-table .success1").replaceWith(html); 
} 

OR

success: function (html) { 
    $("#dynamic-table").find(".success1").replaceWith(html); 
} 

Answer 2

此代码的伟大工程，我已经使用了类似的....试试这个..

function FormSubmit() {  
    $.ajax(
     type: "POST", 
     url: 'success1.php', 
     data: $("#attend_data").serialize(), 
     async: false 
     }).done(function(data) { 
      $("#attend_response").html(data); 
    }); 
} 

Answer 3

我已经更新了下面的代码片段。 ..离线测试代码正在工作：

几个指针：

取而代之的是：request/allot_enrollmentNo_gov.php

试试这个：/request/allot_enrollmentNo_gov.php

通知斜杠(/)这表明阿贾克斯路径必须从具体取决于您的服务器设置的根目录开始。

使用此：下面的$(".success1").html(html);代替$(this).parents(".success1").replaceWith(html);

工作代码：

$("#dynamic-table").on("click", ".submit", function() { 
 
     var rowID = $(this).attr("id"); 
 
     var allottedValue = $(this).parent().find('input').val(); 
 
     alert('Row id = ' + rowID + ' Enrollment no = ' + allottedValue); 
 

 
     var dataString = 'allottedEnroll=' + allottedValue + '&rowid=' + rowID; 
 

 
     // After this line it is not working 
 
     $.ajax({ 
 
      type: "POST", 
 
      url: "/request/allot_enrollmentNo_gov.php", 
 
      data: dataString, 
 
      success: function (html) { 
 
       $(".success1").html(html); 
 
       alert('Response from the POST page = ' + html + '); 
 
      } 
 
     }); 
 
     //$(this).parents(".success1").animate({backgroundColor: "#003"}, "slow").animate({opacity: "hide"}, "slow"); 
 
    });

// HTML to Show Multiple Inputbox for multiple upload with link 
 

 
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script> 
 
<div id="dynamic-table"> 
 
    <div class="success1"> 
 
     <input name="enrollNo" type="text" value="" class="postEnroll"/> 
 
     <br/>&nbsp; 
 
     <a href="#" id="TakeFromDB" class="text-success submit">Allot Enrollment No</a> 
 
    </div> 
 
</div>

Answer 4

现在，它是通过设置家长可以正常使用：

$("#dynamic-table").on("click", ".submit", function() { 
    var rowID = $(this).attr("id"); 
    var rowParent = $(this).parent('.success1'); 
    var allottedValue = rowParent.find('input').val(); 

    var dataString = 'allottedEnroll=' + allottedValue + '&rowid=' + rowID; 
    $.ajax({ 
     type: "POST", 
     url: "request/allot_enrollmentNo_gov.php", 
     data: dataString, 
     success: function (html) { 
      rowParent.replaceWith(html); 
     } 
    }); 
    return false; 
});