Question

您好我是AJAX的初学者，并且在下面的代码中遇到问题，但没有弹出警告信息。
我尝试使用AJAX将数据从表单插入数据库。
此代码成功将数据存储到数据库中。
但它没有弹出警告信息，这是主要问题。

**//index.php**

// following is ajax code in which problem occur. i think.
     $(document).ready(function(){  
          $('#submit').click(function(){            
               $.ajax({  
                    url:"store.php",  
                    method:"POST",  
                    data:$('#add_name').serialize(),  
                    success:function(data)  
                    {  
                         alert(data);  
                         $('#add_name')[0].reset();  
                    }  
               });  
          });  
     });



**//store.php** 
    
     // following is the data base releted query

      $conn = mysqli_connect("localhost", "root", "", "prag");  
      
     $chapter_name = $_POST["chapter_name"];  
     $class = $_POST["class"];  
     $subjects = $_POST["subjects"];  
    
       
                    $sql = "INSERT INTO chapter(class_id,subject_id,name) VALUES('$class','$subjects','$chapter_name')";  
                    mysqli_query($conn, $sql); 
    				echo "Data Inserted";  
    
    
    </pre>

//HTML CODE
//Following are the input fields through which data will store.

<input type="text" name="chapter_name" placeholder="Enter the Chapter name" class="form-control name_list" />
        
<select name="class" id="class" class="form-control">
    <option>Select Class</option>
</select>
        
<select name="subjects" id="subjects" class="form-control">			
</select>
        										
<button type="submit" name="add_chapter" id="submit">Add </button>
<pre>

Answer 1

使用以下代码：

 <!DOCTYPE html>
 <html>
 <head>
    <title></title>
    <script src="https://ajax.googleapis.com/ajax/libs/jquery/1.12.4/jquery.min.js"></script>
    <script type="text/javascript">
              $(document).ready(function(){  
      $('#submit').click(function(e){  
                      e.preventDefault();         
           $.ajax({  
                url:"data.php",  
                method:"post",  
                data:$('#add_name').serialize(),  
                success:function(data)  
                {  
                     alert(data);  
                     $('#add_name')[0].reset();  
                }  
           });  
      });  
 });

    </script>
  </head>
<body>
    <form method="post" id="add_name">
            <input type="text" name="chapter_name" placeholder="Enter the Chapter name" class="form-control name_list" />

<select name="class" id="class" class="form-control">
    <option>Select Class</option>
</select>

<select name="subjects" id="subjects" class="form-control">     
</select>

<button type="submit" name="add_chapter" id="submit">Add </button>
    </form>
</body>
</html>

检查PHP代码：

<?php 
$conn = mysqli_connect("localhost", "root", "", "prag");  


   $chapter_name = $_POST["chapter_name"];  
   $class = $_POST["class_name"];  
   $subjects = $_POST["subject_name"];  

  $sql = "INSERT INTO chapter(class_id,subject_id,name) VALUES('$class','$subjects','$chapter_name')";
  if(mysqli_query($conn, $sql)) {
    echo "Data Inserted";  
  } 
else {
  echo "Data Not Inserted";  
}
  ?>

关于成功的警报消息不适用于AJAX

1 个答案: