Question

我有一个表单提交页面，当提交表单时我需要调用Ajax。但是现在Ajax没有用。我的主页是add_sale.php，Ajax页面是ajx_check_sale.php

我的代码：

add_sale.php

 function validate_form()
     {
     var cust_name= $('#cust_name').val();
     var total= $('#total').val();
      var sale_type= $('#sale_type').val();

     if(sale_type=='return')
     {
        $.ajax({
        type: "POST",
        url: 'ajx_check_sale.php',
        data:'cust_name='+cust_name + '&total=' + total,
        success: function(msg)
         { 

         alert(msg);
          /*if(msg==0)
           {

             alert("Return is greater then sale"); 
             return false;  
           } */
         }
      });
     }
    }
    <form action="" method="post" name="adFrm" onSubmit="return validate_form()">

    </form>

ajx_check_sale.php

 require_once("codelibrary/inc/variables.php");
 require_once("codelibrary/inc/functions.php");
 echo $cust_name=$_POST['cust_name'];
 echo $return=$_POST['total'];

 $cus="select sum(total) as total_sum from customer where id='$cust_id'";
 $cus2=mysql_query($cus);
 $fet=mysql_fetch_array($cus2);
 $total=$fet['total_sum'];

if($return>$total)
 {
     $status=0; 
     echo $status;  
 }
else
 {
      $status=1;    
      echo $status;     
 }

Answer 1

如果您熟悉jquery，则在表单提交事件上使用 event.preventDefault（）函数，以便不会发生表单提交。这是一个示例：

$(form).submit(function(event){
    event.preventDefault();
})

Answer 2

也许像这样......

<script type="text/javascript">

 function validate_form()
 {
 var cust_name= $('#cust_name').val();
 var total= $('#total').val();
 if(sale_type=='return')  //what is variable sale_type here ?please define
 {
    $.ajax({
    type: "POST",dataType:"json",
    url: 'ajx_check_sale.php',
    data:'cust_name='+cust_name + '&total=' + total,
    success: function(msg)
     { 

      if(msg.status==0)
       {

         alert("Return is greater then sale"); 

       } 

       return false; 
     }
  });
 }else{

    return false; //or alert("someth went wrong)

}


}
</script>


<form action="" method="post" name="adFrm" onSubmit="return validate_form()">

</form>

和ajx_check_sale.php

 $data = $_POST;
  //.......

 if($return>$total)
 {
    echo json_encode(array('status'=>0));
 } 
 else
 {

   echo json_encode(array('status'=>1));    
 }

Answer 3

function validate_form(){
    var cust_name= $.trim($('#cust_name').val());//remove spaces
    var total= $.trim($('#total').val());//remove spaces
    if(sale_type=='return')
    {
        $.ajax({
            type: "POST",
            url: 'ajx_check_sale.php',
            data:{ cust_name : cust_name , total: total},//you can use this format, instead of passing the way you were doing
            success : (function(msg){//Success part of the ajax
                alert(msg);
                /*if(msg==0){
                    alert("Return is greater then sale"); 
                    return false;  
                } */
            }),
            error : (function(msg){//This part will run if there is any error from the API
                //error part
            })
        });
    }
}

使用jquery在表单提交时调用Ajax

3 个答案: