php - AJAX 调用问题 - 没有数据传递(这里相对缺乏经验!)

Question

我认为我在这里做了一些非常简单的错误(对于所有经验丰富的程序员)，但是请原谅我，我花了很多时间阅读试图理解 AJAX 调用是如何工作的......

我的变量没有将其插入到recordsale.php 中以插入到SQL 表中。

JavaScript 变量显示在 HTML 页面上，因此我知道它们包含数据。

有人可以提供一些建议吗？

(任何影响我能力的回应都不会受到赞赏 - 我正在尝试:))

我的 html 表单提交按钮看起来像这样

<form action="recordsale.php" method="post">
     <input type="image" src="assets/CompleteSale.png" alt="Submit"> </form>

我的 AJAX Post 功能:

 $( "#submit" ).submit(function( event ) {
alert("hello");
            // Jquery code for making AJAX call to server side script
            $.ajax({
                method: "POST",
                url: "recordsale.php", 
                data: { adultqty: adultticketqty, concessionqty: concessionticketqty, studentqty: studentticketqty, childqty: childticketqty, programqty: programqty, totalsell: totalsaleprice }
            })
           //alert(adultqty + childqty)
        });

记录销售.php

<?php 

$adultticketqty = $_POST['adultqty']; 
$concessionticketqty = $_POST['concessionqty']; 
$studentticketqty = $_POST['studentqty']; 
$childticketqty = $_POST['childqty']; 
$programqty = $_POST['programqty']; 
$totalsaleprice = $_POST['totalsell']; 

//include('phpsqlget.php'); 

/* Attempt MySQL server connection. Assuming you are running MySQL server with default setting (user 'root' with no password) */ 

$link = mysqli_connect("localhost", "root", "speedway", "speedwayticketsales"); 
// Check connection 
if($link === false){ die("ERROR: Could not connect. " . mysqli_connect_error()); } // Attempt insert query execution 
$sql = "INSERT INTO stdsalesrecords (qtyadulttickets, qtyconcessiontickets, qtystudenttickets, qtychildtickets, qtyprograms, totalsell) VALUES ($adultticketqty,$concessionticketqty,$studentticketqty,$childticketqty,$programqty,$totalsaleprice)"; 

if(mysqli_query($link, $sql)){ 
    echo "Records inserted successfully."; } 
else{ echo "ERROR: Could not able to execute $sql. " 

    . mysqli_error($link); } // Close connection mysqli_close($link); 


?>

Answer 1

这就是我通常进行 ajax 提交的方式。还有其他处理表单提交的方法。查看 $(form).serialize() 等。

 $( "#submit" ).submit(function( event ) {
   var adultticketcty = $('selector').val();
   //continue filling out the fields and mirror them in the below dataString.

    var dataString = 'adultqty='+adultticketcty+'&concessionqty='+concessionticketqty;

  //Check if you set the correct values in the datastring.
   console.log(dataString);

            $.ajax({
                type: "POST",
                url: "recordsale.php", 
                data: dataString,
                success: function(data){
                  //check whatever is returned from recordsale.php
                  console.log(data);
                }
            });
        });

然后在你的recordsale.php中执行此操作以查看ajax成功调用中是否返回任何内容:

$adultticketqty = $_POST['adultqty']; 
$concessionticketqty = $_POST['concessionqty'];
echo $adultticketqty;

与问题无关，但我想建议检查 pdo 查询以防止 sql 注入(inject)等。