PHP + Jquery - 通过ajax将值传递给php并检查变量

Question

我不能让下面的PHP + jQuery的工作 - 所有我想要的脚本做的是通过Ajax传递价值,并获得PHP的抓住它,检查它匹配,并且加1分.

这是我写的代码:

<?php
$score = "1";

$userAnswer = $_POST['name'];

if ($_POST['name'] == "145"){
    $score++;
}else{
    //Do nothing
}

echo $score;

?>


<script type="text/javascript">

$(document).ready(function() {

    $("#raaagh").click(function(){

    var value = "145";

    alert(value);

    $.ajax({
        url: 'processing.php', //This is the current doc
        type: "POST",
        data: ({name: value}),
        success: function(){
            location.reload();
       }
    });


    });
});

</script>

<p id="raaagh">Ajax Away</p>

感谢您的帮助,我在两个实例中都将GET更改为POST,并且没有任何乐趣 - 还有其他错误.

Answer 1

首先:不要回到黑暗时代......不要使用相同的脚本来生成HTML并响应ajax请求.

我无法理解你想要做什么......让我改变你的代码,这样至少可以理解并记录正在发生的事情.似乎问题在于您从成功处理程序调用location.reload.

// ajax.php - 如果name参数是145则输出2,否则输出1(????)

<?php
$score = "1";    
$userAnswer = $_POST['name'];    
if ($_POST['name'] == "145"){
    $score++;
}       
echo $score;    
?>

// test.html

<script type="text/javascript">  
$(document).ready(function() {    
    $("#raaagh").click(function(){    
        $.ajax({
            url: 'ajax.php', //This is the current doc
            type: "POST",
            data: ({name: 145}),
            success: function(data){
                // Why were you reloading the page? This is probably your bug
                // location.reload();

                // Replace the content of the clicked paragraph
                // with the result from the ajax call
                $("#raaagh").html(data);
            }
        });        
    });
});

</script>

<p id="raaagh">Ajax Away</p>