为什么Jquery Ajax获取方法实现

时间:2018-09-11 16:30:35

标签: javascript jquery ajax

我想要的是当我按下按钮时弹出甜蜜警报,并且需要调用给定的URL。我只是尝试JQuery ajax get方法。以下代码段无法正常工作。

<script src="https://cdnjs.cloudflare.com/ajax/libs/sweetalert/2.1.0/sweetalert.min.js"></script>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<button onclick="myFunction()">Accept your promotion</button>
<script>
    $.get("http://localhost:8080/update/status/"+pcode+"/ACCEPT", function myFunction(){
        $( ".result" ).html( data );
        swal(
            'Congratulations!',
            'You successfully accept your promotion',
            'success'
        )
    })
</script>

1 个答案:

答案 0 :(得分:1)

onclick事件正在调用名为myFunction()的未定义函数。将get请求包装在函数定义中:

var pcode = "test"; // Not sure where you're defining pcode?

myFunction = function(){ // Define your function
    $.get( "http://localhost:8080/update/status/"+pcode+"/ACCEPT", function(data){
        // Your get request is complete
        $( ".result" ).html( data );
        swal('Congratulations!','You successfully accept your promotion','success');
    });
};

这是一个有效的示例:http://jsfiddle.net/r5uvwj7o/

将网址从"http://echo.jsontest.com/"更改为您的网址"http://localhost:8080/update/status/"+pcode+"/ACCEPT"