Uncaught TypeError: undefined is not a function jquery答案

【问题标题】：Uncaught TypeError: undefined is not a function jqueryUncaught TypeError: undefined is not a function jquery
【发布时间】：2014-11-11 06:43:35
【问题描述】：

我对 ajax 和 jsonp 真的很陌生，并且在调用文件时遇到了问题。该代码有效。但是每次我在同一个脚本中再次调用同一个函数时，它都会显示“未捕获的 TypeError：未定义不是一个函数”。如果该功能只工作一次，它不应该总是工作吗？

这是我的代码示例

var resultAmount = 0;

start = function(teamFile, rowsInDB, ratio_Over_rows, opplastx_gp, callfunction){

    //ajax ONLY calls don't return anything
    (function($) {
    //Connects to the json teamFile
    var url = 'http://xxx.co.uk/football/'+teamFile+'.json?callback=?';
    //Automatic refresh

    $.ajax({
    type: 'GET',
    url: url,
    async: false,
    jsonpCallback: 'jsonCallback',
    contentType: "application/json",
    dataType: 'jsonp',
    success: function(data) {

        if(callfunction == 'mep'){

            resultCount(data, rWin, count);
            resultCount(data, rDraw, count);
            resultCount(data, rLose, count);

            //the total of w/d/l
            resultAmount =  total[rWin] + total[rDraw] + total[rLose] ;

        }else{}

    },
    error: function(e) {
       console.log(e.message);
    }
});
})(jQuery);

}


//Adds the results w, d, l up
 resultCount = function(for_data, result, count_r){

    count_r = 0;

    //Goes through the data
    for(k in for_data){
        //if equals w, d, 1
        if(for_data[k].Results == result){
            //Add 1
            count_r++;
        }else{

        } 
    }
  }

//Then I call the function start twice only one works
console.log(start('ast', 7,5,5, 'mep'));
console.log(start('ars', 7,5,5, 'mep'));

只有第一个函数运行而不是第二个它说“未捕获的类型错误：未定义不是函数”。当我围绕第一个函数运行更改它们时，第二个说“未捕获的类型错误：未定义不是函数”。

如果它有助于我的文件看起来像这样

jsonCallback([{"Brad Guzan":"yes","Jed Steer":"no","Ashley Westwood":"yes","Fabian Delph":"no","Ron Vlaar":"yes","Andreas Weimann":"yes","Gabriel Agbonlahor":"no","Nathan Baker":"yes","Leandro Bacuna":"yes","Karim El Ahmadi":"no","Christian Benteke":"no","Ciaran Clark":"no","Matthew Lowton":"yes","Ryan Bertrand":"yes","Antonio Luna":"no","Marc Albrighton":"yes","Libor Koz\u00e1k":"no","Aleksandar Tonev":"no","Yacouba Sylla":"no","Grant Holt":"yes","Joseph Bennett":"yes","Chris Herd":"no","Jordan Bowery":"no","Jores Okore":"no","Gary Gardner":"no","Daniel Johnson":"no","Nicklas Helenius":"no","Jack Grealish":"no","Janoi Donacien":"no","Callum Robinson":"no","last_gp":"lose","2nd_gp":"lose","3rd_gp":"win","4th_gp":"lose","5th_gp":"lose","Home":"home","Results":"lose"});

【问题讨论】：

你的意思是错误在线console.log(start('ars', 7,5,5, 'mep'));？或者堆栈跟踪还有更多内容吗？
这就是你的全部代码吗？因为如果是的话，你的第二个函数是不完整的，不会编译。因此它不会运行。
您在返回的数据中还缺少一个右数组括号。
是的，它是 console.log(start('ars', 7,5,5, 'mep'));完整的代码有右括号，它只是很多代码，我不得不缩短它以举例说明正在发生的事情
搜索所有代码，确保在其他地方没有start = ，因为这会重新定义您的函数定义。

标签： javascript jquery ajax function jsonp

【解决方案1】：

我发现这两个文件都将“jsonCallback”作为正在读取的文档的函数。每个 jsonCallback 或函数都应该是唯一的。以前发生的事情是冲突，现在是独一无二的。

例子：

function handleData( responseData ) {
    // do what you want with the data
    console.log(responseData);
}

$.ajax({
    url: "hi.php",
    ...
    success: function ( data, status, XHR ) {
        handleData(data);
    }
});

【讨论】：