Question

点击一个链接后，会触发ajax require /material/materialPrintajax，这将返回JSON格式数据，返回的JSON数据已经通过http://jsonlint.com/验证，网站告诉我我的返回JSON是有效的。但是当我遇到时， console.log(jsonResult);始终在firebug控制台中返回null。

stackoverflow中的前一个解决方案并没有解决这个问题，正如我所说：

http://stackoverflow.com/questions/6465468/why-parsejson-returns-null
http://stackoverflow.com/questions/8575479/parsing-json-string-returns-null

依旧......

以下是我的代码：

$.ajax({
    url : $.att.config.BASE_URL + "/material/materialPrintajax",
    type : "POST",
    dataType : "json",
    data:   {
        "choosenCourse" : choosenCourse,
        "choosenDate" : choosenDate,
        "choosenElCode" : choosenElCode
    },
    //read json result & display print data in HTML 
    success : function(re){
        console.log('good!!',re);
        jsonResult = jQuery.parseJSON(re);
        console.log(jsonResult);
    },
    error :function(re) {
        //eg. if return data is not json format
        console.log('error!!',re);
    }

});

如果您需要，这是我返回的JSON（重新）：

{"COOKING":{"A":{"0":{"CLASS_MENU_SYMBOL":"A","STAFF_ID":"3010120001","LAST_NAME_KANJI":"\u6817\u7530","FIRST_NAME_KANJI":"\u6607","STAFF_START_TIME":"09:00","STAFF_END_TIME":"10:00","COURSE_ID":"0","COURSE_NAME_JP":"\u6599\u7406","RESERVE_COUNT":1},"RESERVE_TOTAL":3,"1":{"CLASS_MENU_SYMBOL":"A","STAFF_ID":"3010120012","LAST_NAME_KANJI":"\u9648","FIRST_NAME_KANJI":"\u6167\u5a77","STAFF_START_TIME":"21:00","STAFF_END_TIME":"22:00","COURSE_ID":"0","COURSE_NAME_JP":"\u6599\u7406","RESERVE_COUNT":2}},"B":{"0":{"CLASS_MENU_SYMBOL":"B","STAFF_ID":"3010120001","LAST_NAME_KANJI":"\u6817\u7530","FIRST_NAME_KANJI":"\u6607","STAFF_START_TIME":"13:00","STAFF_END_TIME":"14:00","COURSE_ID":"0","COURSE_NAME_JP":"\u6599\u7406","RESERVE_COUNT":1},"RESERVE_TOTAL":1}},"BREAD":{"B7":{"0":{"CLASS_MENU_SYMBOL":"B7","STAFF_ID":"3010120010","LAST_NAME_KANJI":"\u738b","FIRST_NAME_KANJI":"\u5a77","STAFF_START_TIME":"15:00","STAFF_END_TIME":"17:00","COURSE_ID":"1","COURSE_NAME_JP":"\u9762\u5305","RESERVE_COUNT":1},"RESERVE_TOTAL":1},"B1":{"0":{"CLASS_MENU_SYMBOL":"B1","STAFF_ID":"3010120010","LAST_NAME_KANJI":"\u738b","FIRST_NAME_KANJI":"\u5a77","STAFF_START_TIME":"15:00","STAFF_END_TIME":"17:00","COURSE_ID":"1","COURSE_NAME_JP":"\u9762\u5305","RESERVE_COUNT":1},"RESERVE_TOTAL":1}},"CAKE":{"12":{"0":{"CLASS_MENU_SYMBOL":"12","STAFF_ID":"3010120012","LAST_NAME_KANJI":"\u9648","FIRST_NAME_KANJI":"\u6167\u5a77","STAFF_START_TIME":"09:00","STAFF_END_TIME":"11:00","COURSE_ID":"2","COURSE_NAME_JP":"\u86cb\u7cd5","RESERVE_COUNT":2},"RESERVE_TOTAL":2}}}

Answer 1

jQuery.parseJSON期望JSON字符串作为参数。你传递的不是一个字符串。它已经是一个对象了。 jQuery自动将服务器返回的JSON字符串解析为javascript对象，该对象是传递给success方法的对象。您无需再次致电parseJSON。您可以直接使用该对象。

success : function(re){
    console.log(re.COOKING.A);
},

Answer 2

jQuery在进行AJAX调用时会自动尝试解析返回的数据。

这意味着，您实际上是在尝试解析已经解析过的对象。

解决方案是不要试图解析它并直接使用它。

另见

Documentation page on jQuery.ajax()（dataType和Data Types部分解释了它。）

Answer 3

只需将dataType更改为false

$.ajax({
url : $.att.config.BASE_URL + "/material/materialPrintajax",
type : "POST",
dataType : false,
data:   {
    "choosenCourse" : choosenCourse,
    "choosenDate" : choosenDate,
    "choosenElCode" : choosenElCode
},
//read json result & display print data in HTML 
success : function(re){
    console.log('good!!',re);
    jsonResult = jQuery.parseJSON(re);
    console.log(jsonResult);
},
error :function(re) {
    //eg. if return data is not json format
    console.log('error!!',re);
}

}）;

jQuery.parseJSON始终在firebug控制台中返回null

3 个答案:

另见