JSON.parse 返回错误答案

【问题标题】：JSON.parse returning errorJSON.parse 返回错误
【发布时间】：2016-12-09 13:01:35
【问题描述】：

所以这是我在尝试运行 JSON.parse 之前返回的 JSON.stringify

{"id":"2","name":"<small>L</small>(+)-Amethopterin Hydrate","class":"6.1","subclass":"","packing_group":"III","un":"2811","cas":"133073-73-1","poisons":"","hazardous":"Yes","restricted":"No","epa":"","added_by":"0","carcinogen":null},
{"id":"3","name":"(+)-Biotin 4-Nitrophenyl ester","class":"","subclass":"","packing_group":"","un":"","cas":"33755-53-2","poisons":"","hazardous":"No","restricted":"No","epa":"","added_by":"0","carcinogen":null},
{"id":"4","name":"(+)-Biotin N-hydroxysuccinimide ester","class":"","subclass":"","packing_group":"","un":"","cas":"35013-72-0","poisons":"","hazardous":"No","restricted":"No","epa":"","added_by":"0","carcinogen":null}

当我尝试 JSON.parse 时，我得到Unexpected end of JSON input。而且我不能将它作为 JSON 对象访问，因为它会说 can't define id 或类似的东西。

JSON.parse(this.searchService.searchJson(this.php_url));

this.searchService.searchJson(this.php_url) 基本上就是我的 JSON 字符串。如上所述给出错误。

此外，如果我只是尝试对 3 个元素中的 1 个进行字符串化，它会给我Unexpected token u in JSON at position 0

调用函数：

searchJson(url: any): any 
{   
    let items: any = []; 
    let new_data: any = []; 
    $.getJSON(url , 
        function(data)
        {   
            let temp_items: any = {}; 
            console.log(data);
            $.each(data, function (key, val)
            {   
                new_data.push(JSON.stringify(val));
            }); 
        }); 
    return new_data;
}

【问题讨论】：

介意给我们看一下解析代码吗？
@TheLazySloth 更新了它。返回的基本上是一个字符串化的数组。
就是用逗号写的3个对象..应该是数组什么的..jsfiddle.net/c8743h70
在 JSONLint.com 输入 JSON 并显示错误。将其添加到 [ ] 中，错误消失了。试一试？
@echonax 问题是......我很确定它们在一个数组中，我会用我的调用函数更新

标签： javascript json angular typescript

【解决方案1】：

你必须用[] 包裹它，因为那是一个对象数组：

const data = [{"id":"2","name":"<small>L</small>(+)-Amethopterin Hydrate","class":"6.1","subclass":"","packing_group":"III","un":"2811","cas":"133073-73-1","poisons":"","hazardous":"Yes","restricted":"No","epa":"","added_by":"0","carcinogen":null}, {"id":"3","name":"(+)-Biotin 4-Nitrophenyl ester","class":"","subclass":"","packing_group":"","un":"","cas":"33755-53-2","poisons":"","hazardous":"No","restricted":"No","epa":"","added_by":"0","carcinogen":null}, {"id":"4","name":"(+)-Biotin N-hydroxysuccinimide ester","class":"","subclass":"","packing_group":"","un":"","cas":"35013-72-0","poisons":"","hazardous":"No","restricted":"No","epa":"","added_by":"0","carcinogen":null}]

console.log(data); 将返回 [Object, Object, Object]

或者如果你想作为 JSON 字符串处理，你应该这样做：

const data = '[{"id":"2","name":"<small>L</small>(+)-Amethopterin Hydrate","class":"6.1","subclass":"","packing_group":"III","un":"2811","cas":"133073-73-1","poisons":"","hazardous":"Yes","restricted":"No","epa":"","added_by":"0","carcinogen":null}, {"id":"3","name":"(+)-Biotin 4-Nitrophenyl ester","class":"","subclass":"","packing_group":"","un":"","cas":"33755-53-2","poisons":"","hazardous":"No","restricted":"No","epa":"","added_by":"0","carcinogen":null}, {"id":"4","name":"(+)-Biotin N-hydroxysuccinimide ester","class":"","subclass":"","packing_group":"","un":"","cas":"35013-72-0","poisons":"","hazardous":"No","restricted":"No","epa":"","added_by":"0","carcinogen":null}]'

JSON.parse(data) 也会返回 [Object, Object, Object]

【讨论】：

当我 console.log(data) 返回我[Object {id="2", ...}, Object {id="3", name="(+)-Biotin 4-Nitrophenyl ester", ...}, Object {id="4", ...}] 所以它是 JSON 格式，但由于某种原因我无法访问它。

【解决方案2】：

把调用函数改成这样：

searchAjax(url: any): any
{   
    let new_data: any;
    return $.ajax({
        url: url,
        type: 'post',
        dataType: "json", 
        async: false

    }).responseText;
}

最可能的原因是我的变量是因为异步，我的变量在被调用时为空。

【讨论】：

【解决方案3】：

file Nmae: self.json
    [
          {
            "id": "2",
            "name": "<small>L</small>(+)-Amethopterin Hydrate",
            "class": "6.1",
            "subclass": "",
            "packing_group": "III",
            "un": "2811",
            "cas": "133073-73-1",
            "poisons": "",
            "hazardous": "Yes",
            "restricted": "No",
            "epa": "",
            "added_by": "0",
            "carcinogen": null
          },
          {
            "id": "3",
            "name": "(+)-Biotin 4-Nitrophenyl ester",
            "class": "",
            "subclass": "",
            "packing_group": "",
            "un": "",
            "cas": "33755-53-2",
            "poisons": "",
            "hazardous": "No",
            "restricted": "No",
            "epa": "",
            "added_by": "0",
            "carcinogen": null
          },
          {
            "id": "4",
            "name": "(+)-Biotin N-hydroxysuccinimide ester",
            "class": "",
            "subclass": "",
            "packing_group": "",
            "un": "",
            "cas": "35013-72-0",
            "poisons": "",
            "hazardous": "No",
            "restricted": "No",
            "epa": "",
            "added_by": "0",
            "carcinogen": null
          }
        ]
     $(document).ready(function($) {
            $.ajax({
                url: 'self.json',
                type: 'GET',
                dataType: 'json',
            })
            .done(function(respose) {
                for (var i = 0; i < respose.length; i++) {
                    resText = respose[i].id+' '+respose[i].name+' '+ respose[i].class+' '+respose[i].subclass;
                    console.log(resText);
                };
            })
            .fail(function() {
                console.log("error");
            })
            .always(function() {
                console.log("complete");
            });

        });

输出：

【讨论】：

原来是因为我在做ajax，所以显示的时候变量为null：\
使用 jQuery Ajax
是的，我回答了