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android 用gson 解析数据出错！

le__citron 发布于 2022-10-29 17:15

我用如下方法解析一个json字符串，但是就是运行都这就闪退了，不知道什么原因。

    protected void parseData(String result) {
        // TODO Auto-generated method stub
        Gson gson=new Gson();
        NewData data=gson.fromJson(result,NewData.class);
        System.out.println(data.toString());
         
    }

这是newData类

public class NewData{
    
      public ArrayList  food;
      
       public class kindFood{
            public String Description;
            public String PicUrl;
            public String Title;
            public String Url;
            
            @Override
            public String toString() {
                return "kindFood [Description=" + Description + ", PicUrl=" + PicUrl + ", Title=" + Title + ", Url="
                        + Url + "]";
            }

        }

}

这是json数据

[
    {
        "Description": "",
        "PicUrl": "",
        "Title": "鸡蛋热量(100克)",
        "Url": ""
    },
    {
        "Description": "",
        "PicUrl": "",
        "Title": "鸡蛋    144 大卡",
        "Url": "http://i.boohee.com/mfood/view/910"
    },
    {
        "Description": "",
        "PicUrl": "",
        "Title": "鸡蛋(土鸡)    138 大卡",
        "Url": "http://i.boohee.com/mfood/view/913"
    },
    {
        "Description": "",
        "PicUrl": "",
        "Title": "鸡蛋（煮）    151 大卡",
        "Url": "http://i.boohee.com/mfood/view/1757"
    },
    {
        "Description": "",
        "PicUrl": "",
        "Title": "鸡蛋白(鸡蛋清)    60 大卡",
        "Url": "http://i.boohee.com/mfood/view/914"
    },
    {
        "Description": "",
        "PicUrl": "",
        "Title": "鸡蛋(白皮)    138 大卡",
        "Url": "http://i.boohee.com/mfood/view/911"
    },
    {
        "Description": "",
        "PicUrl": "",
        "Title": "鸡蛋(红皮)    156 大卡",
        "Url": "http://i.boohee.com/mfood/view/912"
    }
]

6 个回答

需要使用到typetoken解析即可

2022-10-31 01:14 回答

dazhi20
你这个json数组的key是food么？而且最好把get set 方法设置了

2022-10-31 01:15 回答

手机用户2502936713

使用gsonformat插件可以快速解决类似问题

public class food{
    /**
     * Description :
     * PicUrl :
     * Title : 鸡蛋热量(100克)
     * Url :
     */
    private String Description;
    private String PicUrl;
    private String Title;
    private String Url;

    public String getDescription() {
        return Description;
    }

    public void setDescription(String Description) {
        this.Description = Description;
    }

    public String getPicUrl() {
        return PicUrl;
    }

    public void setPicUrl(String PicUrl) {
        this.PicUrl = PicUrl;
    }

    public String getTitle() {
        return Title;
    }

    public void setTitle(String Title) {
        this.Title = Title;
    }

    public String getUrl() {
        return Url;
    }

    public void setUrl(String Url) {
        this.Url = Url;
    }
}

然后在添加到arraylist中

2022-10-31 01:16 回答

POPO炮炮_797

首先，将kindFood这个类提出来。
其次，将NewData data=gson.fromJson(result,NewData.class);改成KindFood data=gson.fromJson(result,kindFood[].class);

2022-10-31 17:39 回答

yanghuimin
试一下不要把kindFood作为内部类

2022-10-31 17:39 回答

优雅的鼻涕泡
你可以只保留KindFood类里面注意有getter setter 方法然后
```
Gson gson = new Gson();
List<KindFood> listFood = gson.fromJson(result,new TypeToken<List<KindFood>>(){}.getType());
```
2022-10-31 17:40 回答

手机用户2602925827

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