Question

如何使用google和我的json的gson将复杂的json转换为java对象是这样的：

{
    "error": "200",
    "status": "OK",
    "BarList": {
        "Bar1": {
            "Name": "yash",
            "sex": "male",
            "Type": "barowner",
            "userId": "x25df",
            "ContactNo": "1234567890",
            "zipCode": "110055",
            "Address": "Ghumtarastachaltigali",
            "Email": "nahihairee@gmail.com"
        },
        "Bar2": {
            "Name": "yash",
            "sex": "male",
            "Type": "barowner",
            "userId": "x25df",
            "ContactNo": "1234567890",
            "zipCode": "110055",
            "Address": "Ghumtarastachaltigali",
            "Email": "nahihairee@gmail.com"
        }
    }
}

为了将这个json映射到我的java对象，我已经制作了3个类第一个：BarListResponse - 我这样做了： -

public class BarListResponse {

    @SerializedName("error")
    @Expose(serialize = false)
    String errrocode;

    @Expose(serialize = false)
    String status;

    @SerializedName("data")
    Bar bar_list[];

    public String getErrrocode() {
        return errrocode;
    }

    public void setErrrocode(String errrocode) {
        this.errrocode = errrocode;
    }

    public String getStatus() {
        return status;
    }

    public void setStatus(String status) {
        this.status = status;
    }

    public Bar[] getLst() {
        return bar_list;
    }

    public void setLst(Bar lst[]) {
        this.bar_list = lst;
    }
}

第二个栏目列表：

public class BarList {

    @SerializedName("Bar")
    Bar bar[];

    public Bar[] getBar() {
        return bar;
    }

    public void setBar1(Bar bar[]) {
        this.bar = bar;
    }
}

第三是

public class Bar {

    String Name;
    String sex;
    String type;
    String userId;
    double ContactNo;
    double zipCode;
    String Address;
    String Email;

    public String getName() {
        return Name;
    }
    public void setName(String name) {
        Name = name;
    }
    public String getSex() {
        return sex;
    }
    public void setSex(String sex) {
        this.sex = sex;
    }
    public String getType() {
        return type;
    }
    public void setType(String type) {
        this.type = type;
    }
    public String getUserId() {
        return userId;
    }
    public void setUserId(String userId) {
        this.userId = userId;
    }
    public double getContactNo() {
        return ContactNo;
    }
    public void setContactNo(double contactNo) {
        ContactNo = contactNo;
    }
    public double getZipCode() {
        return zipCode;
    }
    public void setZipCode(double zipCode) {
        this.zipCode = zipCode;
    }
    public String getAddress() {
        return Address;
    }
    public void setAddress(String address) {
        Address = address;
    }
    public String getEmail() {
        return Email;
    }
    public void setEmail(String email) {
        Email = email;
    }
}

从此我想逐一获取每个栏的细节。

请帮助解决此问题。提前谢谢。

Answer 1

您无法将json对象（{...}）映射到java列表，因此您必须为＆＃34; list＆＃34;创建。对象一个自己的类来映射对象。这是非常不理想的，因为你必须为列表中的每个项目创建一个自己的属性，但如果你无法控制json，你必须处理它。

这些类应如下所示 BarListResponse.java ：

public class BarListResponse {

    @SerializedName("error")
    @Expose(serialize = false)
    String errrocode;

    @Expose(serialize = false)
    String status;

    @SerializedName("BarList")
    BarList bar_list;

    public String getErrrocode() {
        return errrocode;
    }

    public void setErrrocode(String errrocode) {
        this.errrocode = errrocode;
    }

    public String getStatus() {
        return status;
    }

    public void setStatus(String status) {
        this.status = status;
    }

    public BarList getLst() {
        return bar_list;
    }

    public void setLst(BarList lst) {
        this.bar_list = lst;
    }
}

<强> BarList.java ：

public class BarList {
    Bar Bar1;
    Bar Bar2;
    ...
}

如果您以某种方式控制json输出，则可以更改＆＃34; BarList＆＃34;到一个json数组：

{
    "error": "200",
    "status": "OK",
    "BarList": [
        {
            "Name": "yash",
            "sex": "male",
            "Type": "barowner",
            "userId": "x25df",
            "ContactNo": "1234567890",
            "zipCode": "110055",
            "Address": "Ghumtarastachaltigali",
            "Email": "nahihairee@gmail.com"
        },
        {
            "Name": "yash",
            "sex": "male",
            "Type": "barowner",
            "userId": "x25df",
            "ContactNo": "1234567890",
            "zipCode": "110055",
            "Address": "Ghumtarastachaltigali",
            "Email": "nahihairee@gmail.com"
        }
    ]
}

并将响应类更改为以下内容：

public class BarListResponse {

    @SerializedName("error")
    @Expose(serialize = false)
    String errrocode;

    @Expose(serialize = false)
    String status;

    @SerializedName("BarList")
    List<Bar> bar_list;

    public String getErrrocode() {
        return errrocode;
    }

    public void setErrrocode(String errrocode) {
        this.errrocode = errrocode;
    }

    public String getStatus() {
        return status;
    }

    public void setStatus(String status) {
        this.status = status;
    }

    public List<Bar> getLst() {
        return bar_list;
    }

    public void setLst(List<Bar> lst) {
        this.bar_list = lst;
    }
}

通过这种方法，您不需要额外的＆＃34; BarList＆＃34;类和列表可以包含＆＃34;无限＆＃34;条目没有改变你的内部类结构。

如何使用gson将复杂的json转换为java对象

1 个答案: