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如何在POJO中使用动态键设置JSON,并在Android JAVA中从中获取数据?

json android java

-1

Aniket · 技术社区 · 7 年前

我正在开发一个电子商务应用程序,我正在从API获取这个JSON数据。

{“status”:0,“data”:{“stores”:{“test-5”:{“name”:“test-5”,“locality”:{“name”:“Some place B”,“id”:2},“cover”:“IMAGE-URL-HERE”},“test-2”:{“name”:“test 2”,“locality”:{“name”:“Some place A”,“id”:2},“cover”:“IMAGE-URL-HERE”},“action”:[“data”,“stores”]}}

我也为这些数据创建了一些POJO

public class PartnerStoreMainPOJO {
    @SerializedName("partnerstore")
    @Expose
    private PartnerStoresPOJO partnerstore;
    /**
     *
     * @return
     * The data
     */
    public PartnerStoresPOJO getPartnerStore() {
        return partnerstore;
    }
    /**
     *
     * @param partnerstore
     * The data
     */
    public void setPartnerStore(PartnerStoresPOJO partnerstore) {
        this.partnerstore = partnerstore;
    }
}

//------------- 公共类PartnerStoresPOJO{

@SerializedName("partnerstoredetail")
@Expose
private Map<String, PartnerStoreDetailPOJO> partnerstoredetail;

/**
 *
 * @return
 * The feeds
 */
public Map<String, PartnerStoreDetailPOJO>  getpartnerstoredetail() {
    return partnerstoredetail;
}
/**
 *
 * @param partnerstoredetail
 * The feeds
 */
public void setpartnerstoredetail(Map<String, PartnerStoreDetailPOJO> partnerstoredetail) {
    this.partnerstoredetail = partnerstoredetail;
}

}

公共类PartnerStoreDetailPOJO{

@SerializedName("partnerstorelocality")
@Expose
private Map<String, PartnerStoreLocalityPOJO> partnerstorelocality;


@SerializedName("cover")
@Expose
private String cover;
@SerializedName("name")
@Expose
private String name;

/**
 * @return The name
 */
public String getName() {
    return name;
}

/**
 * @param name The name
 */
public void setName(String name) {
    this.name = name;
}

/**
 * @return The cover
 */
public String getCover() {
    return cover;
}

/**
 * @param cover The address
 */
public void setCover(String cover) {
    this.cover = cover;
}

public Map<String, PartnerStoreLocalityPOJO> getpartnerstorelocality() {
    return partnerstorelocality;
}

public void setpartnerstorelocality(Map<String, PartnerStoreLocalityPOJO> partnerstorelocality) {
    this.partnerstorelocality = partnerstorelocality;
}

}

public class PartnerStoreLocalityPOJO {
    @SerializedName("name")
    @Expose
    private String name;
    @SerializedName("id")
    @Expose
    private String id;
    /**
     *
     * @return
     * The name
     */
    public String getName() {
        return name;
    }
    /**
     *
     * @param name
     * The name
     */
    public void setName(String name) {
        this.name = name;
    }
    /**
     *
     * @return
     * The id
     */
    public String getId() {
        return id;
    }
    /**
     *
     * @param id
     * The id
     */
    public void setId(String id) {
        this.id = id;
    }
}

//---------------

我正在使用截击库。这是我的java代码-

public void onResultReceived(String response, String tag_json_obj) {
        if (tag_json_obj.equals(LOCALITY_SET)){

            try {
                JSONObject jsonObject=new JSONObject(response);
                String data=jsonObject.getString("data");

            } catch (JSONException e) {
                Log.d("EXCEPTN",e.toString());
                e.printStackTrace();
            }

        }
    }

我正在使用那个数据字符串。

1 回复 | 直到 7 年前

Stuti Kasliwal 7 年前

尝试一次:做相应的更改,它可以为您的查询提供方向。

public void convertJSON(JSONObject jsonObject) {
        try {
            JSONObject object = jsonObject.getJSONObject("data");

            Iterator<String> iter = object.keys();
            while (iter.hasNext()) {
                String key = iter.next();
                Object value = object.get(key);

                JSONObject obj2 = object.getJSONObject(key);
                //set key to POJO

                Iterator<String> iter2 = obj2.keys();
                while (iter2.hasNext()) {
                    String key2 = iter2.next();
                    //....so on
                }
            }
        } catch (JSONException e) {
            e.printStackTrace();
        }
    }