javascript字符串到对象/数组

Question

我正在从 java 代码中接收该字符串： {devices:[3962], contacts:[{_userId='1', _contactId='(+1)1111111', _contactName='Name Surname'}]} 我想将其转换为普通对象，但它不是有效的 JSON 格式。你有什么想法吗？

或者也许有人可以帮助我编写 JAVA 代码。 我想所有的魔法都在这里：

    public void onUpgrade(SQLiteDatabase db, int oldVersion, int newVersion) {
    }
    public String extractDbDump() {
        return "{devices:" + getDevices() + ", contacts:" + getContacts() + "}";
    }
    private String getContacts() {
        List<ContactItem> contactList = new ArrayList<>();
        String selectQuery = "SELECT * FROM " + TABLE_CONTACT;
        SQLiteDatabase db = this.getWritableDatabase();
        Cursor cursor = db.rawQuery(selectQuery, null);
        if (cursor.moveToFirst()) {
            do {
                ContactItem contact = new ContactItem();
                contact.userId = cursor.getString(0);
                contact.contactName = cursor.getString(1);
                contact.contactId = cursor.getString(2);
                contactList.add(contact);
            } while (cursor.moveToNext());
        }
        db.close();
        return "" + contactList.toString();
    }
    private String getDevices() {
        List<String> deviceList = new ArrayList<>();
        String selectQuery = "SELECT * FROM " + TABLE_DEVICE;
        SQLiteDatabase db = this.getWritableDatabase();
        Cursor cursor = db.rawQuery(selectQuery, null);
        if (cursor != null && cursor.getCount() > 0) {
            if (cursor.moveToFirst()) {
                do {
                    deviceList.add(cursor.getString(0));
                } while (cursor.moveToNext());
            }
        }
        db.close();
        return deviceList.toString();
    }
    public class ContactItem {
        public String userId;
        public String contactId;
        public String contactName;
        @Override
        public String toString() {
            return "{" +
                    "_userId='" + userId + '\'' +
                    ", _contactId='" + contactId + '\'' +
                    ", _contactName='" + contactName + '\'' +
                    '}';
        }
    }

Answer 1

我更改了您的 java 并改为传递 JSONObject 和 JSONArray。它将为您的 javascript 程序返回一个 json 对象。

public void onUpgrade(SQLiteDatabase db, int oldVersion, int newVersion) {
    
    public JSONObject extractDbDump() {
        JSONObject json = new JSONObject();
        json.put("devices", getDevices());
        json.put("contacts", getContacts());
        return json;
    }

    private JSONArray getContacts() {
        JSONArray array = new JSONArray();
        String selectQuery = "SELECT * FROM " + TABLE_CONTACT;
        SQLiteDatabase db = this.getWritableDatabase();
        Cursor cursor = db.rawQuery(selectQuery, null);
        if (cursor.moveToFirst()) {
            do {
                JSONObject item = new JSONObject();
                item.put("_userId", cursor.getString(0));
                item.put("_contactId", cursor.getString(1));
                item.put("_contactName",cursor.getString(2));
                array.put(item);

            } while (cursor.moveToNext());
        }
        db.close();
        return array;
    }

    private JSONArray getDevices() {
        JSONArray array = new JSONArray();
        String selectQuery = "SELECT * FROM " + TABLE_DEVICE;
        SQLiteDatabase db = this.getWritableDatabase();
        Cursor cursor = db.rawQuery(selectQuery, null);
        if (cursor != null && cursor.getCount() > 0) {
            if (cursor.moveToFirst()) {
                do {
                    array.put(cursor.getString(0));
                } while (cursor.moveToNext());
            }
        }
        db.close();
        return array;
    }
}

Answer 2

处理此案例的最佳方法是要求Java 开发人员更改格式以使用有效的 JSON 代码进行响应。然后你可以打电话给JSON.parse，这样就可以了。

另一种方法是编写某种代码，使其成为有效的 JSON。为此，您需要将每个键包装到 " 引号中，并将 = 替换为 : 符号。

Answer 3

您可以尝试快速修复字符串。像这样的：

var s = `{numbers:[3962], contacts:[{_userId='1', _contactId='(+1)1111111', _contactName='Name Surname'}]}`;

var obj = eval(s.replace(/=/g, ":").replace(/:\[/g, "=[")); // <-- bad practice !!!

console.log(obj);

但这是个坏主意。我发布它只是作为临时紧急情况的最后手段。不要在严肃的代码中使用它。

您需要修复 Java 代码。

Answer 4

您需要更改以下内容。

1. ContactItem.toString

   public String toString() {
       return "{" +
               "\"_userId\":\"" + userId + '"' +
               ", \"_contactId\":\"" + contactId + '"' +
               ", \"_contactName\":\"" + contactName + '"' +
               '}';
   }
   

2. extractDbDump

public String extractDbDump() {
    return "{\"devices\":" + getDevices() + ", \"contacts\":" + getContacts() + "}";
}

现在结果应该是{"devices":[3962], "contacts":[{"_userId":"1", "_contactId":"(+1)1111111", "_contactName": "Name Surname"}]}

现在你可以运行JSON.parse。

const response = `{"devices":[3962], "contacts":[{"_userId":"1", "_contactId":"(+1)1111111", "_contactName": "Name Surname"}]}`;

console.log(JSON.parse(response));