Question

我正在尝试获取ResultSet from the executeSql库的react-native-sqlite-storage 函数，但得到TypeError: Cannot read property '0' of undefined。

executeSql函数应返回一个Promise<[Transaction, ResultSet]>，但我似乎无法正确地实现销毁。

我的代码：database.ts：

let conn = SQLite.openDatabase({name: DATABASE_NAME, location: 'default'})

export default class Db  {
  static async getConn() {
    return await conn
  }

  static async getTransaction() {
    return new Promise<Transaction>(async (resolve,reject) => {
      const conn = await this.getConn()
      conn.transaction((transaction: Transaction) => {
        resolve(transaction)
      })
    })
  }

  static async executeSql(sqlStatement: string, args?: any[]) {
    return new Promise<ResultSet>(async (resolve,reject) => {
      const transaction = await this.getTransaction()
      const [tr, results] = await transaction.executeSql(sqlStatement, args)  // ??
      resolve(results)
    })
  }
}

已编译：

Db.executeSql = function (sqlStatement, args) {
    return __awaiter(this, void 0, void 0, function () {
        var _this = this;
        return __generator(this, function (_a) {
            return [2 /*return*/, new Promise(function (resolve, reject) { return __awaiter(_this, void 0, void 0, function () {
                    var transaction, _a, tr, results;
                    return __generator(this, function (_b) {
                        switch (_b.label) {
                            case 0: return [4 /*yield*/, this.getTransaction()];
                            case 1:
                                transaction = _b.sent();
                                return [4 /*yield*/, transaction.executeSql(sqlStatement, args)];
                            case 2:
                                _a = _b.sent(), tr = _a[0], results = _a[1];  // <= ERROR HERE: TypeError: Cannot read property '0' of undefined
                                resolve(results);
                                return [2 /*return*/];
                        }
                    });
                }); })];
        });
    });
};

Answer 1

查看代码（https://github.com/andpor/react-native-sqlite-storage/blob/master/lib/sqlite.core.js#L449），transaction.executeSql不会返回Promise。它实际上什么也不会返回。

似乎该方法也接受success和error回调，因此您的代码应类似于：

static async executeSql(sqlStatement: string, args?: any[]) {
    const transaction = await this.getTransaction();
    const [tr, results] = await new Promise<ResultSet>(
      (resolve,reject) =>           
        transaction.executeSql(sqlStatement, args,
          (t, r) => resolve([t, r]), 
          (err) => reject(err)
        )
    );
    return results;
}

TypeScript：从等待的Promise进行数组解构编译并在运行时崩溃

1 个答案: