Question

如果有人可以帮助解决以下打字稿错误，我将不胜感激。

感谢。

我正在使用以下code，其中包含以下内容。它需要能够链接Promise。

line 26:    dbPromise = _db.execute(sql)

当我尝试构建它时，我得到以下内容：

ERROR in ./app/pages/chats/SqlDatabase.ts
(26,9): error TS2322: Type 'Promise<SqlResultSet>' is not assignable to type 'Promise<SqlDatabase>'.
  Type 'SqlResultSet' is not assignable to type 'SqlDatabase'.
    Property '_db' is missing in type 'SqlResultSet'.

我正在使用Typescript version 2.0.3.

代码：

import { isBrowser } from './platform';
import { SqlResultSet } from './SqlResultSet';

export class SqlDatabase {

  constructor(private _db: any) { }

  static open(name: string, initStatements: string[] = []): Promise<SqlDatabase> {
    let dbPromise = isBrowser()
      .then(browser => {
        const openDatabase = browser ? openBrowserDatabase : openCordovaDatabase;
        return openDatabase(name);
      });
    if (initStatements.length === 0) {
      return dbPromise;
    }
    let _db: SqlDatabase;
    // execute the first statement and capture the _db
    dbPromise.then(db => {
      _db = db;
      return db.execute(initStatements.shift());
    });
    // execute all the other statements (if any) sequentially
    for (let sql of initStatements) {
      dbPromise.then(() => {
        dbPromise = _db.execute(sql)
      });
    }
    // resolve the _db only after all statements have completed
    return new Promise((resolve, reject) => {
      console.log('resolve: ', resolve);
      dbPromise.then(() => resolve(_db)).catch(reject);
    });
  }

  execute(statement: string, params: any[] = []): Promise<SqlResultSet> {
    console.log('execute: ' + statement);
    return new Promise((resolve, reject) => {
      this._db.transaction(tx => tx.executeSql(statement, params, (tx, resultSet) => {
        console.log('execute: resolve: ', resultSet);
        resolve(resultSet);
      }, (tx, error) => {
        reject(error)
      }));
    });
  }
}

declare var sqlitePlugin: any;

function openCordovaDatabase(name: string): Promise<SqlDatabase> {
  return new Promise((resolve, reject) => {
    if (typeof sqlitePlugin === 'undefined') {
      reject(new Error('[ionix-sqlite] sqlitePlugin global object not found; did you install a Cordova SQLite plugin?'));
    }
    const db = sqlitePlugin.openDatabase({
      name: name,
      location: 'default'
    });
    console.info('[ionix-sqlite] using Cordova sqlitePlugin');
    resolve(new SqlDatabase(db));
  });
}

declare function openDatabase(name: string, version: string, desc: string, size: number): any;

function openBrowserDatabase(name: string): Promise<SqlDatabase> {
  return new Promise((resolve, reject) => {
    try {
      const db = openDatabase(name, '1.0', name, -1);
      console.info('[ionix-sqlite] using WebSQL');
      resolve(new SqlDatabase(db));
    } catch (error) {
      reject(error);
    }
  });
}

Answer 1

错误是由于您尝试更改dbPromise的类型首次宣布时：

let dbPromise = isBrowser()
      .then(browser => {
        const openDatabase = browser ? openBrowserDatabase : openCordovaDatabase;
        return openDatabase(name);
      });

编译器将类型引用到Promise<SqlDatabase>（基于错误），但之后您尝试分配其他内容：

dbPromise = _db.execute(sql)

类型Promise<SqlResultSet>。
您可以这样解决：

let dbPromise: Promise<any> = ...

或者你可以有两个不同的承诺变量（即：dbPromise: Promise<SqlDatabase>和resultPromise: Promise<SqlResultSet>），这对我来说听起来更好。

打字稿链接承诺

1 个答案: