React native 表的id被外键的id替换
我在react原生项目中使用Expo的SQLite模块。我有两个表、消息和用户:React native 表的id被外键的id替换,react-native,sqlite,expo,React Native,Sqlite,Expo,我在react原生项目中使用Expo的SQLite模块。我有两个表、消息和用户: CREATE TABLE IF NOT EXISTS messages ( id text not null, sender_id integer, thread_id integer, body text, foreign key (sender_id) references users (id), foreign key (thread_id) reference
CREATE TABLE IF NOT EXISTS messages (
id text not null,
sender_id integer,
thread_id integer,
body text,
foreign key (sender_id) references users (id),
foreign key (thread_id) references threads (id)
);
CREATE TABLE IF NOT EXISTS users (
id integer primary key,
first_name text,
last_name text,
email text
);
如果我插入一些信息:
INSERT INTO messages (id, sender_id, thread_id, body) values ('xeieoef-ee, 3, 1, 'test');
INSERT INTO messages (id, sender_id, thread_id, body) values ('ttrefzr-ry, 3, 1, 'payload');
我希望通过比较线程id来获取所有消息,包括其用户的数据。这就是我的疑问:
select * from messages, users where thread_id = 2 AND messages.sender_id = users.id;
但是,这会导致消息id和用户id相同:
[
{
"body": "test",
"email": "userthree@gmail.com",
"first_name": "userThreeF",
"id": 3,
"last_name": "threeUserL",
"sender_id": 3,
"thread_id": 1,
},
{
"body": "payload",
"email": "userthree@gmail.com",
"first_name": "userThreeF",
"id": 3,
"last_name": "threeUserL",
"sender_id": 3,
"thread_id": 1,
},
]
邮件id不是要有自己的id,而是发件人的id。我在这里做错了什么
更新代码
我在应用程序中有一个按钮,onpress会将数据发送到MessageStore的功能
sendText = () => {
const {MessageStore, UserStore, SocketStore} = this.props;
const data = {
id: uuid.v4(),
sender_id: UserStore.userInfo.user.id,
thread_id: 1,
body: this.state.text,
}
MessageStore.addMessageToDB(data);
}
在存储中,addMessageToDB将消息添加到数据库中,它再次调用getMessageFromDatabase来获取所有消息
@action addMessageToDB = (payload) => {
db.transaction(
tx => {
tx.executeSql(
`INSERT INTO messages
(id, sender_id, thread_id, body, status) values (?, ?, ?, ?, ?);`,
[payload.id, payload.sender_id, payload.thread_id, payload.body, "pending"],
(tx, results) => {
console.log('message insertion success')
},
(tx, error) => console.log('message insertion error', error)
);
this.getMessageFromDatabase(payload.thread_id);
}
)
}
@action getMessageFromDatabase = (payload) => {
console.log('>>> getMessageFromDatabase payload', payload);
db.transaction(
tx => {
tx.executeSql(
`select * from messages inner join users on messages.sender_id=users.id where thread_id = ?;`, [payload],
(tx, {rows}) => {
console.log('inner join success', rows._array);
},
(tx, error) => console.log('inner join error', error),
);
}
)
}
使用解决方案编辑:
两个表都有一个“id”列,因此用户id将覆盖消息id
这是有效的:选择messages.*,users.first\u name,users.last\u name,users.email from messages internal join users on messages.sender\u id=users.id其中thread\u id= 很抱歉,但这有相同的结果。我们需要查看一些代码,我不认为你在这里发布的内容有任何问题嗨!我已经用相关代码更新了问题。你能看一下吗?看不出有什么问题。你能举一个简单的例子在snack.expo.io上复制这个问题吗?很有趣。我尝试在snack中复制代码,得到了相同的结果。有趣的是。我尝试在snack中复制代码,得到了相同的结果。这是@Kakar是的!就这样。不知道这里怎么了。