MySQL获取结果"/>
从MySQL获取结果
我的后端由Api和DB组成。当我想从数据库获得响应时,我已将输出延迟了1个查询。API (我认为api可以。首先开始读取数据库)
app.post('/api/query', (req, res) => {
console.log(`\n Query input : ${JSON.stringify(req.body)}`);
let queryInput = (Object.values(req.body).join(' '));
if(!dbApi.checkArray(queryInput)){ //If array is not made from clear strings
res.json(dbApi.queryFromUser(queryInput));
}
else{
res.json(dbApi.queryOutput);
}
});
app.listen(dbConfig.server.port, () =>
console.log(`Server running on port ${dbConfig.server.port}`));
DB
queryOutput = [];
const receivingQuery =(queryInput) => {
db.query(queryInput, (err, result) =>{
if(err) throw err+' : '+queryInput;
queryOutput = result;
console.log("\nQuery output "+ JSON.stringify(queryOutput)); //Output (result) is ok
});
return queryOutput //Here is Output from previous query (sends to API)
}
module.exports = {
queryOutput: queryOutput,
queryFromUser: receivingQuery,
}
我尝试了回调方法,并重写了两次。但是我没有足够的技能来解决它。
回答如下:如果要返回查询结果,只需执行以下操作:
1)将query
方法添加到数据库模块:
function query(sql, args = []) {
return new Promise(function(resolve, reject) {
db.query(sql, args, (err, result) => {
if (err) return reject(err);
resolve(result);
});
});
}
// extra feature, getting user by id
async function getUserById(id) {
const result = await query('SELECT * FROM users WHER id = ? LIMIT 1', [id]);
if (Array.isArray(result) && result[0]) return result[0];
return null;
}
module.exports = {
query,
getUserById, // export user by id
queryOutput: queryOutput,
queryFromUser: receivingQuery,
}
2)使用它(通过异步和等待):
app.post('/api/query', async (req, res) => {
try {
console.log('Query input:', req.body);
let queryInput = Object.values(req.body).join(' ');
const result = await dbApi.query(queryInput);
res.json(result);
}
catch (error) {
console.error(error);
res.status(500).json({message: 'Please try again soon'});
}
});
app.get('/api/users/:id', async (req, res) => {
try {
const user = await dbApi.getUserById(req.params.id);
if (!user) return res.status(404).json({message: 'User not found'});
res.status(200).json(user);
}
catch (error) {
console.error(error);
res.status(500).json({message: 'Please try again soon'});
}
});
app.listen(dbConfig.server.port, () =>
console.log('Server running on port', dbConfig.server.port));
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从MySQL获取结果
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