【发布时间】:2015-10-20 00:53:44
【问题描述】:
我正面临这样一种情况,即我是 Moongose 和 MongoDB 的新手,我需要在 MySQL 中编写一些看起来像这样的东西:
SELECT * FROM table_name WHERE case_id = $case_id AND calculation_id = $calculation_id;
这就是我现在在Node.js 中的声明:
app.get('/positions/:case_id/:calculation_id', function (req, res) {
Positions.find()
.where('case_id')
.equals(req.params.case_id)
.exec(function (err, records) {
res.send(records);
});
})
添加一个新的 .where 真的没有任何交易,那我该怎么办?
示例行:
{"_id":"55e993ad29149e61931b826c","calculation_id":22,"case_id":"2","position_id":"33660","blockline":"B510L1 ","repairmethod":"20 ","guidenumber":"N/A ","amount":"6","hours":"0","starmutation":"N","text":"BLATNÍK P L V-Z","originalpartnumber":"","originalpartprice":"0","manufacturercode":"","quality":"","suppliercode":"","id":4},
所以在这一行中,case_id = 2 并且计算 ID 为 22。
如果我打开网址positions/2/22,
app.get('/positions/:case_id/:calculation_id', function (req, res) {
Positions.find()
.where('case_id').equals(req.params.case_id)
.where('calculation_id').equals(req.params.calculation_id)
.exec(function (err, records) {
res.send(records);
});
})
我没有得到任何结果:(
【问题讨论】:
-
我读到这个:mongoosejs.com/docs/queries.html 但还是没明白
标签: mongodb mongoose mongodb-query