var footballerIdToPassive = "qqqqq";
var footballerlevelToPassive = ["3052002","3052003"];
db.goals.find({ "footballer": footballerIdToPassive, "footballerlevel": { $in: [footballerlevelToPassive]}}))
因为它除了在目标 db shell 中什么也没有带来,当我对此进行计数时,计数为 1
{ "footballer": "qqqqq", "footballerlevel": "3052002" }
也为
var footballerlevelToPassive = "3052002";
这个,有效。但我不能做多个。我该如何完成这项工作?
我需要使用每一个,我需要更新每一个
最佳答案
footballerlevelToPassive
已经是一个数组,查询时不需要再用另一个数组包裹,直接用$in
as
var footballerIdToPassive = "qqqqq";
var footballerlevelToPassive = ["3052002","3052003"];
db.goals.find({
"footballer": footballerIdToPassive,
"footballerlevel": { "$in": footballerlevelToPassive }
})
相当于查询
db.goals.find({
"footballer": footballerIdToPassive,
"$or": [
{ "footballerlevel": "3052002" },
{ "footballerlevel": "3052003" }
]
})
快速演示
关于javascript - $in 用于多个值,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/43225932/