英文:
MongoDB , $project a field depending on the existance of another field
问题
如标题所示，我想要仅在“x”字段不存在时投影“y”字段。
英文:
As the title says, i want to project for example the "y" field only if "x" field does not exsist.
[
{
"_id": "1",
"y": "project"
},
{
"_id": "2",
"x": "has x",
"y": "don't project"
}
]
Thanks in advance.
答案1
得分: 0
如果我理解正确，您可以在$project阶段中使用$cond来决定是否投影字段：
在这里，如果x存在，那么将移除（不投影）字段y，否则将显示y的值。
db.collection.aggregate([
{
"$project": {
"_id": 1,
"y": {
"$cond": {
"if": {"$gt": ["$x", null]},
"then": "$$REMOVE",
"else": "$y"
}
}
}
}
])
示例在这里
注意，您也可以投影字段x。查看此示例。
英文:
If I've understood correctly you can use $cond into a $project stage to decide if project or not a field:
Here if x exists then remove (does not project) the field y, otherwise the y value is shown.
db.collection.aggregate([
{
"$project": {
"_id": 1,
"y": {
"$cond": {
"if": {"$gt": ["$x",null]},
"then": "$$REMOVE",
"else": "$y"
}
}
}
}
])
Example here
Note that also you can project the field x. Check this example
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