我如何在Mongo Shell中将其作为一个命令运行

Question

db.people.update(
    { "age": "Thirty Two" }, { age: 32 }, { upsert: false }
)

db.people.update(
    { "age": "Fifty-Five" }, { age: 55 }, { upsert: false }
)

db.people.update(
    { "age": "Twenty" }, { age: 20 }, { upsert: false }
)

我刚在想：

db.people.update(
    [{ "age": "Thirty Two" }, { age: 32 }, { upsert: false }],
    [{ "age": "Fifty-Five" }, { age: 55 }, { upsert: false }],
    [{ "age": "Twenty" }, { age: 20 }, { upsert: false }]
)

但这没用..我知道我真的很累，无法想到...

Answer 1

也许这就是您想要的。

db.eval(function(name, options) {
    var coll = db.getCollection(name);
    for (var x in options) {
        var option = options[x];
        coll.update(option[0], option[1], option[2]);
    }
},
"people",
[
    [{ "age": "Thirty Two" }, { age: 32 }, { upsert: false }],
    [{ "age": "Fifty-Five" }, { age: 55 }, { upsert: false }],
    [{ "age": "Twenty" }, { age: 20 }, { upsert: false }]
]);

Answer 2

为了详细说明我的意见并提出db.eval()的替代方案，“无操作员”更新将整个匹配的文档替换为更新文档（保留_id除外）：

> db.people.find()
{ "_id" : 0, "name" : "Barney Big", "age" : 32 }
{ "_id" : 1, "name" : "Sam Small", "age" : 26 }
{ "_id" : 2, "name" : "Mindy Medium", "age" : 26 }
> db.people.update({ "age" : 32 }, { "age" : 45 })
> db.people.find()
{ "_id" : 0, "age" : 45 }
{ "_id" : 1, "name" : "Sam Small", "age" : 26 }
{ "_id" : 2, "name" : "Mindy Medium", "age" : 26 }

仅更改一个字段意味着使用$set这样的更新运算符 ：

> db.people.find()
{ "_id" : 0, "name" : "Barney Big", "age" : 32 }
{ "_id" : 1, "name" : "Sam Small", "age" : 26 }
{ "_id" : 2, "name" : "Mindy Medium", "age" : 26 }
> db.people.update({ "age" : 32 }, { "$set" : { "age" : 45 } })
> db.people.find()
{ "_id" : 0, "name" : "Barney Big", "age" : 45 }
{ "_id" : 1, "name" : "Sam Small", "age" : 26 }
{ "_id" : 2, "name" : "Mindy Medium", "age" : 26 }

默认情况下，更新仅影响一个匹配的文档。 选项multi=true将导致更新影响所有匹配的文档：

> db.people.find()
{ "_id" : 0, "name" : "Barney Big", "age" : 32 }
{ "_id" : 1, "name" : "Larry Large", "age" : 32 }
{ "_id" : 2, "name" : "Sam Small", "age" : 26 }
{ "_id" : 3, "name" : "Mindy Medium", "age" : 26 }
> db.people.update({ "age" : 32 }, { "$set" : { "age" : 45 } })
> db.people.find()
{ "_id" : 0, "name" : "Barney Big", "age" : 45 }
{ "_id" : 1, "name" : "Larry Large", "age" : 32 }
{ "_id" : 2, "name" : "Sam Small", "age" : 26 }
{ "_id" : 3, "name" : "Mindy Medium", "age" : 26 }
> db.people.update({ "age" : 26 }, { "$set" : { "age" : 23 } }, { "multi" : true })
> db.people.find()
{ "_id" : 0, "name" : "Barney Big", "age" : 45 }
{ "_id" : 1, "name" : "Larry Large", "age" : 32 }
{ "_id" : 2, "name" : "Sam Small", "age" : 23 }
{ "_id" : 3, "name" : "Mindy Medium", "age" : 23 }

因此，要在外壳程序中执行您在问题中询问的更新，我将使用for循环，但省却了db.eval ，而多次调用db.people.update()

> updates = [
    [{ "age": "Thirty Two" }, { "$set" : { "age" : 32 } }],
    [{ "age": "Fifty-Five" }, { "$set" : { "age" : 55 } }],
    [{ "age": "Twenty" }, { "$set" : { "age" : 20 } }]
]
> updates.forEach(function(pair) {
    db.people.update(pair[0], pair[1])
})

如果您有数百个更新，并且正在使用MongoDB 2.6或更高版本，请查看批量操作 。 db.collection.update()文档中有更多信息。