Question

我想用脚本初始化一个mongoDB因为我有一些我需要在开头有值的集合所以我按照说明操作：http://michelebusta.com/the-little-things-5-initialize-a-local-mongo-db/

我创建了一个（.js）文件，我加载它并且它可以工作，

class TeamSerializer(serializers.ModelSerializer):
    class Meta:
        model = Team
        fields = ('id', 'name', 'members','services_used')

问题是我有这样的ID：＆＃34; _id＆＃34; ：ObjectId（＆＃34; 572b8601e6352994d76bb13e＆＃34;），我希望有一个自动增量ID（_id：1），请你帮帮我。

db.serviceSpeedList.insert({
_id: 0,
name:"DHL", service:"DHL Ground"});

db.serviceSpeedList.insert({
_id: 1,
name:"DHL", service:"DHL Second Day Service"});

db.serviceSpeedList.insert({
_id: 2,
name:"DHL", service:"DHL Express"});

db.serviceSpeedList.insert({
_id: 3,
name:"DHL", service:"DHL International Express"});

我用过

{
"_id" : ObjectId("572b8601e6352994d76bb139"),
"name" : "DHL",
"service" : "DHL Ground"
}

{
"_id" : ObjectId("572b8601e6352994d76bb13a"),
"name" : "DHL",
"service" : "DHL Second Day Service"
}

当我在shell中单独使用它时，它可以工作，但当我把它放在一个脚本中并使用load（＆＃34; D：\ initMongo.js＆＃34;）;它不起作用:(请帮帮我

Answer 1

您可以将_id设置为任何其他属性，如果您有js文件，请考虑使用var i = 0;并在每个插入语句中增加该值。

db.serviceSpeedList.insert({ "_id" : i++, name:"DHL", service:"DHL Ground"});
db.serviceSpeedList.insert({"_id" : i++, name:"DHL", service:"DHL Second Day Service"});

按照流程

> var i = 0;
> db.serviceSpeedList.insert({ "_id" : i++, name:"DHL", service:"DHL Ground"});
WriteResult({ "nInserted" : 1 })
> db.serviceSpeedList.findOne()
{ "_id" : 0, "name" : "DHL", "service" : "DHL Ground" }
> db.serviceSpeedList.insert({"_id" : i++, name:"DHL", service:"DHL Second Day S
ervice"});
WriteResult({ "nInserted" : 1 })
> db.serviceSpeedList.find()
{ "_id" : 0, "name" : "DHL", "service" : "DHL Ground" }
{ "_id" : 1, "name" : "DHL", "service" : "DHL Second Day Service" }
>

Answer 2

您必须在mongodb数据库中执行自动增量序列才能执行此操作。可以找到完整的例子：https://docs.mongodb.org/manual/tutorial/create-an-auto-incrementing-field/

使用自动增量ID初始化MongoDB的脚本

2 个答案: