如果需要从变量中创建regexp，那么这是一种更好的方法:https://stackoverflow.com/a/10728069/309514

然后你可以这样做:

var string = "SomeStringToFind";
var regex = new RegExp(["^", string, "$"].join(""), "i");
// Creates a regex of: /^SomeStringToFind$/i
db.stuff.find( { foo: regex } );

这样做的好处是更加程序化，或者如果您经常重用它，则可以通过提前编译它来提高性能。

正如你在mongo docs中看到的那样——自3.2版以来$text索引默认情况下是不区分大小写的:https://docs.mongodb.com/manual/core/index-text/#text-index-case-insensitivity

创建一个文本索引并在查询中使用$text操作符。

使用Mongoose对我来说很管用:

var find = function(username, next){
    User.find({'username': {$regex: new RegExp('^' + username, 'i')}}, function(err, res){
        if(err) throw err;
        next(null, res);
    });
}

这些已经用于字符串搜索进行了测试

{'_id': /.*CM.*/}               ||find _id where _id contains   ->CM
{'_id': /^CM/}                  ||find _id where _id starts     ->CM
{'_id': /CM$/}                  ||find _id where _id ends       ->CM

{'_id': /.*UcM075237.*/i}       ||find _id where _id contains   ->UcM075237, ignore upper/lower case
{'_id': /^UcM075237/i}          ||find _id where _id starts     ->UcM075237, ignore upper/lower case
{'_id': /UcM075237$/i}          ||find _id where _id ends       ->UcM075237, ignore upper/lower case

博士TL;

正确的方法做到这一点在mongo

不使用RegExp

使用mongodb的内置索引，搜索

第一步:

db.articles.insert(
   [
     { _id: 1, subject: "coffee", author: "xyz", views: 50 },
     { _id: 2, subject: "Coffee Shopping", author: "efg", views: 5 },
     { _id: 3, subject: "Baking a cake", author: "abc", views: 90  },
     { _id: 4, subject: "baking", author: "xyz", views: 100 },
     { _id: 5, subject: "Café Con Leche", author: "abc", views: 200 },
     { _id: 6, subject: "Сырники", author: "jkl", views: 80 },
     { _id: 7, subject: "coffee and cream", author: "efg", views: 10 },
     { _id: 8, subject: "Cafe con Leche", author: "xyz", views: 10 }
   ]
)
 

第二步:

需要在你想要搜索的任何TEXT字段上创建索引，没有索引查询将会非常慢

db.articles.createIndex( { subject: "text" } )

第三步:

db.articles.find( { $text: { $search: "coffee",$caseSensitive :true } } )  //FOR SENSITIVITY
db.articles.find( { $text: { $search: "coffee",$caseSensitive :false } } ) //FOR INSENSITIVITY


 

我也遇到过类似的问题，这对我很有帮助:

  const flavorExists = await Flavors.findOne({
    'flavor.name': { $regex: flavorName, $options: 'i' },
  });