mgo regular expression doesn't work

Question

Now I have some documents, each of which has a key path and value like \A\, \B\, \A\C\, \A\C\D\, \A\E\, \A\E\F\.

I want to find the ones which have only 1 segment. It means the result should be \A\ and \B\. I use Regular Expression /^\\[^\\]*\\$/, which works fine in MongoDB terminal. But when I tried to apply it to Go programs it doesn't work.

Go codes:

var nodeList []NodeEntry // NodeEntry would match every field of one document
err = c.Find(bson.M{"path": bson.M{"$regex": bson.RegEx{"^\\[^\\]*\\$", ""}}}).All(&nodeList)
fmt.Println(nodeList)

Output:

[]

It's so strange, and then I found out that any Regex with \\ would produce an empty result.

So is it a bug of mgo?

(I don't know if it's inappropriate, but I've also posted this question on the mgo.users mailing list.)

Try with `\\\` or `\\\\` as some other languages support in this way when they are placed inside string `""` — Sabuj Hassan, Apr 10 '14 at 17:29
@SabujHassan I'm sorry, is it the editor problem? I don't really know what you're trying to say. — Melkor, Apr 11 '14 at 02:55

score 6 · Accepted Answer · answered Apr 10 '14 at 22:44

6

In Go, the backslash (\) is the escape character of an interpreted string literal (using "..." as enclosures). In your case, you´d rather want to use a raw string literal (using `...` as enclosures).

Let's look at this piece of code:

package main

import "fmt"

func main() {
    fmt.Println("^\\[^\\]*\\$")
    fmt.Println(`^\\[^\\]*\\$`)
}

Result:

^\[^\]*\$
^\\[^\\]*\\$

You can see that it is the second option that is the regex string you desire. So, to solve your problem, just enclose your regex string in backticks instead of quotes:

err = c.Find(bson.M{"path": bson.M{"$regex": bson.RegEx{`^\\[^\\]*\\$`, ""}}}).All(&nodeList)

Go spec reference: http://golang.org/ref/spec#String_literals

answered Apr 10 '14 at 22:44

ANisus

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Yes it works, but what if I want my regex to be some strings concatenating with string variables? For example, I already have a string variable whose value is `\A\ `, and I want to find `path` with 2 segments, ie `\A\C\ ` and `\A\E\ `, so the regex should be `\`^\`+variable+\`[^\\]*\\$\``, but actually inside `variable`, it's `\A\ `, not `\\A\\ `, and manually add a `\ ` in the front and end doesn't work. I tried to replace `\ ` with `\\ `, but it still doesn't work. – Melkor Apr 11 '14 at 02:05
Oh sorry, I made a mistake. It works well if I replace `\ ` with `\\ `. But still it's a little bit strange, because what I've stored in mongodb is not `\\ ` but `\ `, so why do I need to use `\\ ` to find it? – Melkor Apr 11 '14 at 02:18
@Melkor: You need to use `\\ ` because perl style regex also uses backslash as an escape character. So, in regex `\\ ` means `the character \ ` in the same way that `\.` means `the character .` unlike `.` which means `any character`. – ANisus Apr 11 '14 at 03:51
yes I understand it, I later found out that in order to match `\ `, I need to pass `\\\\ `. `\\\\ ` will be translated to `\\ ` in golang, and `\\ ` will be translated to `\ ` in regex searching query. – Melkor Apr 11 '14 at 04:41

score 1 · Answer 2 · answered Jun 22 '18 at 16:21

1

To add to @sandun-priyanka solution, if you want to make it case insensitive:

selector:= bson.M{"title": bson.M{"$regex": `(?i)`+wordOffset}}

answered Jun 22 '18 at 16:21

Oras

1,036
1
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score 0 · Answer 3 · answered Mar 12 '18 at 19:12

0

Simply use this

   wordOffset := anyregular expression OR text to filter

   selector:= bson.M{"title": bson.M{"$regex": wordOffset}}

answered Mar 12 '18 at 19:12

Sandun Priyanka

591
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mgo regular expression doesn't work

3 Answers3