文件和目录正则表达式

Question

I am trying to create a regEx for file and directory path validation. I have implemented this, but its failing 1 of the conditions, that it should not allow ie multiple slashes together. Also, no other special character should not be allowed

var x = /^(\\|\/){1}([a-zA-Z0-9\s\-_\@\-\^!#$%&]*?(\\|\/)?)+(\.[a-z\/\/]+)?$/i

• test 1 -> / (should pass)
• test 2 -> /asdf (should pass)
• test 3 -> /asdf/scd.csv (should pass)
• test 4 -> //asdf (should fail, currently passing)
• test 5 -> /asd/ads/c.csv/ (should pass)
• test 6 -> asd/asfd/a (should fail)

Can suggestion how to solve this?

Answer 1

The path //asdf is valid on LINUX, UNIX, iOS, and Android, so your code already works. However, if it is important for some reason to invalidate that particular set of valid paths, simply substitute a plus sign in place of the an asterisk after the [az...] character group. That will cause invalidation of multiple path separators with no intervening characters.

It is probably useful to comment on larger issues with the regex approach and details.

1) You can use [\\/] instead of (\\|/), however both will allow false positives on every combination of operating system and file system. (Those that require forward slash should exclude backslashes as a separator and vice versa.)

2) The character range [a-zA-Z0-9\\s-_\\@-\\^!#$%&] in the question is not the permissible character range for directory path elements for any known combination of operating system and file system. For instance, a period is valid in directory names for most.

3) Permissible character ranges are not portable. (The most reliable way to test path validation is to touch the file name on the actual file system, meaning actually instantiate an empty file and capture any indications of instantiation failure.)

4) You don't want or need a question mark after your asterisk or after your second (\\|/) group. They don't create a bug, but they waste either compilation or run time, and they obfuscate your regex purpose.

5) You also need to repeat the character range just before the extension or rearrange like the example below.

6) You don't need to add the AZ range to the az range if you use \\i as a flag at the end of the regex.

7) It appears from the list of desired results that relative paths are to be filtered out, but there is no explicit mention of that as a rule for the solution.

With hesitation, this code is provided to demonstrate a few of the above improvements.

 // This code is not production worthy // for reasons (1) through (3) given // above and is provided only for the // purpose of clarifying points made. var re = /^([\\\\/][a-z0-9\\s\\-_\\@\\-\\^!#$%&]*)+(\\.[az][a-z0-9]+)?$/i console.log( [ '/', '/asdf', '/asdf/scd.csv', '//asdf', '/asd/ads/c.csv/', 'asd/asfd/a' ].map(RegExp.prototype.test, re)) 

Answer 2

Try using /^(\\/|([\\\\/][\\w\\s@^!#$%&-]+)+(\\.[az]+[\\\\/]?)?)$/i instead, which forces at least one character to match between each slash:

 var regex = /^(\\/|([\\\\/][\\w\\s@^!#$%&-]+)+(\\.[az]+[\\\\/]?)?)$/i console.log([ '/', '/asdf', '/asdf/scd.csv', '//asdf', '/asd/ads/c.csv/', 'asd/asfd/a' ].map(RegExp.prototype.test, regex)) 

Answer 3

((\/[\w\s\.@^!#$%&-]+)+\/?)|\/[\w\.\s@^!#$%&-]*

This was tested to match your sample input,
BUT on np++ (ie perl-regex flavor), because I have no experience with javascript.
Therefor here the same in flavor-indpendent prose.

"(slash and character many times, followed by optional slash)
or
slash and zero or more characters".

Note1: I added explicit "." to allowed characters.
Note2: I assume your "\\/" means, "explicit slash, not backslash".