javascript - Regular expression for arabic and english and some limited special characters -

i need regular expression detect

1- english letter or

2- arabic letter or

3- both or

4- number in between ( or start or end place dose not matter)

5-maximum length can 30.

and make sure dose not have special characters except (they can appear where)

@ , = % $ # & _

my solution: use regular expression:

    var regex = /^([a-za-z0-9\s@,=%$#&_\u0600-\u06ff]).{0,30}$/;     var result = regex.test('happy- - - /*  */ % ! ~');     console.log(result);//prints true , should print false because of ! , other special characters not allowed 

the problem

my solution not correct

my references arabic use check here # 13 in list second reference third reference

u+0600–u+06ff

https://jsfiddle.net/shareefhiasat/bdm99b1x/

you need quantify group, not dot.
here regex arabic (unicode 9) characters.
note you've included english characters, did mean latin ?

^(?:[a-za-z0-9\s@,=%$#&_\u0600-\u06ff\u0750-\u077f\u08a0-\u08ff\ufb50-\ufdcf\ufdf0-\ufdff\ufe70-\ufeff]|(?:\ud802[\ude60-\ude9f]|\ud83b[\ude00-\udeff])){0,30}$

https://jsfiddle.net/pydbqxgb/

expanded

 ^       (?:       [a-za-z0-9\s@,=%$#&_\u0600-\u06ff\u0750-\u077f\u08a0-\u08ff\ufb50-\ufdcf\ufdf0-\ufdff\ufe70-\ufeff]     |         (?:            \ud802 [\ude60-\ude9f]          |  \ud83b [\ude00-\udeff]        )  ){0,30}  $  

---

an alternative, uses basic non-control (whitespace only) latin.

^(?:[\u0009-\u000d\u001c-\u007e\u0600-\u06ff\u0750-\u077f\u08a0-\u08ff\ufb50-\ufdcf\ufdf0-\ufdff\ufe70-\ufeff]|(?:\ud802[\ude60-\ude9f]|\ud83b[\ude00-\udeff])){0,30}$

https://jsfiddle.net/st60dyve/

expanded

 ^       (?:       [\u0009-\u000d\u001c-\u007e\u0600-\u06ff\u0750-\u077f\u08a0-\u08ff\ufb50-\ufdcf\ufdf0-\ufdff\ufe70-\ufeff]     |         (?:            \ud802 [\ude60-\ude9f]          |  \ud83b [\ude00-\udeff]        )  ){0,30}  $