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Need regex for mobile number start with 61 to 99 with 10 digit number

Requirement is mobile number should start with 61 to 99
like 61xxxxxxxx, 62xxxxxxxxx... , 99xxxxxxxxxx
Need regular expression to match this case.
If mobile no is start with 0 or 11,12 or anything less than 61 then it should be invalid
Mobile no is max 10 digits, no country code needed.

You're probably better off using whatever programming tool you have to evaluate whether the first 2 digits are in range, far simpler and probably performant too. However, if you strictly want to use regex, this will do-
(?:6[1-9]|[7-9][0-9])\d{8}$
Here's the demo
It essentially, checks the first digit, if it's a 6, the next digit should be in range [1-9], if it's a 7, 8 or 9 (i.e range [7-9]), the next digit can be in range [0-9]. Then there should be 8 digits that follow.
Ofcourse, this above is a simple and easy to understand solution. Essentially checking each first digit and then matching the next. However if your regex flavor supports negative lookbehind, you could probably shorten this a bit more (sacrificing readability for brevity) but I do prefer this.

You could generate the prefix for the numbers and add a pattern for the remaining 8 digits.
Something like this
const regexp = new RegExp('('+[...Array(39).keys()].map(key => key + 61).join('|') + ')\\d{8,8}')

Regex to select all characters that do not match a pattern

I'm weak with regexes but have put together the following regex which selects when my pattern is met, the problem is that i need to select any characters that do not fit the pattern.
/^\d{1,2}[ ]\d{1,2}[ ]\d{1,2}[ ][AB]/i
Correct pattern is:
## ## ## A|B aka [0 < x <= 90]*space*[0 < x <= 90] [0 < x <= 90] [A|B]
EG:
12 34 56 A → good
12 34 56 B → good
12 34 5.6 A → bad - select .
12 34 5.6 C → bad - select . and C
1A 23 45 6 → bad - select A and 6
Edit:
As my impression was that regex is used to perform validation of both characters and pattern/sequence at the same time. The simple question is how to select characters that do not fit the category of non negative numbers, spaces and distinct characters.

Answer 1
Brief
This isn't really realizable with 1 regex due to the nature of the regex. This answer provides a regex that will capture the last incorrect entry. For multiple incorrect entries, a loop must be used. You can correct the incorrect entries by running some code logic on the resulting captured groups to determine why it isn't valid.
My ultimate suggestion would be to split the string by a known delimiter (in this case the space character and then using some logic (or even a small regex) to determine why it's incorrect and how to fix it, as seen in Answer 2.
Non-matches
The following logic is applied in my second answer.
For any users wondering what I did to catch incorrect matches: At the most basic level, all this regex is doing is adding |(.*) to every subsection of the regex. Some sections required additional changes for catching specific invalid string formats, but the |(.*) or slight modifications of this will likely solve anyone else's issues.
Other modifications include:
Using opposite tokens
For example: Matching a digit
Original regex: \d
Opposite regex \D
For example: Matching a digit or whitepace
Original regex: [\d\s]
Opposite regex: [^\d\s]
Note [\D\S] is incorrect as it matches both sets of characters, thus, any non-whitespace or non-digit character (since non-whitespace includes digits and non-digits include whitespace, both will be matched)
Negative lookaheads
For example: Catching up to 31 days in a month
Original regex \b(?:[0-2]?\d|3[01])\b
Opposite regex: \b(?![0-2]?\d\b|3[01]\b)\d+\b
Code
First, creating a more correct regex that also ensures 0 < x <= 90 as per the OP's question.
^(?:(?:[0-8]?\d|90) ){3}[AB]$
See regex in use here
^(?:(?:(?:[0-8]?\d|90) |(\S*) ?)){3}(?:[AB]|(.*))$
Note: This regex uses the mi flags (multiline - assuming input is in that format, and case-insensitive)
Other Formats
Realistically, this following regex would be ideal. Unfortunately, JavaScript doesn't support some of the tokens used in the regex, but I feel it may be useful to the OP or other users that see this question.
See regex in use here
^(?:(?:(?:[0-8]?\d|90) |(?<n>\S*?) |(?<n>\S*?) ?)){3}(?:(?<n>\S*) )?(?:[AB]|(.*))$
Results
Input
The first section (sections separated by the extra newline/break) shows valid strings, while the second shows invalid strings.
0 45 90 A
0 45 90 B
-1 45 90 A
0 45 91 A
12 34 5.6 A
12 34 56 C
1A 23 45 6
11 1A 12 12 A
12 12 A
12 12 A
Output
0 45 90 A VALID
0 45 90 B VALID
-1 45 90 A INVALID: -1
0 45 91 A INVALID: 91
12 34 5.6 A INVALID: 5.6
12 34 56 C INVALID: C
1A 23 45 6 INVALID: 1A, 6
11 1A 12 12 A INVALID: 12 A
12 12 A INVALID: (missing value)
12 12 A INVALID: A, (missing value)
Note: The last entry shows an odd output, but that's due to a limitation with JavaScript's regex engine. The Other Formats section describes this and another method to use to properly catch these cases (using a different regex engine)
Explanation
This uses a simple | (OR) and captures the incorrect matches into a capture group.
^ Assert position at the start of the line
(?:(?:(?:[0-8]?\d|90) |(\S*) ?)){3} Match the following exactly 3 times
(?:(?:[0-8]?\d|90) |(.+)) Match either of the following
(?:[0-8]?\d|90) Match either of the following, followed by a space character literally
[0-8]?\d Match between zero and one of the characters in the set 0-8 (a digit between 0 and 8), followed by any digit
90 Match 90 literally
(\S*) ? Capture any non-whitespace character one or more times into capture group 1, followed by zero or one space character literally
(?:[AB]|(.*)) Match either of the following
[AB] Match any character present in the set (A or B)
(.*) Capture any character any number of times into capture group 2
$ Assert position at the end of the line
Answer 2
Brief
This method splits the string on the given delimiter and tests each section for the proper set of characters. It outputs a message if the value is incorrect. You would likely replace the console outputs with whatever logic you want use.
Code
var arr = [
"0 45 90 A",
"0 45 90 B",
"-1 45 90 A",
"0 45 91 A",
"12 34 5.6 A",
"12 34 56 C",
"1A 23 45 6",
"11 1A 12 12 A",
"12 12 A",
"12 12 A"
];
arr.forEach(function(e) {
var s = e.split(" ");
var l = s.pop();
var numElements = 3;
var maxNum = 90;
var syntaxErrors = [];
if(s.length != numElements) {
syntaxErrors.push(`Invalid number of elements: Number = ${numElements}, Given = ${s.length}`);
}
s.forEach(function(v) {
if(v.match(/\D/)) {
syntaxErrors.push(`Invalid value "${v}" exists`);
} else if(!v.length) {
syntaxErrors.push(`An empty value or double space exists`);
} else if(Number(v) > maxNum) {
syntaxErrors.push(`Value greater than ${maxNum} exists: ${v}`);
}
});
if(l.match(/[^AB]/)) {
syntaxErrors.push(`Last element ${l} in "${e}" is invalid`);
}
if(syntaxErrors.length) {
console.log(`"${e}" [\n\t${syntaxErrors.join('\n\t')}\n]`);
} else {
console.log(`No errors found in "${e}"`);
}
});

Regex match number between 1 and 31 with or without leading 0

I want to setup some validation on an <input> to prevent the user from entering wrong characters. For this I am using ng-pattern. It currently disables the user from entering wrong characters, but I also noticed this is not the expected behavior so I am also planning on creating a directive.
I am using
AngularJS: 1.6.1
What should the regex match
Below are the requirements for the regex string:
Number 0x to xx (example 01 to 93)
Number x to xx (example 9 to 60)
Characters are not allowed
Special characters are not allowed
Notice:
the 'x' is variable and could be any number between 0 and 100.
The number on the place of 'x' is variable so if it is possible to create a string that is easily changeable that would be appreciated!
What I tried
A few regex strings I tried where:
1) ^0*([0-9]\d{1,2})$
--> Does match 01 but not 1
--> Does match 32 where it shouldn't
2) ^[1-9][0-9]?$|^31$
--> Does match 1 but not 01
--> Does match 32 where it shouldn't
For testing I am using https://regex101.com/tests.
What am I missing in my attempts?

If your aim is to match 0 to 100, here's a way, based on the previous solution.
\b(0?[1-9]|[1-9][0-9]|100)\b
Basically, there's 3 parts to that match...
0?[1-9] Addresses numbers 1 to 9, by mentionning that 0 migh be present
[1-9][0-9] covers number 10 to 99, the [1-9] representing the tens
100 covers for 100
Here's an example of it
Where you to require to set the higher boundary to 42, the middle part of the expression would become [1-3][0-9] (covering 10 to 39) and the last part would become 4[0-2] (covering 40 to 42) like so:
\b(0?[1-9]|[1-3][0-9]|4[0-2])\b

This should work:
^(0?[1-9]|[12][0-9]|3[01])$
https://regex101.com/r/BYSDwz/1

Expression regular for check phone numbers at word level

I'm trying to write a RegEx to test if a number is valid and for valid I mean any number that matches country calling codes but also where the format of telephone numbers is standardized by ITU-T in the recommendation E.164. This specifies that the entire number should be 15 digits or shorter, and begin with a country prefix as said here so I did this:
^\+\d{2}|\d{3}([0-9])\d{7}$
But it's not working. In my case (VE numbers can't match the RegEx since this one are validated in another way) this input is valid:
+1420XXXXXXXXXXX // Slovakia - X is a digit and could be more, tough, 5 minimum
001420XXXXXXXXXX // Slovakia - I've changed from + to 00
420XXXXXXXXXXXXX // Slovakia - I've removed the 00 o + but number still being valid
+40XXXXXXXXXXXXX // Romania
Invalid numbers are the one that doesn't match the RegEx and the one started with +58 since they are from VE. So, resuming, a valid number should have:
+XX|+XXX plus 12|11 digits (5 minimum) where XX|XXX is the country code and then since maximum is 15 digits then should be 12 or 11 digits depending on the country format
Can any help me with this? It's a one I called complex

Few strange things going on with your regexp:
\d is shorthand for [0-9] - fine to use both, but I'm wondering why they're mixed
what you are searching with you OR (|) is "something that starts with +XX" i.e. plus and two numbers (^\+\d{2}) OR "something that ends with XXXXXXXXXXX" i.e. 11 numbers (\d{3}([0-9])\d{7}$)
You need to group (with brackets) the OR choices, otherwise it is everything to the left or everything to the right (simplistically)
^\+(\d{2}|\d{3})([0-9])\d{7}$
There is, however, another way of giving the number of occurrences : {m,n} means occurs between m and n times. So you could say ^\+\d{7,15}$ (where 7 is your minimum 5 + the minimum country code of 2).
To really do this, however, you might want to take a look here (https://code.google.com/p/libphonenumber/ 1) where there is a complete validation and formatting for all phone numbers available as javascript.

Regular Expression Phone Number Validation

I wrote this regular expression for the Lebanese phone number basically it should start with
00961 or +961 which is the international code then the area code which
could be either any digit from 0 to 9 or cellular code "70" or "76" or
"79" then a 6 digit number exactly
I have coded the following reg ex without the 6 digit part :
^(([0][0]|[+])([9][6][1])([0-9]{1}|[7][0]|[7][1]|[7][6]|[7][8]))$
when i want to add code to ensure only 6 digits more are allowed to the expression:
^(([0][0]|[+])([9][6][1])([0-9]{1}|[7][0]|[7][1]|[7][6]|[7][8])([0-9]{6}))$
It Seems to accept 5 or 6 digits not 6 digits exactly
i am having difficulty finding whats wrong

use this regex ((00)|(\+))961((\d)|(7[0168]))\d{6}

Ths is what I would use.
/^(00|\+)961(\d|7[069])\d{6}$/
00 or +
961
a 1-digit number or 70 or 76 or 79
a 6-digit number

The [0-9]{1} will match also the cellular codes 7x since 7 is between 0 and 9. This means that a "5 digit cellular number" will match on a 7 and six more digits.

Try
/^(00961|\+961)([0-9]|70|76|79)\d{6}$/.test( phonenumber );
//^ start of string
// ^^^^^^^^^^^^^ 00961 or +0961
// ^^^^^^^^^^^^^^^^ a digit 0 to 9 or 70 or 76 or 79
// ^^^^^ 6 digits
// ^ end of string

The cellar code is forming a trap, as #ellak points out:
/^((00)|(\+))961((\d)|(7[0168]))\d{6}$/.test("009617612345"); // true
Here the code should breaks like this: 00 961 76 12345,
but the RegEx practically breaks it like this: 00 961 7 612345, because 7 is matched in \d, and the rest is combined, exactly in 6 digits, and matched.
I'm not sure if this is actually valid, but I guess this is not what you want, otherwise the RegEx in your question should work.
Here's a kinda long RegEx that avoids the trap:
/^(00|\+)961([0-68-9]\d{6}|7[234579]\d{5}|7[0168]\d{6})$/
A few test result:
/(00|\+)961([0-68-9]\d{6}|7[234579]\d{5}|7[0168]\d{6})/.test("009617012345")
false
/(00|\+)961([0-68-9]\d{6}|7[234579]\d{5}|7[0168]\d{6})/.test("009618012345")
true
/(00|\+)961([0-68-9]\d{6}|7[234579]\d{5}|7[0168]\d{6})/.test("009617612345")
false
/(00|\+)961([0-68-9]\d{6}|7[234579]\d{5}|7[0168]\d{6})/.test("0096176123456")
true

Just recently, the Lebanese Ministry of Telecommunication has changed area codes on the IMS. So the current Regex matcher becomes:
^(00|\+)961[ -]?(2[1245789]|7[0168]|8[16]|\d)[ -]?\d{6}$
Prefix: 00 OR +
Country code: 961
Area code: 1-digit or 2-digits; including 2*, 7*, 8*..., OR a single digit for Ogero numbers on the old IMS network starting with 0*, and finally older mobile lines starting with 03.
The 6-digit number
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