Develop Reference

I need to filter hyphen(-) and space(" ") in my string, what am I doing wrong?

This is the code I am using (but it only works for hyphens not for spaces)
console.log(
"some - text".replace(/\s\-/g, '').toLowerCase()
)
Can anyone point out what is wrong in this RegEx ?

The regex you have removes the space and the dash, leaving the second space
You need other regex to remove all spaces
console.log(
"some - text".replace(/\s\-/g, '') // space and dash
)
console.log(
"some - text".replace(/\s-\s/g, '') // space before and after a dash
)
console.log(
"some - text".replace(/\s+\-\s+/g, '') // any whitespace before and after
)
console.log(
"so- me - te-xt".replace(/\s|-/g, '') // all spaces and all dashes, can be written as a acharacter class [\s-]
)

Use\W for non-word character:
console.log("some - text".replace(/\W/g, '').toLowerCase())

Your regex is matching "(1 occurrence of some space)(definitely followed by 1 occurrence of ad ash)". You want it to match any occurrence of those, so use a character class in your regex (see: MDN WebDocs: JavaScript Character Classes, with brackets, /[\s\-]/g, like so:
console.log(
"some - text".replace(/[\s-]/g, '').toLowerCase()
)
Want to replace another character? Just throw it into the class and you're good to go.

My main syntax checker condition is not passing all my string combinations

This should pass the condition:
syntax_search = (){return 0;}
syntax_search = ( ) { fsf return 0;}
syntax_search = ( ) { return 0; }
syntax_search = (){ return; }
syntax_search = (){ if(x){ sdfsdf } return 0;}
syntax_search = (){ char x[20]; return };
It is not passing all the combinations above, What is the right way?
if( /^\s*(\s*)\s*{[\s\S]*\s+return\s*[0-9]*\s*;\s*}\s*/.test(syntax_search) )

You regular expression contains many unneeded complexities and there are some characters that need escaping such as { and }.
Anyway you can use this modified version of your regex and it should work.
^\s*\(\s*\)\s*\{(.*(return\s*\d*\s*;)\s*)\}\s*;?$
// ^
// |
// There was a ? here
Regex 101 Demo

Some issues:
As M42 pointed out, you need to escape the curly brackets
The parentheses at the begining also need to be escaped (otherwise you are defining a capture group)
"return" is required by the expression. Your first 2 test cases don't contain the word return and will fail. Is that on purpose?
Same as #3 for ;.
[\s\S]* Anything which is a space and everything which isn't. Replace by a dot .* If you need to also match a newline, use [^]*
This regex is not anchored to the end of the string so it will allow invalid strings. (You can put anything you want after the last }
/^\s*(\s*)\s*{[^]return\s\d*\s*;\s*}\s*$/

Get text between two rounded brackets

How can I retrieve the word my from between the two rounded brackets in the following sentence using a regex in JavaScript?
"This is (my) simple text"

console.log(
"This is (my) simple text".match(/\(([^)]+)\)/)[1]
);
\( being opening brace, ( — start of subexpression, [^)]+ — anything but closing parenthesis one or more times (you may want to replace + with *), ) — end of subexpression, \) — closing brace. The match() returns an array ["(my)","my"] from which the second element is extracted.

var txt = "This is (my) simple text";
re = /\((.*)\)/;
console.log(txt.match(re)[1]);​
jsFiddle example

You may also try a non-regex method (of course if there are multiple such brackets, it will eventually need looping, or regex)
init = txt.indexOf('(');
fin = txt.indexOf(')');
console.log(txt.substr(init+1,fin-init-1))

For anyone looking to return multiple texts in multiple brackets
var testString = "(Charles) de (Gaulle), (Paris) [CDG]"
var reBrackets = /\((.*?)\)/g;
var listOfText = [];
var found;
while(found = reBrackets.exec(testString)) {
listOfText.push(found[1]);
};

Use this to get text between closest ( and ):
const string = "This is (my) (simple (text)"
console.log( string.match(/\(([^()]*)\)/)[1] )
console.log( string.match(/\(([^()]*)\)/g).map(function($0) { return $0.substring(1,$0.length-1) }) )
Results: my and ["my","text"].
EXPLANATION
--------------------------------------------------------------------------------
\( '('
--------------------------------------------------------------------------------
( group and capture to \1:
--------------------------------------------------------------------------------
[^()]* any character except: '(', ')' (0 or
more times (matching the most amount
possible))
--------------------------------------------------------------------------------
) end of \1
--------------------------------------------------------------------------------
\) ')'

to get fields from string formula.
var txt = "{{abctext/lsi)}} = {{abctext/lsi}} + {{adddddd}} / {{ttttttt}}";
re = /\{{(.*?)\}}/g;
console.log(txt.match(re));

to return multiple items within rounded brackets
var res2=str.split(/(|)/);
var items=res2.filter((ele,i)=>{
if(i%2!==0) {
return ele;
}
});

Match exactly 10 digits with unknown boundaries?

Is there any regex I can use to match blocks of exactly 10 digits? For instance, I have this:
/\d{10}(?!\d+)/g
And this matches 2154358383 when given 2154358383 fine, but is also matches 1213141516 when given 12345678910111213141516, which I don't want.
What I think I need is a look-behind assertion (in addition to my lookahead already in there), that checks to make sure the character preceding the match is not an integer, but I can't figure out how to do that.
I tried
/(?:[^\d]+)\d{10}(?!\d+)/g
But that broke my first match of 2154358383, which is bad.
How can I write this to only match groups of 10 integers (no more, no less) with unknown boundaries?
I should also note that I'm trying to extract these out of a much larger string, so ^ and $ are out of the question.

This should work: ([^\d]|^)\d{10}([^\d]|$)

Could you do something like:
([^\d]|^)(\d{10})([^\d]|$)
In other words, the beginning of the string or a non-digit, ten digits, then the end of the string or a non-digit. That should solve the cases you looked for above.
You can use the regex like this:
var regex = /([^\d]|^)(\d{10})([^\d]|$)/;
var match = regex.exec(s);
var digits = match[2];

This should match numbers at the beginning of the string (the ^) or in the middle/end (the [^\d] and the (?!\d). If you care about the exact match and not just that it matches in the first place, you'll need to grab the first group in the match.
/(?:[^\d]|^)(\d{10})(?!\d)/g
This would be easier if JavaScript regular expressions supported lookbehind.

What about the next?
perl -nle 'print if /(\b|\D)(\d{10})(\D|\b)/' <<EOF
123456789
x123456789
123456789x
1234567890
x1234567890
1234567890x
12345678901
x12345678901
x12345678901x
EOF
will print only
1234567890
x1234567890
1234567890x

I know you said "no ^" but maybe it's okay if you use it like this?:
rx = /(?:^|\D)(\d{10})(?!\d)/g
Here's a quick test:
> val = '1234567890 sdkjsdkjfsl 2234567890 323456789000 4234567890'
'1234567890 sdkjsdkjfsl 2234567890 323456789000 4234567890'
> rx.exec(val)[1]
'1234567890'
> rx.exec(val)[1]
'2234567890'
> rx.exec(val)[1]
'4234567890'

Try this
var re = /(?:^|[^\d])(\d{10})(?:$|[^\d])/g
re.exec ( "2154358383")
//["2154358383", "2154358383"]
re.exec ( "12345678910111213141516" )
//null
re.exec ( "abc1234567890def" )
//["c1234567890d", "1234567890"]
val = '1234567890 sdkjsdkjfsl 2234567890 323456789000 4234567890';
re.exec ( val )
//["1234567890 ", "1234567890"]
re.exec ( val )
//[" 2234567890 ", "2234567890"]
re.exec ( val )
//[" 4234567890", "4234567890"]
re.exec ( val )
//null

Simple with lookbehind:
/(?<!\d)\d{10}(?!\d)/g

i would cheat and do something like
if (myvar.toString().substring(1, 10) = "1234567890") ....
:)

Javascript regex assistance

I have the following javascript regex...
.replace("/^(a-zA-Z\-)/gi", '');
It isn't complete... in fact it's wrong. Essentially what I need to do is take a string like "XJ FC-X35 (1492)" and remove the () and it's contents and the whitespace before the parenthesis.

replace(/^(.*?)\s?\(([^)]+)\)$/gi, "$1$2")
Takes XJ FC-X35 (1492) and returns XJ FC-X351492.
Remove the $2 to turn XJ FC-X35 (1492) into XJ FC-X35, if that's what you wanted instead.
Long explanation
^ // From the start of the string
( // Capture a group ($1)
.*? // That contains 0 or more elements (non-greedy)
) // Finish group $1
\s? // Followed by 0 or 1 whitespace characters
\( // Followed by a "("
( // Capture a group ($2)
[ // That contains any characters in the following set
^) // Not a ")"
]+ // One or more times
) // Finish group $2
\)$ // Followed by a ")" followed by the end of the string.

Try this:
x = "XJ FC-X35 (1492)"
x.replace(/\s*\(.*?\)/,'');