How to trim multiple characters?

up vote
6
down vote

favorite

I have a string as follows

const example = ' ( some string ()() here ) ';

If I trim the string with

example.trim()

it will give me the output: ( some string ()() here )

But I want the output some string ()() here. How to achieve that?

const example = ' ( some string ()() here ) ';
console.log(example.trim());

edited 2 hours ago

asked 2 hours ago

sertsedat

2,1141231

add a commentÂ |Â

up vote
6
down vote

favorite

I have a string as follows

const example = ' ( some string ()() here ) ';

If I trim the string with

example.trim()

it will give me the output: ( some string ()() here )

But I want the output some string ()() here. How to achieve that?

const example = ' ( some string ()() here ) ';
console.log(example.trim());

edited 2 hours ago

asked 2 hours ago

sertsedat

2,1141231

add a commentÂ |Â

up vote
6
down vote

favorite

I have a string as follows

const example = ' ( some string ()() here ) ';

If I trim the string with

example.trim()

it will give me the output: ( some string ()() here )

But I want the output some string ()() here. How to achieve that?

const example = ' ( some string ()() here ) ';
console.log(example.trim());

edited 2 hours ago

asked 2 hours ago

sertsedat

2,1141231

I have a string as follows

const example = ' ( some string ()() here ) ';

If I trim the string with

example.trim()

it will give me the output: ( some string ()() here )

But I want the output some string ()() here. How to achieve that?

const example = ' ( some string ()() here ) ';
console.log(example.trim());

const example = ' ( some string ()() here ) ';
console.log(example.trim());

const example = ' ( some string ()() here ) ';
console.log(example.trim());

javascript string trim

edited 2 hours ago

asked 2 hours ago

sertsedat

2,1141231

edited 2 hours ago

asked 2 hours ago

sertsedat

2,1141231

edited 2 hours ago

asked 2 hours ago

sertsedat

2,1141231

asked 2 hours ago

sertsedat

2,1141231

asked 2 hours ago

sertsedat

2,1141231

add a commentÂ |Â

5 Answers
5

active

oldest

votes

up vote
4
down vote

accepted

You can use a regex for leading and trailing space/brackets:

function grabText(str) 
 return str.replace(/^s+(s+(.*)s+)s+$/g,"$1");


console.log(
 grabText(' ( some (string) here ) ')
)
console.log(
 grabText(' ( some string ()() here ) ')
)

edited 2 hours ago

answered 2 hours ago

mplungjan

84k20120179

thanks for this, but this is a bad assumption. the string inside can also contain (). That's why I'm trying to trim. Just updated my question to have better clarity. Sorry about that
â€“Â sertsedat
2 hours ago

Well, by definition trim only removes whitespace. Occasionally you'd find some trim functionality that does more - actually the other day I found out C# does that, where you can define which characters to remove from the beginning and the end but the generally understood definition of trim is for whitespace only.
â€“Â vlaz
2 hours ago

Please see my updated example which includes inner brackets
â€“Â mplungjan
2 hours ago

1

Well, you can write your own function that does it without regex: take parameters for what to strip, go from the beginning and end of the string and skip any strippable character, then return the inner part. But it'd be a bit of a wasted effort, as regex+replace is readily available.
â€“Â vlaz
2 hours ago

1

@sertsedat - my latest version handles both your cases
â€“Â mplungjan
2 hours ago

Â |Â
show 2 more comments

up vote
3
down vote

You can use the regex to get the matched string, The below regex, matches the first character followed by characters or whitespaces and ending with a alphanumberic character

const example = ' ( some (string) ()()here ) ';
console.log(example.match(/(w[ws.(.*)]+)w/g));

edited 2 hours ago

answered 2 hours ago

Shubham Khatri

66.9k1376117

@mplungjan, yes that won't work. Ahh, OP updates his question
â€“Â Shubham Khatri
2 hours ago

well this also works. sorry about that. I guess you didn't see edited question. thanks for the answer
â€“Â sertsedat
1 hour ago

add a commentÂ |Â

up vote
1
down vote

 const str = ' ( some ( string ) here ) '.replace(/^s+(s+(.*)s+)s+$/g,'$1');
 console.log(str);

edited 1 hour ago

answered 2 hours ago

khaled_bhar

114118

1

Why trim after using a regex? You can let the regex do the trim too
â€“Â mplungjan
1 hour ago

yes yes :) I was writing an other solution so it is fixed now
â€“Â khaled_bhar
1 hour ago

Your solution is now identical to mine - exactly the same
â€“Â mplungjan
1 hour ago

add a commentÂ |Â

up vote
0
down vote

You could use a recursive method and specify the number of times you wish to trim the string. This will also work with things other than round brackets, for instance, square brackets:

const example = ' ( some string ()() here ) ';
const exampleTwo = ' [ This, is [some] text ] ';

function trim_factor(str, times) 
 if(times == 0) 
 return str;
 
 str = str.trim();
 return str.charAt(0) + trim_factor(str.substr(1, str.length-2), times-1) + str.charAt(str.length-1);


console.log(trim_factor(example, 2));
console.log(trim_factor(exampleTwo, 2));

edited 1 hour ago

answered 2 hours ago

Nick Parsons

1,603415

add a commentÂ |Â

up vote
-2
down vote

You should to replace the "(" and ")" characters as in the following example:

var str = " ( Some string ) ";
str.trim();
str = str.replace("(", "");
str = str.replace(")", "");

answered 2 hours ago

Sh. Pavel

769317

add a commentÂ |Â

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5 Answers
5

active

oldest

votes

5 Answers
5

active

oldest

votes

up vote
4
down vote

accepted

You can use a regex for leading and trailing space/brackets:

function grabText(str) 
 return str.replace(/^s+(s+(.*)s+)s+$/g,"$1");


console.log(
 grabText(' ( some (string) here ) ')
)
console.log(
 grabText(' ( some string ()() here ) ')
)

edited 2 hours ago

answered 2 hours ago

mplungjan

84k20120179

thanks for this, but this is a bad assumption. the string inside can also contain (). That's why I'm trying to trim. Just updated my question to have better clarity. Sorry about that
â€“Â sertsedat
2 hours ago

Well, by definition trim only removes whitespace. Occasionally you'd find some trim functionality that does more - actually the other day I found out C# does that, where you can define which characters to remove from the beginning and the end but the generally understood definition of trim is for whitespace only.
â€“Â vlaz
2 hours ago

Please see my updated example which includes inner brackets
â€“Â mplungjan
2 hours ago

1

Well, you can write your own function that does it without regex: take parameters for what to strip, go from the beginning and end of the string and skip any strippable character, then return the inner part. But it'd be a bit of a wasted effort, as regex+replace is readily available.
â€“Â vlaz
2 hours ago

1

@sertsedat - my latest version handles both your cases
â€“Â mplungjan
2 hours ago

Â |Â
show 2 more comments

up vote
4
down vote

accepted

You can use a regex for leading and trailing space/brackets:

function grabText(str) 
 return str.replace(/^s+(s+(.*)s+)s+$/g,"$1");


console.log(
 grabText(' ( some (string) here ) ')
)
console.log(
 grabText(' ( some string ()() here ) ')
)

edited 2 hours ago

answered 2 hours ago

mplungjan

84k20120179

thanks for this, but this is a bad assumption. the string inside can also contain (). That's why I'm trying to trim. Just updated my question to have better clarity. Sorry about that
â€“Â sertsedat
2 hours ago

Well, by definition trim only removes whitespace. Occasionally you'd find some trim functionality that does more - actually the other day I found out C# does that, where you can define which characters to remove from the beginning and the end but the generally understood definition of trim is for whitespace only.
â€“Â vlaz
2 hours ago

Please see my updated example which includes inner brackets
â€“Â mplungjan
2 hours ago

1

Well, you can write your own function that does it without regex: take parameters for what to strip, go from the beginning and end of the string and skip any strippable character, then return the inner part. But it'd be a bit of a wasted effort, as regex+replace is readily available.
â€“Â vlaz
2 hours ago

1

@sertsedat - my latest version handles both your cases
â€“Â mplungjan
2 hours ago

Â |Â
show 2 more comments

up vote
4
down vote

accepted

You can use a regex for leading and trailing space/brackets:

function grabText(str) 
 return str.replace(/^s+(s+(.*)s+)s+$/g,"$1");


console.log(
 grabText(' ( some (string) here ) ')
)
console.log(
 grabText(' ( some string ()() here ) ')
)

edited 2 hours ago

answered 2 hours ago

mplungjan

84k20120179

You can use a regex for leading and trailing space/brackets:

function grabText(str) 
 return str.replace(/^s+(s+(.*)s+)s+$/g,"$1");


console.log(
 grabText(' ( some (string) here ) ')
)
console.log(
 grabText(' ( some string ()() here ) ')
)

function grabText(str) 
 return str.replace(/^s+(s+(.*)s+)s+$/g,"$1");


console.log(
 grabText(' ( some (string) here ) ')
)
console.log(
 grabText(' ( some string ()() here ) ')
)

function grabText(str) 
 return str.replace(/^s+(s+(.*)s+)s+$/g,"$1");


console.log(
 grabText(' ( some (string) here ) ')
)
console.log(
 grabText(' ( some string ()() here ) ')
)

edited 2 hours ago

answered 2 hours ago

mplungjan

84k20120179

edited 2 hours ago

answered 2 hours ago

mplungjan

84k20120179

answered 2 hours ago

mplungjan

84k20120179

answered 2 hours ago

mplungjan

84k20120179

thanks for this, but this is a bad assumption. the string inside can also contain (). That's why I'm trying to trim. Just updated my question to have better clarity. Sorry about that
â€“Â sertsedat
2 hours ago

Well, by definition trim only removes whitespace. Occasionally you'd find some trim functionality that does more - actually the other day I found out C# does that, where you can define which characters to remove from the beginning and the end but the generally understood definition of trim is for whitespace only.
â€“Â vlaz
2 hours ago

Please see my updated example which includes inner brackets
â€“Â mplungjan
2 hours ago

1

Well, you can write your own function that does it without regex: take parameters for what to strip, go from the beginning and end of the string and skip any strippable character, then return the inner part. But it'd be a bit of a wasted effort, as regex+replace is readily available.
â€“Â vlaz
2 hours ago

1

@sertsedat - my latest version handles both your cases
â€“Â mplungjan
2 hours ago

Â |Â
show 2 more comments

thanks for this, but this is a bad assumption. the string inside can also contain (). That's why I'm trying to trim. Just updated my question to have better clarity. Sorry about that
â€“Â sertsedat
2 hours ago

Well, by definition trim only removes whitespace. Occasionally you'd find some trim functionality that does more - actually the other day I found out C# does that, where you can define which characters to remove from the beginning and the end but the generally understood definition of trim is for whitespace only.
â€“Â vlaz
2 hours ago

Please see my updated example which includes inner brackets
â€“Â mplungjan
2 hours ago

1

Well, you can write your own function that does it without regex: take parameters for what to strip, go from the beginning and end of the string and skip any strippable character, then return the inner part. But it'd be a bit of a wasted effort, as regex+replace is readily available.
â€“Â vlaz
2 hours ago

1

@sertsedat - my latest version handles both your cases
â€“Â mplungjan
2 hours ago

thanks for this, but this is a bad assumption. the string inside can also contain (). That's why I'm trying to trim. Just updated my question to have better clarity. Sorry about that
â€“Â sertsedat
2 hours ago

Well, by definition trim only removes whitespace. Occasionally you'd find some trim functionality that does more - actually the other day I found out C# does that, where you can define which characters to remove from the beginning and the end but the generally understood definition of trim is for whitespace only.
â€“Â vlaz
2 hours ago

Please see my updated example which includes inner brackets
â€“Â mplungjan
2 hours ago

Well, you can write your own function that does it without regex: take parameters for what to strip, go from the beginning and end of the string and skip any strippable character, then return the inner part. But it'd be a bit of a wasted effort, as regex+replace is readily available.
â€“Â vlaz
2 hours ago

@sertsedat - my latest version handles both your cases
â€“Â mplungjan
2 hours ago

Â |Â
show 2 more comments

up vote
3
down vote

You can use the regex to get the matched string, The below regex, matches the first character followed by characters or whitespaces and ending with a alphanumberic character

const example = ' ( some (string) ()()here ) ';
console.log(example.match(/(w[ws.(.*)]+)w/g));

edited 2 hours ago

answered 2 hours ago

Shubham Khatri

66.9k1376117

@mplungjan, yes that won't work. Ahh, OP updates his question
â€“Â Shubham Khatri
2 hours ago

well this also works. sorry about that. I guess you didn't see edited question. thanks for the answer
â€“Â sertsedat
1 hour ago

add a commentÂ |Â

up vote
3
down vote

You can use the regex to get the matched string, The below regex, matches the first character followed by characters or whitespaces and ending with a alphanumberic character

const example = ' ( some (string) ()()here ) ';
console.log(example.match(/(w[ws.(.*)]+)w/g));

edited 2 hours ago

answered 2 hours ago

Shubham Khatri

66.9k1376117

@mplungjan, yes that won't work. Ahh, OP updates his question
â€“Â Shubham Khatri
2 hours ago

well this also works. sorry about that. I guess you didn't see edited question. thanks for the answer
â€“Â sertsedat
1 hour ago

add a commentÂ |Â

up vote
3
down vote

You can use the regex to get the matched string, The below regex, matches the first character followed by characters or whitespaces and ending with a alphanumberic character

const example = ' ( some (string) ()()here ) ';
console.log(example.match(/(w[ws.(.*)]+)w/g));

edited 2 hours ago

answered 2 hours ago

Shubham Khatri

66.9k1376117

You can use the regex to get the matched string, The below regex, matches the first character followed by characters or whitespaces and ending with a alphanumberic character

const example = ' ( some (string) ()()here ) ';
console.log(example.match(/(w[ws.(.*)]+)w/g));

const example = ' ( some (string) ()()here ) ';
console.log(example.match(/(w[ws.(.*)]+)w/g));

const example = ' ( some (string) ()()here ) ';
console.log(example.match(/(w[ws.(.*)]+)w/g));

edited 2 hours ago

answered 2 hours ago

Shubham Khatri

66.9k1376117

edited 2 hours ago

answered 2 hours ago

Shubham Khatri

66.9k1376117

answered 2 hours ago

Shubham Khatri

66.9k1376117

answered 2 hours ago

Shubham Khatri

66.9k1376117

@mplungjan, yes that won't work. Ahh, OP updates his question
â€“Â Shubham Khatri
2 hours ago

well this also works. sorry about that. I guess you didn't see edited question. thanks for the answer
â€“Â sertsedat
1 hour ago

add a commentÂ |Â

@mplungjan, yes that won't work. Ahh, OP updates his question
â€“Â Shubham Khatri
2 hours ago

well this also works. sorry about that. I guess you didn't see edited question. thanks for the answer
â€“Â sertsedat
1 hour ago

@mplungjan, yes that won't work. Ahh, OP updates his question
â€“Â Shubham Khatri
2 hours ago

well this also works. sorry about that. I guess you didn't see edited question. thanks for the answer
â€“Â sertsedat
1 hour ago

add a commentÂ |Â

up vote
1
down vote

 const str = ' ( some ( string ) here ) '.replace(/^s+(s+(.*)s+)s+$/g,'$1');
 console.log(str);

edited 1 hour ago

answered 2 hours ago

khaled_bhar

114118

1

Why trim after using a regex? You can let the regex do the trim too
â€“Â mplungjan
1 hour ago

yes yes :) I was writing an other solution so it is fixed now
â€“Â khaled_bhar
1 hour ago

Your solution is now identical to mine - exactly the same
â€“Â mplungjan
1 hour ago

add a commentÂ |Â

up vote
1
down vote

 const str = ' ( some ( string ) here ) '.replace(/^s+(s+(.*)s+)s+$/g,'$1');
 console.log(str);

edited 1 hour ago

answered 2 hours ago

khaled_bhar

114118

1

Why trim after using a regex? You can let the regex do the trim too
â€“Â mplungjan
1 hour ago

yes yes :) I was writing an other solution so it is fixed now
â€“Â khaled_bhar
1 hour ago

Your solution is now identical to mine - exactly the same
â€“Â mplungjan
1 hour ago

add a commentÂ |Â

up vote
1
down vote

 const str = ' ( some ( string ) here ) '.replace(/^s+(s+(.*)s+)s+$/g,'$1');
 console.log(str);

edited 1 hour ago

answered 2 hours ago

khaled_bhar

114118

 const str = ' ( some ( string ) here ) '.replace(/^s+(s+(.*)s+)s+$/g,'$1');
 console.log(str);

 const str = ' ( some ( string ) here ) '.replace(/^s+(s+(.*)s+)s+$/g,'$1');
 console.log(str);

 const str = ' ( some ( string ) here ) '.replace(/^s+(s+(.*)s+)s+$/g,'$1');
 console.log(str);

edited 1 hour ago

answered 2 hours ago

khaled_bhar

114118

edited 1 hour ago

answered 2 hours ago

khaled_bhar

114118

answered 2 hours ago

khaled_bhar

114118

answered 2 hours ago

khaled_bhar

114118

1

Why trim after using a regex? You can let the regex do the trim too
â€“Â mplungjan
1 hour ago

yes yes :) I was writing an other solution so it is fixed now
â€“Â khaled_bhar
1 hour ago

Your solution is now identical to mine - exactly the same
â€“Â mplungjan
1 hour ago

add a commentÂ |Â

1

Why trim after using a regex? You can let the regex do the trim too
â€“Â mplungjan
1 hour ago

yes yes :) I was writing an other solution so it is fixed now
â€“Â khaled_bhar
1 hour ago

Your solution is now identical to mine - exactly the same
â€“Â mplungjan
1 hour ago

Why trim after using a regex? You can let the regex do the trim too
â€“Â mplungjan
1 hour ago

yes yes :) I was writing an other solution so it is fixed now
â€“Â khaled_bhar
1 hour ago

Your solution is now identical to mine - exactly the same
â€“Â mplungjan
1 hour ago

add a commentÂ |Â

up vote
0
down vote

You could use a recursive method and specify the number of times you wish to trim the string. This will also work with things other than round brackets, for instance, square brackets:

const example = ' ( some string ()() here ) ';
const exampleTwo = ' [ This, is [some] text ] ';

function trim_factor(str, times) 
 if(times == 0) 
 return str;
 
 str = str.trim();
 return str.charAt(0) + trim_factor(str.substr(1, str.length-2), times-1) + str.charAt(str.length-1);


console.log(trim_factor(example, 2));
console.log(trim_factor(exampleTwo, 2));

edited 1 hour ago

answered 2 hours ago

Nick Parsons

1,603415

add a commentÂ |Â

up vote
0
down vote

You could use a recursive method and specify the number of times you wish to trim the string. This will also work with things other than round brackets, for instance, square brackets:

const example = ' ( some string ()() here ) ';
const exampleTwo = ' [ This, is [some] text ] ';

function trim_factor(str, times) 
 if(times == 0) 
 return str;
 
 str = str.trim();
 return str.charAt(0) + trim_factor(str.substr(1, str.length-2), times-1) + str.charAt(str.length-1);


console.log(trim_factor(example, 2));
console.log(trim_factor(exampleTwo, 2));

edited 1 hour ago

answered 2 hours ago

Nick Parsons

1,603415

add a commentÂ |Â

up vote
0
down vote

You could use a recursive method and specify the number of times you wish to trim the string. This will also work with things other than round brackets, for instance, square brackets:

const example = ' ( some string ()() here ) ';
const exampleTwo = ' [ This, is [some] text ] ';

function trim_factor(str, times) 
 if(times == 0) 
 return str;
 
 str = str.trim();
 return str.charAt(0) + trim_factor(str.substr(1, str.length-2), times-1) + str.charAt(str.length-1);


console.log(trim_factor(example, 2));
console.log(trim_factor(exampleTwo, 2));

edited 1 hour ago

answered 2 hours ago

Nick Parsons

1,603415

You could use a recursive method and specify the number of times you wish to trim the string. This will also work with things other than round brackets, for instance, square brackets:

const example = ' ( some string ()() here ) ';
const exampleTwo = ' [ This, is [some] text ] ';

function trim_factor(str, times) 
 if(times == 0) 
 return str;
 
 str = str.trim();
 return str.charAt(0) + trim_factor(str.substr(1, str.length-2), times-1) + str.charAt(str.length-1);


console.log(trim_factor(example, 2));
console.log(trim_factor(exampleTwo, 2));

const example = ' ( some string ()() here ) ';
const exampleTwo = ' [ This, is [some] text ] ';

function trim_factor(str, times) 
 if(times == 0) 
 return str;
 
 str = str.trim();
 return str.charAt(0) + trim_factor(str.substr(1, str.length-2), times-1) + str.charAt(str.length-1);


console.log(trim_factor(example, 2));
console.log(trim_factor(exampleTwo, 2));

const example = ' ( some string ()() here ) ';
const exampleTwo = ' [ This, is [some] text ] ';

function trim_factor(str, times) 
 if(times == 0) 
 return str;
 
 str = str.trim();
 return str.charAt(0) + trim_factor(str.substr(1, str.length-2), times-1) + str.charAt(str.length-1);


console.log(trim_factor(example, 2));
console.log(trim_factor(exampleTwo, 2));

edited 1 hour ago

answered 2 hours ago

Nick Parsons

1,603415

edited 1 hour ago

answered 2 hours ago

Nick Parsons

1,603415

answered 2 hours ago

Nick Parsons

1,603415

answered 2 hours ago

Nick Parsons

1,603415

add a commentÂ |Â

up vote
-2
down vote

You should to replace the "(" and ")" characters as in the following example:

var str = " ( Some string ) ";
str.trim();
str = str.replace("(", "");
str = str.replace(")", "");

answered 2 hours ago

Sh. Pavel

769317

add a commentÂ |Â

up vote
-2
down vote

You should to replace the "(" and ")" characters as in the following example:

var str = " ( Some string ) ";
str.trim();
str = str.replace("(", "");
str = str.replace(")", "");

answered 2 hours ago

Sh. Pavel

769317

add a commentÂ |Â

up vote
-2
down vote

You should to replace the "(" and ")" characters as in the following example:

var str = " ( Some string ) ";
str.trim();
str = str.replace("(", "");
str = str.replace(")", "");

answered 2 hours ago

Sh. Pavel

769317

You should to replace the "(" and ")" characters as in the following example:

var str = " ( Some string ) ";
str.trim();
str = str.replace("(", "");
str = str.replace(")", "");

var str = " ( Some string ) ";
str.trim();
str = str.replace("(", "");
str = str.replace(")", "");

var str = " ( Some string ) ";
str.trim();
str = str.replace("(", "");
str = str.replace(")", "");

answered 2 hours ago

Sh. Pavel

769317

answered 2 hours ago

Sh. Pavel

769317

answered 2 hours ago

Sh. Pavel

769317

answered 2 hours ago

Sh. Pavel

769317

add a commentÂ |Â

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