replace in javascript to cut off part of string

I have this string, how can I make it until .html?

string

https://www.lazada.com.ph/products/fujitsu-standard-aa-rechargeable-battery-2000mah-4pcs-i237968048-s311807989.html?spm=a2o4k.searchlist.list.47.7f1a484d4gpCFd&search=1

make it into

https://www.lazada.com.ph/products/fujitsu-standard-aa-rechargeable-battery-2000mah-4pcs-i237968048-s311807989.html

edited Nov 20 at 7:37

KarelG

4,01432443

asked Nov 20 at 7:15

Micheal

213

4

The posted question does not appear to include any attempt at all to solve the problem. StackOverflow expects you to try to solve your own problem first, as your attempts help us to better understand what you want. Please edit the question to show what you've tried, so as to illustrate a specific roadblock you're running into a Minimal, Complete, and Verifiable example. For more information, please see How to Ask and take the tour.
– CertainPerformance
Nov 20 at 7:15

@CertainPerformance but I have no clue how to start?
– Micheal
Nov 20 at 7:16

Check How do I parse a URL into hostname and path in javascript?
– Mohammad
Nov 20 at 7:22

1

Michael, it is not appropriate to suddenly change the scope of your question. Now all of those answers are invalid.
– KarelG
Nov 20 at 7:31

@KarelG ok then.
– Micheal
Nov 20 at 7:39

add a comment |

I have this string, how can I make it until .html?

string

https://www.lazada.com.ph/products/fujitsu-standard-aa-rechargeable-battery-2000mah-4pcs-i237968048-s311807989.html?spm=a2o4k.searchlist.list.47.7f1a484d4gpCFd&search=1

make it into

https://www.lazada.com.ph/products/fujitsu-standard-aa-rechargeable-battery-2000mah-4pcs-i237968048-s311807989.html

edited Nov 20 at 7:37

KarelG

4,01432443

asked Nov 20 at 7:15

Micheal

213

4

The posted question does not appear to include any attempt at all to solve the problem. StackOverflow expects you to try to solve your own problem first, as your attempts help us to better understand what you want. Please edit the question to show what you've tried, so as to illustrate a specific roadblock you're running into a Minimal, Complete, and Verifiable example. For more information, please see How to Ask and take the tour.
– CertainPerformance
Nov 20 at 7:15

@CertainPerformance but I have no clue how to start?
– Micheal
Nov 20 at 7:16

Check How do I parse a URL into hostname and path in javascript?
– Mohammad
Nov 20 at 7:22

1

Michael, it is not appropriate to suddenly change the scope of your question. Now all of those answers are invalid.
– KarelG
Nov 20 at 7:31

@KarelG ok then.
– Micheal
Nov 20 at 7:39

add a comment |

I have this string, how can I make it until .html?

string

https://www.lazada.com.ph/products/fujitsu-standard-aa-rechargeable-battery-2000mah-4pcs-i237968048-s311807989.html?spm=a2o4k.searchlist.list.47.7f1a484d4gpCFd&search=1

make it into

https://www.lazada.com.ph/products/fujitsu-standard-aa-rechargeable-battery-2000mah-4pcs-i237968048-s311807989.html

edited Nov 20 at 7:37

KarelG

4,01432443

asked Nov 20 at 7:15

Micheal

213

I have this string, how can I make it until .html?

string

https://www.lazada.com.ph/products/fujitsu-standard-aa-rechargeable-battery-2000mah-4pcs-i237968048-s311807989.html?spm=a2o4k.searchlist.list.47.7f1a484d4gpCFd&search=1

make it into

https://www.lazada.com.ph/products/fujitsu-standard-aa-rechargeable-battery-2000mah-4pcs-i237968048-s311807989.html

javascript ecmascript-6

edited Nov 20 at 7:37

KarelG

4,01432443

asked Nov 20 at 7:15

Micheal

213

edited Nov 20 at 7:37

KarelG

4,01432443

asked Nov 20 at 7:15

Micheal

213

edited Nov 20 at 7:37

KarelG

4,01432443

edited Nov 20 at 7:37

KarelG

4,01432443

edited Nov 20 at 7:37

KarelG

4,01432443

asked Nov 20 at 7:15

Micheal

213

asked Nov 20 at 7:15

Micheal

213

asked Nov 20 at 7:15

Micheal

213

4

The posted question does not appear to include any attempt at all to solve the problem. StackOverflow expects you to try to solve your own problem first, as your attempts help us to better understand what you want. Please edit the question to show what you've tried, so as to illustrate a specific roadblock you're running into a Minimal, Complete, and Verifiable example. For more information, please see How to Ask and take the tour.
– CertainPerformance
Nov 20 at 7:15

@CertainPerformance but I have no clue how to start?
– Micheal
Nov 20 at 7:16

Check How do I parse a URL into hostname and path in javascript?
– Mohammad
Nov 20 at 7:22

1

Michael, it is not appropriate to suddenly change the scope of your question. Now all of those answers are invalid.
– KarelG
Nov 20 at 7:31

@KarelG ok then.
– Micheal
Nov 20 at 7:39

add a comment |

4

The posted question does not appear to include any attempt at all to solve the problem. StackOverflow expects you to try to solve your own problem first, as your attempts help us to better understand what you want. Please edit the question to show what you've tried, so as to illustrate a specific roadblock you're running into a Minimal, Complete, and Verifiable example. For more information, please see How to Ask and take the tour.
– CertainPerformance
Nov 20 at 7:15

@CertainPerformance but I have no clue how to start?
– Micheal
Nov 20 at 7:16

Check How do I parse a URL into hostname and path in javascript?
– Mohammad
Nov 20 at 7:22

1

Michael, it is not appropriate to suddenly change the scope of your question. Now all of those answers are invalid.
– KarelG
Nov 20 at 7:31

@KarelG ok then.
– Micheal
Nov 20 at 7:39

The posted question does not appear to include any attempt at all to solve the problem. StackOverflow expects you to try to solve your own problem first, as your attempts help us to better understand what you want. Please edit the question to show what you've tried, so as to illustrate a specific roadblock you're running into a Minimal, Complete, and Verifiable example. For more information, please see How to Ask and take the tour.
– CertainPerformance
Nov 20 at 7:15

@CertainPerformance but I have no clue how to start?
– Micheal
Nov 20 at 7:16

Check How do I parse a URL into hostname and path in javascript?
– Mohammad
Nov 20 at 7:22

Michael, it is not appropriate to suddenly change the scope of your question. Now all of those answers are invalid.
– KarelG
Nov 20 at 7:31

@KarelG ok then.
– Micheal
Nov 20 at 7:39

add a comment |

4 Answers
4

active

oldest

votes

Maybe you're looking for something like this example:

var str = 'https://www.lazada.com.ph/products/fujitsu-standard-aa-rechargeable-battery-2000mah-4pcs-i237968048-s311807989.html?spm=a2o4k.searchlist.list.47.7f1a484d4gpCFd&search=1';



console.log(str.match('^.+.html')[0]);

I've used regex with the pattern ^.+.html to match the url. The regex means: From the beginning of the string to .html

If you want to split the url to 2 parts, you could try:

var str = 'https://www.lazada.com.ph/products/fujitsu-standard-aa-rechargeable-battery-2000mah-4pcs-i237968048-s311807989.html?spm=a2o4k.searchlist.list.47.7f1a484d4gpCFd&search=1';



var parts = str.split(/?/);



console.log(parts[0]);

console.log('?' + parts[1]);

edited Nov 20 at 7:36

answered Nov 20 at 7:19

Foo

There is a reason why there is something called as URL API I rather prefer to see users using this instead of that regex route because this answer is not useful if you have to include other extensions as well (eg .php or .asp).
– KarelG
Nov 20 at 7:26

what I have other string around the url? then it should catch the https as prefix, also how do I replace it?, I don't want to print out the url only, I want the entire text.
– Micheal
Nov 20 at 7:29

add a comment |

You can use String.prototype.split()

var url =  "https://www.lazada.com.ph/products/fujitsu-standard-aa-rechargeable-battery-2000mah-4pcs-i237968048-s311807989.html?spm=a2o4k.searchlist.list.47.7f1a484d4gpCFd&search=1";





console.log(url.split('?', 1)[0])

answered Nov 20 at 7:26

Gildas.Tambo

16.3k33663

add a comment |

You can use the URL API:

var url = new URL(string);

url.search = "";

url.hash = "";

return url.href;

answered Nov 20 at 7:24

Bergi

363k58540866

nice but what is the next step? I want to replace the cleaned url back into its location.
– Micheal
Nov 20 at 7:30

@Micheal What do you mean by "its location"? You had a string for the url, right? Now you have the new url string.
– Bergi
Nov 20 at 7:32

sorry updated my question just now. It's not only the string, the urls are within a document string.
– Micheal
Nov 20 at 7:38

add a comment |

This works fine according to your requirment

let url = "https://www.lazada.com.ph/products/fujitsu-standard-aa-rechargeable- 

         battery-2000mah-4pcs-i237968048-s311807989.html?spm=a2o4k.searchlist.list.47.7f1a484d4gpCFd&search=1"





url.substring(0, <nbsp> url.indexOf('.html')<nbsp> + <nbsp>5)

edited Nov 20 at 7:37

chŝdk

23.5k52852

answered Nov 20 at 7:29

this is yash

14310

add a comment |

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4 Answers
4

active

oldest

votes

4 Answers
4

active

oldest

votes

Maybe you're looking for something like this example:

var str = 'https://www.lazada.com.ph/products/fujitsu-standard-aa-rechargeable-battery-2000mah-4pcs-i237968048-s311807989.html?spm=a2o4k.searchlist.list.47.7f1a484d4gpCFd&search=1';



console.log(str.match('^.+.html')[0]);

I've used regex with the pattern ^.+.html to match the url. The regex means: From the beginning of the string to .html

If you want to split the url to 2 parts, you could try:

var str = 'https://www.lazada.com.ph/products/fujitsu-standard-aa-rechargeable-battery-2000mah-4pcs-i237968048-s311807989.html?spm=a2o4k.searchlist.list.47.7f1a484d4gpCFd&search=1';



var parts = str.split(/?/);



console.log(parts[0]);

console.log('?' + parts[1]);

edited Nov 20 at 7:36

answered Nov 20 at 7:19

Foo

There is a reason why there is something called as URL API I rather prefer to see users using this instead of that regex route because this answer is not useful if you have to include other extensions as well (eg .php or .asp).
– KarelG
Nov 20 at 7:26

what I have other string around the url? then it should catch the https as prefix, also how do I replace it?, I don't want to print out the url only, I want the entire text.
– Micheal
Nov 20 at 7:29

add a comment |

Maybe you're looking for something like this example:

var str = 'https://www.lazada.com.ph/products/fujitsu-standard-aa-rechargeable-battery-2000mah-4pcs-i237968048-s311807989.html?spm=a2o4k.searchlist.list.47.7f1a484d4gpCFd&search=1';



console.log(str.match('^.+.html')[0]);

I've used regex with the pattern ^.+.html to match the url. The regex means: From the beginning of the string to .html

If you want to split the url to 2 parts, you could try:

var str = 'https://www.lazada.com.ph/products/fujitsu-standard-aa-rechargeable-battery-2000mah-4pcs-i237968048-s311807989.html?spm=a2o4k.searchlist.list.47.7f1a484d4gpCFd&search=1';



var parts = str.split(/?/);



console.log(parts[0]);

console.log('?' + parts[1]);

edited Nov 20 at 7:36

answered Nov 20 at 7:19

Foo

There is a reason why there is something called as URL API I rather prefer to see users using this instead of that regex route because this answer is not useful if you have to include other extensions as well (eg .php or .asp).
– KarelG
Nov 20 at 7:26

what I have other string around the url? then it should catch the https as prefix, also how do I replace it?, I don't want to print out the url only, I want the entire text.
– Micheal
Nov 20 at 7:29

add a comment |

Maybe you're looking for something like this example:

var str = 'https://www.lazada.com.ph/products/fujitsu-standard-aa-rechargeable-battery-2000mah-4pcs-i237968048-s311807989.html?spm=a2o4k.searchlist.list.47.7f1a484d4gpCFd&search=1';



console.log(str.match('^.+.html')[0]);

I've used regex with the pattern ^.+.html to match the url. The regex means: From the beginning of the string to .html

If you want to split the url to 2 parts, you could try:

var str = 'https://www.lazada.com.ph/products/fujitsu-standard-aa-rechargeable-battery-2000mah-4pcs-i237968048-s311807989.html?spm=a2o4k.searchlist.list.47.7f1a484d4gpCFd&search=1';



var parts = str.split(/?/);



console.log(parts[0]);

console.log('?' + parts[1]);

edited Nov 20 at 7:36

answered Nov 20 at 7:19

Foo

Maybe you're looking for something like this example:

var str = 'https://www.lazada.com.ph/products/fujitsu-standard-aa-rechargeable-battery-2000mah-4pcs-i237968048-s311807989.html?spm=a2o4k.searchlist.list.47.7f1a484d4gpCFd&search=1';



console.log(str.match('^.+.html')[0]);

I've used regex with the pattern ^.+.html to match the url. The regex means: From the beginning of the string to .html

If you want to split the url to 2 parts, you could try:

var str = 'https://www.lazada.com.ph/products/fujitsu-standard-aa-rechargeable-battery-2000mah-4pcs-i237968048-s311807989.html?spm=a2o4k.searchlist.list.47.7f1a484d4gpCFd&search=1';



var parts = str.split(/?/);



console.log(parts[0]);

console.log('?' + parts[1]);

var str = 'https://www.lazada.com.ph/products/fujitsu-standard-aa-rechargeable-battery-2000mah-4pcs-i237968048-s311807989.html?spm=a2o4k.searchlist.list.47.7f1a484d4gpCFd&search=1';



console.log(str.match('^.+.html')[0]);

var str = 'https://www.lazada.com.ph/products/fujitsu-standard-aa-rechargeable-battery-2000mah-4pcs-i237968048-s311807989.html?spm=a2o4k.searchlist.list.47.7f1a484d4gpCFd&search=1';



console.log(str.match('^.+.html')[0]);

var str = 'https://www.lazada.com.ph/products/fujitsu-standard-aa-rechargeable-battery-2000mah-4pcs-i237968048-s311807989.html?spm=a2o4k.searchlist.list.47.7f1a484d4gpCFd&search=1';



var parts = str.split(/?/);



console.log(parts[0]);

console.log('?' + parts[1]);

var str = 'https://www.lazada.com.ph/products/fujitsu-standard-aa-rechargeable-battery-2000mah-4pcs-i237968048-s311807989.html?spm=a2o4k.searchlist.list.47.7f1a484d4gpCFd&search=1';



var parts = str.split(/?/);



console.log(parts[0]);

console.log('?' + parts[1]);

edited Nov 20 at 7:36

answered Nov 20 at 7:19

Foo

edited Nov 20 at 7:36

answered Nov 20 at 7:19

Foo

answered Nov 20 at 7:19

Foo

answered Nov 20 at 7:19

Foo

There is a reason why there is something called as URL API I rather prefer to see users using this instead of that regex route because this answer is not useful if you have to include other extensions as well (eg .php or .asp).
– KarelG
Nov 20 at 7:26

what I have other string around the url? then it should catch the https as prefix, also how do I replace it?, I don't want to print out the url only, I want the entire text.
– Micheal
Nov 20 at 7:29

add a comment |

There is a reason why there is something called as URL API I rather prefer to see users using this instead of that regex route because this answer is not useful if you have to include other extensions as well (eg .php or .asp).
– KarelG
Nov 20 at 7:26

what I have other string around the url? then it should catch the https as prefix, also how do I replace it?, I don't want to print out the url only, I want the entire text.
– Micheal
Nov 20 at 7:29

There is a reason why there is something called as URL API I rather prefer to see users using this instead of that regex route because this answer is not useful if you have to include other extensions as well (eg .php or .asp).
– KarelG
Nov 20 at 7:26

what I have other string around the url? then it should catch the https as prefix, also how do I replace it?, I don't want to print out the url only, I want the entire text.
– Micheal
Nov 20 at 7:29

add a comment |

You can use String.prototype.split()

var url =  "https://www.lazada.com.ph/products/fujitsu-standard-aa-rechargeable-battery-2000mah-4pcs-i237968048-s311807989.html?spm=a2o4k.searchlist.list.47.7f1a484d4gpCFd&search=1";





console.log(url.split('?', 1)[0])

answered Nov 20 at 7:26

Gildas.Tambo

16.3k33663

add a comment |

You can use String.prototype.split()

var url =  "https://www.lazada.com.ph/products/fujitsu-standard-aa-rechargeable-battery-2000mah-4pcs-i237968048-s311807989.html?spm=a2o4k.searchlist.list.47.7f1a484d4gpCFd&search=1";





console.log(url.split('?', 1)[0])

answered Nov 20 at 7:26

Gildas.Tambo

16.3k33663

add a comment |

You can use String.prototype.split()

var url =  "https://www.lazada.com.ph/products/fujitsu-standard-aa-rechargeable-battery-2000mah-4pcs-i237968048-s311807989.html?spm=a2o4k.searchlist.list.47.7f1a484d4gpCFd&search=1";





console.log(url.split('?', 1)[0])

answered Nov 20 at 7:26

Gildas.Tambo

16.3k33663

You can use String.prototype.split()

var url =  "https://www.lazada.com.ph/products/fujitsu-standard-aa-rechargeable-battery-2000mah-4pcs-i237968048-s311807989.html?spm=a2o4k.searchlist.list.47.7f1a484d4gpCFd&search=1";





console.log(url.split('?', 1)[0])

var url =  "https://www.lazada.com.ph/products/fujitsu-standard-aa-rechargeable-battery-2000mah-4pcs-i237968048-s311807989.html?spm=a2o4k.searchlist.list.47.7f1a484d4gpCFd&search=1";





console.log(url.split('?', 1)[0])

var url =  "https://www.lazada.com.ph/products/fujitsu-standard-aa-rechargeable-battery-2000mah-4pcs-i237968048-s311807989.html?spm=a2o4k.searchlist.list.47.7f1a484d4gpCFd&search=1";





console.log(url.split('?', 1)[0])

answered Nov 20 at 7:26

Gildas.Tambo

16.3k33663

answered Nov 20 at 7:26

Gildas.Tambo

16.3k33663

answered Nov 20 at 7:26

Gildas.Tambo

16.3k33663

answered Nov 20 at 7:26

Gildas.Tambo

16.3k33663

add a comment |

You can use the URL API:

var url = new URL(string);

url.search = "";

url.hash = "";

return url.href;

answered Nov 20 at 7:24

Bergi

363k58540866

nice but what is the next step? I want to replace the cleaned url back into its location.
– Micheal
Nov 20 at 7:30

@Micheal What do you mean by "its location"? You had a string for the url, right? Now you have the new url string.
– Bergi
Nov 20 at 7:32

sorry updated my question just now. It's not only the string, the urls are within a document string.
– Micheal
Nov 20 at 7:38

add a comment |

You can use the URL API:

var url = new URL(string);

url.search = "";

url.hash = "";

return url.href;

answered Nov 20 at 7:24

Bergi

363k58540866

nice but what is the next step? I want to replace the cleaned url back into its location.
– Micheal
Nov 20 at 7:30

@Micheal What do you mean by "its location"? You had a string for the url, right? Now you have the new url string.
– Bergi
Nov 20 at 7:32

sorry updated my question just now. It's not only the string, the urls are within a document string.
– Micheal
Nov 20 at 7:38

add a comment |

You can use the URL API:

var url = new URL(string);

url.search = "";

url.hash = "";

return url.href;

answered Nov 20 at 7:24

Bergi

363k58540866

You can use the URL API:

var url = new URL(string);

url.search = "";

url.hash = "";

return url.href;

answered Nov 20 at 7:24

Bergi

363k58540866

answered Nov 20 at 7:24

Bergi

363k58540866

answered Nov 20 at 7:24

Bergi

363k58540866

answered Nov 20 at 7:24

Bergi

363k58540866

nice but what is the next step? I want to replace the cleaned url back into its location.
– Micheal
Nov 20 at 7:30

@Micheal What do you mean by "its location"? You had a string for the url, right? Now you have the new url string.
– Bergi
Nov 20 at 7:32

sorry updated my question just now. It's not only the string, the urls are within a document string.
– Micheal
Nov 20 at 7:38

add a comment |

nice but what is the next step? I want to replace the cleaned url back into its location.
– Micheal
Nov 20 at 7:30

@Micheal What do you mean by "its location"? You had a string for the url, right? Now you have the new url string.
– Bergi
Nov 20 at 7:32

sorry updated my question just now. It's not only the string, the urls are within a document string.
– Micheal
Nov 20 at 7:38

nice but what is the next step? I want to replace the cleaned url back into its location.
– Micheal
Nov 20 at 7:30

@Micheal What do you mean by "its location"? You had a string for the url, right? Now you have the new url string.
– Bergi
Nov 20 at 7:32

sorry updated my question just now. It's not only the string, the urls are within a document string.
– Micheal
Nov 20 at 7:38

add a comment |

This works fine according to your requirment

let url = "https://www.lazada.com.ph/products/fujitsu-standard-aa-rechargeable- 

         battery-2000mah-4pcs-i237968048-s311807989.html?spm=a2o4k.searchlist.list.47.7f1a484d4gpCFd&search=1"





url.substring(0, <nbsp> url.indexOf('.html')<nbsp> + <nbsp>5)

edited Nov 20 at 7:37

chŝdk

23.5k52852

answered Nov 20 at 7:29

this is yash

14310

add a comment |

This works fine according to your requirment

let url = "https://www.lazada.com.ph/products/fujitsu-standard-aa-rechargeable- 

         battery-2000mah-4pcs-i237968048-s311807989.html?spm=a2o4k.searchlist.list.47.7f1a484d4gpCFd&search=1"





url.substring(0, <nbsp> url.indexOf('.html')<nbsp> + <nbsp>5)

edited Nov 20 at 7:37

chŝdk

23.5k52852

answered Nov 20 at 7:29

this is yash

14310

add a comment |

This works fine according to your requirment

let url = "https://www.lazada.com.ph/products/fujitsu-standard-aa-rechargeable- 

         battery-2000mah-4pcs-i237968048-s311807989.html?spm=a2o4k.searchlist.list.47.7f1a484d4gpCFd&search=1"





url.substring(0, <nbsp> url.indexOf('.html')<nbsp> + <nbsp>5)

edited Nov 20 at 7:37

chŝdk

23.5k52852

answered Nov 20 at 7:29

this is yash

14310

This works fine according to your requirment

let url = "https://www.lazada.com.ph/products/fujitsu-standard-aa-rechargeable- 

         battery-2000mah-4pcs-i237968048-s311807989.html?spm=a2o4k.searchlist.list.47.7f1a484d4gpCFd&search=1"





url.substring(0, <nbsp> url.indexOf('.html')<nbsp> + <nbsp>5)

edited Nov 20 at 7:37

chŝdk

23.5k52852

answered Nov 20 at 7:29

this is yash

14310

edited Nov 20 at 7:37

chŝdk

23.5k52852

edited Nov 20 at 7:37

chŝdk

23.5k52852

edited Nov 20 at 7:37

chŝdk

23.5k52852

answered Nov 20 at 7:29

this is yash

14310

answered Nov 20 at 7:29

this is yash

14310

answered Nov 20 at 7:29

this is yash

14310

add a comment |

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