how to extract only specific value from google search link? [duplicate]
This question already has an answer here:
Get query string parameters with jQuery
9 answers
I want to extract "start" value from a google search link. How can i do this in jquery. This below link having value of start=20 and i only want 20 from this string.
https://www.google.com/search?q=python+code&ei=yGQsXL7yIMnQwQLis4CoAQ&start=20&sa=N&ved=0ahUKEwi-9tOmxs7fAhVJaFAKHeIZABU4ChDy0wMIew&biw=1242&bih=569
jquery
marked as duplicate by Tim Biegeleisen, Community♦ Jan 2 at 7:24
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
This question already has an answer here:
Get query string parameters with jQuery
9 answers
I want to extract "start" value from a google search link. How can i do this in jquery. This below link having value of start=20 and i only want 20 from this string.
https://www.google.com/search?q=python+code&ei=yGQsXL7yIMnQwQLis4CoAQ&start=20&sa=N&ved=0ahUKEwi-9tOmxs7fAhVJaFAKHeIZABU4ChDy0wMIew&biw=1242&bih=569
jquery
marked as duplicate by Tim Biegeleisen, Community♦ Jan 2 at 7:24
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
Duplicate of stackoverflow.com/q/19491336/5052778
– Shehroz Ahmed
Jan 2 at 7:23
add a comment |
This question already has an answer here:
Get query string parameters with jQuery
9 answers
I want to extract "start" value from a google search link. How can i do this in jquery. This below link having value of start=20 and i only want 20 from this string.
https://www.google.com/search?q=python+code&ei=yGQsXL7yIMnQwQLis4CoAQ&start=20&sa=N&ved=0ahUKEwi-9tOmxs7fAhVJaFAKHeIZABU4ChDy0wMIew&biw=1242&bih=569
jquery
This question already has an answer here:
Get query string parameters with jQuery
9 answers
I want to extract "start" value from a google search link. How can i do this in jquery. This below link having value of start=20 and i only want 20 from this string.
https://www.google.com/search?q=python+code&ei=yGQsXL7yIMnQwQLis4CoAQ&start=20&sa=N&ved=0ahUKEwi-9tOmxs7fAhVJaFAKHeIZABU4ChDy0wMIew&biw=1242&bih=569
This question already has an answer here:
Get query string parameters with jQuery
9 answers
jquery
jquery
asked Jan 2 at 7:18
Amir shahAmir shah
1066
1066
marked as duplicate by Tim Biegeleisen, Community♦ Jan 2 at 7:24
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Tim Biegeleisen, Community♦ Jan 2 at 7:24
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
Duplicate of stackoverflow.com/q/19491336/5052778
– Shehroz Ahmed
Jan 2 at 7:23
add a comment |
1
Duplicate of stackoverflow.com/q/19491336/5052778
– Shehroz Ahmed
Jan 2 at 7:23
1
1
Duplicate of stackoverflow.com/q/19491336/5052778
– Shehroz Ahmed
Jan 2 at 7:23
Duplicate of stackoverflow.com/q/19491336/5052778
– Shehroz Ahmed
Jan 2 at 7:23
add a comment |
1 Answer
1
active
oldest
votes
var url = 'https://www.google.com/search?q=python+code&ei=yGQsXL7yIMnQwQLis4CoAQ&start=20&sa=N&ved=0ahUKEwi-9tOmxs7fAhVJaFAKHeIZABU4ChDy0wMIew&biw=1242&bih=569';
function getParameterByName(name, url) {
if (!url) url = window.location.href;
name = name.replace(/[[]]/g, '\$&');
var regex = new RegExp('[?&]' + name + '(=([^&#]*)|&|#|$)'),
results = regex.exec(url);
if (!results) return null;
if (!results[2]) return '';
return decodeURIComponent(results[2].replace(/+/g, ' '));
}
var start = getParameterByName('start', url);
console.log(start);
up1 for the quick and easy answer. thanks
– Amir shah
Jan 2 at 7:35
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
var url = 'https://www.google.com/search?q=python+code&ei=yGQsXL7yIMnQwQLis4CoAQ&start=20&sa=N&ved=0ahUKEwi-9tOmxs7fAhVJaFAKHeIZABU4ChDy0wMIew&biw=1242&bih=569';
function getParameterByName(name, url) {
if (!url) url = window.location.href;
name = name.replace(/[[]]/g, '\$&');
var regex = new RegExp('[?&]' + name + '(=([^&#]*)|&|#|$)'),
results = regex.exec(url);
if (!results) return null;
if (!results[2]) return '';
return decodeURIComponent(results[2].replace(/+/g, ' '));
}
var start = getParameterByName('start', url);
console.log(start);
up1 for the quick and easy answer. thanks
– Amir shah
Jan 2 at 7:35
add a comment |
var url = 'https://www.google.com/search?q=python+code&ei=yGQsXL7yIMnQwQLis4CoAQ&start=20&sa=N&ved=0ahUKEwi-9tOmxs7fAhVJaFAKHeIZABU4ChDy0wMIew&biw=1242&bih=569';
function getParameterByName(name, url) {
if (!url) url = window.location.href;
name = name.replace(/[[]]/g, '\$&');
var regex = new RegExp('[?&]' + name + '(=([^&#]*)|&|#|$)'),
results = regex.exec(url);
if (!results) return null;
if (!results[2]) return '';
return decodeURIComponent(results[2].replace(/+/g, ' '));
}
var start = getParameterByName('start', url);
console.log(start);
up1 for the quick and easy answer. thanks
– Amir shah
Jan 2 at 7:35
add a comment |
var url = 'https://www.google.com/search?q=python+code&ei=yGQsXL7yIMnQwQLis4CoAQ&start=20&sa=N&ved=0ahUKEwi-9tOmxs7fAhVJaFAKHeIZABU4ChDy0wMIew&biw=1242&bih=569';
function getParameterByName(name, url) {
if (!url) url = window.location.href;
name = name.replace(/[[]]/g, '\$&');
var regex = new RegExp('[?&]' + name + '(=([^&#]*)|&|#|$)'),
results = regex.exec(url);
if (!results) return null;
if (!results[2]) return '';
return decodeURIComponent(results[2].replace(/+/g, ' '));
}
var start = getParameterByName('start', url);
console.log(start);
var url = 'https://www.google.com/search?q=python+code&ei=yGQsXL7yIMnQwQLis4CoAQ&start=20&sa=N&ved=0ahUKEwi-9tOmxs7fAhVJaFAKHeIZABU4ChDy0wMIew&biw=1242&bih=569';
function getParameterByName(name, url) {
if (!url) url = window.location.href;
name = name.replace(/[[]]/g, '\$&');
var regex = new RegExp('[?&]' + name + '(=([^&#]*)|&|#|$)'),
results = regex.exec(url);
if (!results) return null;
if (!results[2]) return '';
return decodeURIComponent(results[2].replace(/+/g, ' '));
}
var start = getParameterByName('start', url);
console.log(start);
answered Jan 2 at 7:27
AnjiAnji
17912
17912
up1 for the quick and easy answer. thanks
– Amir shah
Jan 2 at 7:35
add a comment |
up1 for the quick and easy answer. thanks
– Amir shah
Jan 2 at 7:35
up1 for the quick and easy answer. thanks
– Amir shah
Jan 2 at 7:35
up1 for the quick and easy answer. thanks
– Amir shah
Jan 2 at 7:35
add a comment |
1
Duplicate of stackoverflow.com/q/19491336/5052778
– Shehroz Ahmed
Jan 2 at 7:23