how to extract only specific value from google search link? [duplicate]












0
















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







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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
















0
















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







share|improve this 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














0












0








0









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







share|improve this question















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






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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














  • 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












1 Answer
1






active

oldest

votes


















1














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);





share|improve this answer
























  • up1 for the quick and easy answer. thanks

    – Amir shah
    Jan 2 at 7:35


















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1














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);





share|improve this answer
























  • up1 for the quick and easy answer. thanks

    – Amir shah
    Jan 2 at 7:35
















1














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);





share|improve this answer
























  • up1 for the quick and easy answer. thanks

    – Amir shah
    Jan 2 at 7:35














1












1








1







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);





share|improve this answer













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);






share|improve this answer












share|improve this answer



share|improve this answer










answered Jan 2 at 7:27









AnjiAnji

17912




17912













  • 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





up1 for the quick and easy answer. thanks

– Amir shah
Jan 2 at 7:35





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