how to embed the php image path in javascript












-1















var arrow2 ="<?php echo base_url() ?>/assets/images/apply_coupon/arrow-2.png" ;

var arrow1 ="<?php echo base_url() ?>/assets/images/apply_coupon/arrow-1.png" ;

img_array= new Array($arrow2,$arrow1);


I have created one img_array, and I have passed the php variable into it is it the correct format or not?






javascript php arrays







share|improve this question




















  • 1





    var arrow2 and arrow1 both are JS variable. Why did u make it as php variable? Change it to => img_array= new Array( arrow2, arrow1);

    – Aravind Bhat K
    Nov 23 '18 at 6:53


















-1















var arrow2 ="<?php echo base_url() ?>/assets/images/apply_coupon/arrow-2.png" ;

var arrow1 ="<?php echo base_url() ?>/assets/images/apply_coupon/arrow-1.png" ;

img_array= new Array($arrow2,$arrow1);


I have created one img_array, and I have passed the php variable into it is it the correct format or not?






javascript php arrays







share|improve this question




















  • 1





    var arrow2 and arrow1 both are JS variable. Why did u make it as php variable? Change it to => img_array= new Array( arrow2, arrow1);

    – Aravind Bhat K
    Nov 23 '18 at 6:53
















-1












-1








-1








var arrow2 ="<?php echo base_url() ?>/assets/images/apply_coupon/arrow-2.png" ;

var arrow1 ="<?php echo base_url() ?>/assets/images/apply_coupon/arrow-1.png" ;

img_array= new Array($arrow2,$arrow1);


I have created one img_array, and I have passed the php variable into it is it the correct format or not?






javascript php arrays







share|improve this question
















var arrow2 ="<?php echo base_url() ?>/assets/images/apply_coupon/arrow-2.png" ;

var arrow1 ="<?php echo base_url() ?>/assets/images/apply_coupon/arrow-1.png" ;

img_array= new Array($arrow2,$arrow1);


I have created one img_array, and I have passed the php variable into it is it the correct format or not?






javascript php arrays




javascript php arrays






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 23 '18 at 6:54









Nick

37.2k132443




37.2k132443










asked Nov 23 '18 at 6:48









Kavya SenthilKavya Senthil

14




14








  • 1





    var arrow2 and arrow1 both are JS variable. Why did u make it as php variable? Change it to => img_array= new Array( arrow2, arrow1);

    – Aravind Bhat K
    Nov 23 '18 at 6:53
















  • 1





    var arrow2 and arrow1 both are JS variable. Why did u make it as php variable? Change it to => img_array= new Array( arrow2, arrow1);

    – Aravind Bhat K
    Nov 23 '18 at 6:53










1




1





var arrow2 and arrow1 both are JS variable. Why did u make it as php variable? Change it to => img_array= new Array( arrow2, arrow1);

– Aravind Bhat K
Nov 23 '18 at 6:53







var arrow2 and arrow1 both are JS variable. Why did u make it as php variable? Change it to => img_array= new Array( arrow2, arrow1);

– Aravind Bhat K
Nov 23 '18 at 6:53














1 Answer
1






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oldest

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change variable names used for making new array



`var img_array= new Array(arrow2, arrow1);



or



var img_array= [arrow2, arrow1];



`



or replace the var arrow1 to $arrow1 if you want to use dollar($) sign






share|improve this answer























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    change variable names used for making new array



    `var img_array= new Array(arrow2, arrow1);



    or



    var img_array= [arrow2, arrow1];



    `



    or replace the var arrow1 to $arrow1 if you want to use dollar($) sign






    share|improve this answer




























      0














      change variable names used for making new array



      `var img_array= new Array(arrow2, arrow1);



      or



      var img_array= [arrow2, arrow1];



      `



      or replace the var arrow1 to $arrow1 if you want to use dollar($) sign






      share|improve this answer


























        0












        0








        0







        change variable names used for making new array



        `var img_array= new Array(arrow2, arrow1);



        or



        var img_array= [arrow2, arrow1];



        `



        or replace the var arrow1 to $arrow1 if you want to use dollar($) sign






        share|improve this answer













        change variable names used for making new array



        `var img_array= new Array(arrow2, arrow1);



        or



        var img_array= [arrow2, arrow1];



        `



        or replace the var arrow1 to $arrow1 if you want to use dollar($) sign







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Nov 23 '18 at 7:09









        AkhileshAkhilesh

        626




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