Accessing a class name that is stored in the Jquery variable












1














There are three pictures on the site.



<img src="images/kep1.jpg" class="kep1">

<img src="images/kep2.jpg" class="kep2">

<img src="images/kep4.jpg" class="kep4">


Click function to magnify one page at a time in the middle of the page page, but I'll store the necessary data in a variable.



var pos;

    var t;

    var h;

    var pic;

$("img").click(function(){

            pos = $(this).css("position");

            pic = $(this).attr("class");

            t = $(this).css("top");

            h = $(this).css("height");

            $(".szurkeDiv").show();

            $(this).animate(

                {

                    "position" : "fixed",

                    "top" : "50%",

                    "left" : "50%",

                    "margin-top" : "-300px",

                    "margin-left" : "-400px",

                    "width" : "800px",

                    "height" : "600px"

                }, 1000, function(){}

            );

        });


Then I want you to click on the image in the middle of the page to animate it back to its original location. I would like to access the PIC variable but not succeed.



$(pic).animate(

                {

                    "position" : pos,

                    "width" : "150px",

                    "margin-top" : "0px",

                    "margin-left" : "0px",

                    "left" : "5px",

                    "height" : h,

                    "top" : t

                }, 1000, function(){}

            );


How can I do this well, refer to the PIC variable?



Thanks for the help!!






jquery html css







share|improve this question
























  • Possible duplicate of jquery select elements by class using name from variable
    – Heretic Monkey
    Nov 12 '18 at 17:11










  • jQuery is not a language, it's a library for JavaScript. It doesn't do any magic, it's written in JavaScript code. You should learn the basics of JavaScript before trying to learn how to use any third-party library. I'm not trying to sound judgmental here, just save yourself some trouble and sit through a proper lesson.
    – Jacque Goupil
    Nov 12 '18 at 17:13












  • I'm just learning this. This was the home lesson! :-)
    – PengeDroid
    Nov 12 '18 at 17:40
















1














There are three pictures on the site.



<img src="images/kep1.jpg" class="kep1">

<img src="images/kep2.jpg" class="kep2">

<img src="images/kep4.jpg" class="kep4">


Click function to magnify one page at a time in the middle of the page page, but I'll store the necessary data in a variable.



var pos;

    var t;

    var h;

    var pic;

$("img").click(function(){

            pos = $(this).css("position");

            pic = $(this).attr("class");

            t = $(this).css("top");

            h = $(this).css("height");

            $(".szurkeDiv").show();

            $(this).animate(

                {

                    "position" : "fixed",

                    "top" : "50%",

                    "left" : "50%",

                    "margin-top" : "-300px",

                    "margin-left" : "-400px",

                    "width" : "800px",

                    "height" : "600px"

                }, 1000, function(){}

            );

        });


Then I want you to click on the image in the middle of the page to animate it back to its original location. I would like to access the PIC variable but not succeed.



$(pic).animate(

                {

                    "position" : pos,

                    "width" : "150px",

                    "margin-top" : "0px",

                    "margin-left" : "0px",

                    "left" : "5px",

                    "height" : h,

                    "top" : t

                }, 1000, function(){}

            );


How can I do this well, refer to the PIC variable?



Thanks for the help!!






jquery html css







share|improve this question
























  • Possible duplicate of jquery select elements by class using name from variable
    – Heretic Monkey
    Nov 12 '18 at 17:11










  • jQuery is not a language, it's a library for JavaScript. It doesn't do any magic, it's written in JavaScript code. You should learn the basics of JavaScript before trying to learn how to use any third-party library. I'm not trying to sound judgmental here, just save yourself some trouble and sit through a proper lesson.
    – Jacque Goupil
    Nov 12 '18 at 17:13












  • I'm just learning this. This was the home lesson! :-)
    – PengeDroid
    Nov 12 '18 at 17:40














1












1








1







There are three pictures on the site.



<img src="images/kep1.jpg" class="kep1">

<img src="images/kep2.jpg" class="kep2">

<img src="images/kep4.jpg" class="kep4">


Click function to magnify one page at a time in the middle of the page page, but I'll store the necessary data in a variable.



var pos;

    var t;

    var h;

    var pic;

$("img").click(function(){

            pos = $(this).css("position");

            pic = $(this).attr("class");

            t = $(this).css("top");

            h = $(this).css("height");

            $(".szurkeDiv").show();

            $(this).animate(

                {

                    "position" : "fixed",

                    "top" : "50%",

                    "left" : "50%",

                    "margin-top" : "-300px",

                    "margin-left" : "-400px",

                    "width" : "800px",

                    "height" : "600px"

                }, 1000, function(){}

            );

        });


Then I want you to click on the image in the middle of the page to animate it back to its original location. I would like to access the PIC variable but not succeed.



$(pic).animate(

                {

                    "position" : pos,

                    "width" : "150px",

                    "margin-top" : "0px",

                    "margin-left" : "0px",

                    "left" : "5px",

                    "height" : h,

                    "top" : t

                }, 1000, function(){}

            );


How can I do this well, refer to the PIC variable?



Thanks for the help!!






jquery html css







share|improve this question















There are three pictures on the site.



<img src="images/kep1.jpg" class="kep1">

<img src="images/kep2.jpg" class="kep2">

<img src="images/kep4.jpg" class="kep4">


Click function to magnify one page at a time in the middle of the page page, but I'll store the necessary data in a variable.



var pos;

    var t;

    var h;

    var pic;

$("img").click(function(){

            pos = $(this).css("position");

            pic = $(this).attr("class");

            t = $(this).css("top");

            h = $(this).css("height");

            $(".szurkeDiv").show();

            $(this).animate(

                {

                    "position" : "fixed",

                    "top" : "50%",

                    "left" : "50%",

                    "margin-top" : "-300px",

                    "margin-left" : "-400px",

                    "width" : "800px",

                    "height" : "600px"

                }, 1000, function(){}

            );

        });


Then I want you to click on the image in the middle of the page to animate it back to its original location. I would like to access the PIC variable but not succeed.



$(pic).animate(

                {

                    "position" : pos,

                    "width" : "150px",

                    "margin-top" : "0px",

                    "margin-left" : "0px",

                    "left" : "5px",

                    "height" : h,

                    "top" : t

                }, 1000, function(){}

            );


How can I do this well, refer to the PIC variable?



Thanks for the help!!






jquery html css




jquery html css






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 12 '18 at 16:58

























asked Nov 12 '18 at 16:52









PengeDroid

135




135












  • Possible duplicate of jquery select elements by class using name from variable
    – Heretic Monkey
    Nov 12 '18 at 17:11










  • jQuery is not a language, it's a library for JavaScript. It doesn't do any magic, it's written in JavaScript code. You should learn the basics of JavaScript before trying to learn how to use any third-party library. I'm not trying to sound judgmental here, just save yourself some trouble and sit through a proper lesson.
    – Jacque Goupil
    Nov 12 '18 at 17:13












  • I'm just learning this. This was the home lesson! :-)
    – PengeDroid
    Nov 12 '18 at 17:40


















  • Possible duplicate of jquery select elements by class using name from variable
    – Heretic Monkey
    Nov 12 '18 at 17:11










  • jQuery is not a language, it's a library for JavaScript. It doesn't do any magic, it's written in JavaScript code. You should learn the basics of JavaScript before trying to learn how to use any third-party library. I'm not trying to sound judgmental here, just save yourself some trouble and sit through a proper lesson.
    – Jacque Goupil
    Nov 12 '18 at 17:13












  • I'm just learning this. This was the home lesson! :-)
    – PengeDroid
    Nov 12 '18 at 17:40
















Possible duplicate of jquery select elements by class using name from variable
– Heretic Monkey
Nov 12 '18 at 17:11




Possible duplicate of jquery select elements by class using name from variable
– Heretic Monkey
Nov 12 '18 at 17:11












jQuery is not a language, it's a library for JavaScript. It doesn't do any magic, it's written in JavaScript code. You should learn the basics of JavaScript before trying to learn how to use any third-party library. I'm not trying to sound judgmental here, just save yourself some trouble and sit through a proper lesson.
– Jacque Goupil
Nov 12 '18 at 17:13






jQuery is not a language, it's a library for JavaScript. It doesn't do any magic, it's written in JavaScript code. You should learn the basics of JavaScript before trying to learn how to use any third-party library. I'm not trying to sound judgmental here, just save yourself some trouble and sit through a proper lesson.
– Jacque Goupil
Nov 12 '18 at 17:13














I'm just learning this. This was the home lesson! :-)
– PengeDroid
Nov 12 '18 at 17:40




I'm just learning this. This was the home lesson! :-)
– PengeDroid
Nov 12 '18 at 17:40












2 Answers
2






active

oldest

votes


















2














try it with $('.' + pic) instead of $(pic).



See https://jsfiddle.net/g15v3c4z/



var pic = $('p').attr("class");



$('.' + pic).html('world!');





share|improve this answer





















  • Thanks!!! :-)))
    – PengeDroid
    Nov 12 '18 at 17:44



















0














Use plain CSS for the animation: 






var img = $("img")



img.on("click", function(){

  $(this).toggleClass("zoom");

})

img {

    transition: all 0.5s ease;



}



.zoom {

  transform:scale(2);

  transform-origin: 0 0;

}

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

<div class="container">

  <img src="https://via.placeholder.com/350" alt="">

    <img src="https://via.placeholder.com/350" alt="">



  <img src="https://via.placeholder.com/350" alt="">



</div>








share|improve this answer





















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    2 Answers
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    2 Answers
    2






    active

    oldest

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    active

    oldest

    votes






    active

    oldest

    votes









    2














    try it with $('.' + pic) instead of $(pic).



    See https://jsfiddle.net/g15v3c4z/



    var pic = $('p').attr("class");
    

    
$('.' + pic).html('world!');





    share|improve this answer





















    • Thanks!!! :-)))
      – PengeDroid
      Nov 12 '18 at 17:44
















    2














    try it with $('.' + pic) instead of $(pic).



    See https://jsfiddle.net/g15v3c4z/



    var pic = $('p').attr("class");
    

    
$('.' + pic).html('world!');





    share|improve this answer





















    • Thanks!!! :-)))
      – PengeDroid
      Nov 12 '18 at 17:44














    2












    2








    2






    try it with $('.' + pic) instead of $(pic).



    See https://jsfiddle.net/g15v3c4z/



    var pic = $('p').attr("class");
    

    
$('.' + pic).html('world!');





    share|improve this answer












    try it with $('.' + pic) instead of $(pic).



    See https://jsfiddle.net/g15v3c4z/



    var pic = $('p').attr("class");
    

    
$('.' + pic).html('world!');






    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered Nov 12 '18 at 17:02









    Andreas

    443




    443












    • Thanks!!! :-)))
      – PengeDroid
      Nov 12 '18 at 17:44


















    • Thanks!!! :-)))
      – PengeDroid
      Nov 12 '18 at 17:44
















    Thanks!!! :-)))
    – PengeDroid
    Nov 12 '18 at 17:44




    Thanks!!! :-)))
    – PengeDroid
    Nov 12 '18 at 17:44













    0














    Use plain CSS for the animation: 






    var img = $("img")
    

    
img.on("click", function(){
    
  $(this).toggleClass("zoom");
    
})

    img {
    
    transition: all 0.5s ease;
    

    
}
    

    
.zoom {
    
  transform:scale(2);
    
  transform-origin: 0 0;
    
}

    <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
    
<div class="container">
    
  <img src="https://via.placeholder.com/350" alt="">
    
    <img src="https://via.placeholder.com/350" alt="">
    

    
  <img src="https://via.placeholder.com/350" alt="">
    

    
</div>








    share|improve this answer


























      0














      Use plain CSS for the animation: 






      var img = $("img")
      

      
img.on("click", function(){
      
  $(this).toggleClass("zoom");
      
})

      img {
      
    transition: all 0.5s ease;
      

      
}
      

      
.zoom {
      
  transform:scale(2);
      
  transform-origin: 0 0;
      
}

      <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
      
<div class="container">
      
  <img src="https://via.placeholder.com/350" alt="">
      
    <img src="https://via.placeholder.com/350" alt="">
      

      
  <img src="https://via.placeholder.com/350" alt="">
      

      
</div>








      share|improve this answer
























        0












        0








        0






        Use plain CSS for the animation: 






        var img = $("img")
        

        
img.on("click", function(){
        
  $(this).toggleClass("zoom");
        
})

        img {
        
    transition: all 0.5s ease;
        

        
}
        

        
.zoom {
        
  transform:scale(2);
        
  transform-origin: 0 0;
        
}

        <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
        
<div class="container">
        
  <img src="https://via.placeholder.com/350" alt="">
        
    <img src="https://via.placeholder.com/350" alt="">
        

        
  <img src="https://via.placeholder.com/350" alt="">
        

        
</div>








        share|improve this answer












        Use plain CSS for the animation: 






        var img = $("img")
        

        
img.on("click", function(){
        
  $(this).toggleClass("zoom");
        
})

        img {
        
    transition: all 0.5s ease;
        

        
}
        

        
.zoom {
        
  transform:scale(2);
        
  transform-origin: 0 0;
        
}

        <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
        
<div class="container">
        
  <img src="https://via.placeholder.com/350" alt="">
        
    <img src="https://via.placeholder.com/350" alt="">
        

        
  <img src="https://via.placeholder.com/350" alt="">
        

        
</div>








        var img = $("img")
        

        
img.on("click", function(){
        
  $(this).toggleClass("zoom");
        
})

        img {
        
    transition: all 0.5s ease;
        

        
}
        

        
.zoom {
        
  transform:scale(2);
        
  transform-origin: 0 0;
        
}

        <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
        
<div class="container">
        
  <img src="https://via.placeholder.com/350" alt="">
        
    <img src="https://via.placeholder.com/350" alt="">
        

        
  <img src="https://via.placeholder.com/350" alt="">
        

        
</div>





        var img = $("img")
        

        
img.on("click", function(){
        
  $(this).toggleClass("zoom");
        
})

        img {
        
    transition: all 0.5s ease;
        

        
}
        

        
.zoom {
        
  transform:scale(2);
        
  transform-origin: 0 0;
        
}

        <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
        
<div class="container">
        
  <img src="https://via.placeholder.com/350" alt="">
        
    <img src="https://via.placeholder.com/350" alt="">
        

        
  <img src="https://via.placeholder.com/350" alt="">
        

        
</div>






        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Nov 12 '18 at 17:24









        Ihsan S

        324




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