shown image only after mouse hover 2 times?
0
It's very simple code
<img src="images/Card.png" id=I1 class="img"/>
<img src="images/Card.png" id=I2 class="img"/>
<img src="images/Card.png" id=I3 class="img"/>
<img src="images/Card.png" id=I4 class="img"/>
<img src="images/Capture1.JPG" id="myImage1" style="display: none ; " />
<img src="images/Capture2.JPG" id="myImage2" style="display: none ; " />
<img src="images/Capture3.JPG" id="myImage3" style="display: none ; " />
<img src="images/Capture4.JPG" id="myImage4" style="display: none ; " />
<script>
$('.img').mouseover(function() {
var GetId = this.id.substring(1);
$("#I" + GetId).hover(
function () { $("#myImage" + GetId ).show(); },
function () { $("#myImage" + GetId ).hide(); } );
});
</script>
It's working fine, but it have one problem you need put mouse over 2 times to display image.
Google the problem but no solution.
Can any one help?
Thank you
jquery html onmouseover
asked Nov 21 '18 at 14:45
Green_Crocodile
156
add a comment |
0
It's very simple code
<img src="images/Card.png" id=I1 class="img"/>
<img src="images/Card.png" id=I2 class="img"/>
<img src="images/Card.png" id=I3 class="img"/>
<img src="images/Card.png" id=I4 class="img"/>
<img src="images/Capture1.JPG" id="myImage1" style="display: none ; " />
<img src="images/Capture2.JPG" id="myImage2" style="display: none ; " />
<img src="images/Capture3.JPG" id="myImage3" style="display: none ; " />
<img src="images/Capture4.JPG" id="myImage4" style="display: none ; " />
<script>
$('.img').mouseover(function() {
var GetId = this.id.substring(1);
$("#I" + GetId).hover(
function () { $("#myImage" + GetId ).show(); },
function () { $("#myImage" + GetId ).hide(); } );
});
</script>
It's working fine, but it have one problem you need put mouse over 2 times to display image.
Google the problem but no solution.
Can any one help?
Thank you
jquery html onmouseover
asked Nov 21 '18 at 14:45
Green_Crocodile
156
$().hover()is an binding event listeners, just like your mouseover is binding an event listener. So of course it takes two times. Also, this logic is going to bind the hover event listener every time the image is mouseover, which is a bad idea.
– Taplar
Nov 21 '18 at 14:46
add a comment |
0
0
0
It's very simple code
<img src="images/Card.png" id=I1 class="img"/>
<img src="images/Card.png" id=I2 class="img"/>
<img src="images/Card.png" id=I3 class="img"/>
<img src="images/Card.png" id=I4 class="img"/>
<img src="images/Capture1.JPG" id="myImage1" style="display: none ; " />
<img src="images/Capture2.JPG" id="myImage2" style="display: none ; " />
<img src="images/Capture3.JPG" id="myImage3" style="display: none ; " />
<img src="images/Capture4.JPG" id="myImage4" style="display: none ; " />
<script>
$('.img').mouseover(function() {
var GetId = this.id.substring(1);
$("#I" + GetId).hover(
function () { $("#myImage" + GetId ).show(); },
function () { $("#myImage" + GetId ).hide(); } );
});
</script>
It's working fine, but it have one problem you need put mouse over 2 times to display image.
Google the problem but no solution.
Can any one help?
Thank you
jquery html onmouseover
asked Nov 21 '18 at 14:45
Green_Crocodile
156
It's very simple code
<img src="images/Card.png" id=I1 class="img"/>
<img src="images/Card.png" id=I2 class="img"/>
<img src="images/Card.png" id=I3 class="img"/>
<img src="images/Card.png" id=I4 class="img"/>
<img src="images/Capture1.JPG" id="myImage1" style="display: none ; " />
<img src="images/Capture2.JPG" id="myImage2" style="display: none ; " />
<img src="images/Capture3.JPG" id="myImage3" style="display: none ; " />
<img src="images/Capture4.JPG" id="myImage4" style="display: none ; " />
<script>
$('.img').mouseover(function() {
var GetId = this.id.substring(1);
$("#I" + GetId).hover(
function () { $("#myImage" + GetId ).show(); },
function () { $("#myImage" + GetId ).hide(); } );
});
</script>
It's working fine, but it have one problem you need put mouse over 2 times to display image.
Google the problem but no solution.
Can any one help?
Thank you
jquery html onmouseover
jquery html onmouseover
asked Nov 21 '18 at 14:45
Green_Crocodile
156
asked Nov 21 '18 at 14:45
Green_Crocodile
156
asked Nov 21 '18 at 14:45
Green_Crocodile
156
asked Nov 21 '18 at 14:45
Green_Crocodile
156
asked Nov 21 '18 at 14:45
Green_Crocodile
156
156
$().hover()is an binding event listeners, just like your mouseover is binding an event listener. So of course it takes two times. Also, this logic is going to bind the hover event listener every time the image is mouseover, which is a bad idea.
– Taplar
Nov 21 '18 at 14:46
add a comment |
$().hover()is an binding event listeners, just like your mouseover is binding an event listener. So of course it takes two times. Also, this logic is going to bind the hover event listener every time the image is mouseover, which is a bad idea.
– Taplar
Nov 21 '18 at 14:46
$().hover() is an binding event listeners, just like your mouseover is binding an event listener. So of course it takes two times. Also, this logic is going to bind the hover event listener every time the image is mouseover, which is a bad idea.– Taplar
Nov 21 '18 at 14:46
$().hover() is an binding event listeners, just like your mouseover is binding an event listener. So of course it takes two times. Also, this logic is going to bind the hover event listener every time the image is mouseover, which is a bad idea.– Taplar
Nov 21 '18 at 14:46
add a comment |
2 Answers
2
active
oldest
votes
1
You can simplify your logic, and fix your issue in the process, by using a data field on the mouseover images.
$('.img')
.on('mouseenter', function(){
$('#'+ $(this).data('target')).show();
})
.on('mouseleave', function(){
$('#'+ $(this).data('target')).hide();
});<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<img src="images/Card.png" id=I1 class="img" data-target="myImage1" />
<img src="images/Card.png" id=I2 class="img" data-target="myImage2" />
<img src="images/Card.png" id=I3 class="img" data-target="myImage3" />
<img src="images/Card.png" id=I4 class="img" data-target="myImage4" />
<img src="images/Capture1.JPG" id="myImage1" style="display: none ; " alt="image1" />
<img src="images/Capture2.JPG" id="myImage2" style="display: none ; " alt="image2" />
<img src="images/Capture3.JPG" id="myImage3" style="display: none ; " alt="image3" />
<img src="images/Capture4.JPG" id="myImage4" style="display: none ; " alt="image4" />answered Nov 21 '18 at 14:52
Taplar
15.6k21529
Thank you Taplar. Your code working perfectly.
– Green_Crocodile
Nov 21 '18 at 15:37
add a comment |
0
IF you can group the images together should do this by using pure css. If not use Taplar's answer.
Example:
html:
<!-- repeat for all your images -->
<div class="image-wrapper">
<img class="img">
<img class="image-on-hover">
</div>
CSS:
.image-on-hover{
display:none
}
.img:hover ~.image-on-hover{
display: block
}
Explanation:
The :hover Selector on CSS will be triggered when an element is hovered. With the ~ selector you can select a child element on the same level (you could call it "neighbour-element"). Since we don't want to display all neighbouring hover-images, we wrap the pairs of image and hover-images together
Different option if your grouping is exactly like the example:
.image-on-hover{
display: none;
}
/*repeat for every image pair */
#I1:hover ~ #myImage1{
display:block
}
answered Nov 21 '18 at 14:49
Heady
306318
This only works if they are grouped together, no? The OP is not grouping what is being hovered with what is being shown on hover.
– Taplar
Nov 21 '18 at 14:53
Yes, i don't know the entire use case. It really depends on how these containers are actually grouped. It can also work if the structure is different though. Would have to see the actual html
– Heady
Nov 21 '18 at 14:56
e.g. if the grouping is exactly like the example you can do this:.hover-image{ display: none}; #I1: hover ~ #myImage1{ display: block}and so on
– Heady
Nov 21 '18 at 14:56
Unfortunatly I can't group the images together. But Taplar's code it what I am looking for. but thank you for another options.
– Green_Crocodile
Nov 21 '18 at 15:39
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
You can simplify your logic, and fix your issue in the process, by using a data field on the mouseover images.
$('.img')
.on('mouseenter', function(){
$('#'+ $(this).data('target')).show();
})
.on('mouseleave', function(){
$('#'+ $(this).data('target')).hide();
});<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<img src="images/Card.png" id=I1 class="img" data-target="myImage1" />
<img src="images/Card.png" id=I2 class="img" data-target="myImage2" />
<img src="images/Card.png" id=I3 class="img" data-target="myImage3" />
<img src="images/Card.png" id=I4 class="img" data-target="myImage4" />
<img src="images/Capture1.JPG" id="myImage1" style="display: none ; " alt="image1" />
<img src="images/Capture2.JPG" id="myImage2" style="display: none ; " alt="image2" />
<img src="images/Capture3.JPG" id="myImage3" style="display: none ; " alt="image3" />
<img src="images/Capture4.JPG" id="myImage4" style="display: none ; " alt="image4" />answered Nov 21 '18 at 14:52
Taplar
15.6k21529
Thank you Taplar. Your code working perfectly.
– Green_Crocodile
Nov 21 '18 at 15:37
add a comment |
1
You can simplify your logic, and fix your issue in the process, by using a data field on the mouseover images.
$('.img')
.on('mouseenter', function(){
$('#'+ $(this).data('target')).show();
})
.on('mouseleave', function(){
$('#'+ $(this).data('target')).hide();
});<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<img src="images/Card.png" id=I1 class="img" data-target="myImage1" />
<img src="images/Card.png" id=I2 class="img" data-target="myImage2" />
<img src="images/Card.png" id=I3 class="img" data-target="myImage3" />
<img src="images/Card.png" id=I4 class="img" data-target="myImage4" />
<img src="images/Capture1.JPG" id="myImage1" style="display: none ; " alt="image1" />
<img src="images/Capture2.JPG" id="myImage2" style="display: none ; " alt="image2" />
<img src="images/Capture3.JPG" id="myImage3" style="display: none ; " alt="image3" />
<img src="images/Capture4.JPG" id="myImage4" style="display: none ; " alt="image4" />answered Nov 21 '18 at 14:52
Taplar
15.6k21529
Thank you Taplar. Your code working perfectly.
– Green_Crocodile
Nov 21 '18 at 15:37
add a comment |
1
1
1
You can simplify your logic, and fix your issue in the process, by using a data field on the mouseover images.
$('.img')
.on('mouseenter', function(){
$('#'+ $(this).data('target')).show();
})
.on('mouseleave', function(){
$('#'+ $(this).data('target')).hide();
});<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<img src="images/Card.png" id=I1 class="img" data-target="myImage1" />
<img src="images/Card.png" id=I2 class="img" data-target="myImage2" />
<img src="images/Card.png" id=I3 class="img" data-target="myImage3" />
<img src="images/Card.png" id=I4 class="img" data-target="myImage4" />
<img src="images/Capture1.JPG" id="myImage1" style="display: none ; " alt="image1" />
<img src="images/Capture2.JPG" id="myImage2" style="display: none ; " alt="image2" />
<img src="images/Capture3.JPG" id="myImage3" style="display: none ; " alt="image3" />
<img src="images/Capture4.JPG" id="myImage4" style="display: none ; " alt="image4" />answered Nov 21 '18 at 14:52
Taplar
15.6k21529
You can simplify your logic, and fix your issue in the process, by using a data field on the mouseover images.
$('.img')
.on('mouseenter', function(){
$('#'+ $(this).data('target')).show();
})
.on('mouseleave', function(){
$('#'+ $(this).data('target')).hide();
});<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<img src="images/Card.png" id=I1 class="img" data-target="myImage1" />
<img src="images/Card.png" id=I2 class="img" data-target="myImage2" />
<img src="images/Card.png" id=I3 class="img" data-target="myImage3" />
<img src="images/Card.png" id=I4 class="img" data-target="myImage4" />
<img src="images/Capture1.JPG" id="myImage1" style="display: none ; " alt="image1" />
<img src="images/Capture2.JPG" id="myImage2" style="display: none ; " alt="image2" />
<img src="images/Capture3.JPG" id="myImage3" style="display: none ; " alt="image3" />
<img src="images/Capture4.JPG" id="myImage4" style="display: none ; " alt="image4" />$('.img')
.on('mouseenter', function(){
$('#'+ $(this).data('target')).show();
})
.on('mouseleave', function(){
$('#'+ $(this).data('target')).hide();
});<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<img src="images/Card.png" id=I1 class="img" data-target="myImage1" />
<img src="images/Card.png" id=I2 class="img" data-target="myImage2" />
<img src="images/Card.png" id=I3 class="img" data-target="myImage3" />
<img src="images/Card.png" id=I4 class="img" data-target="myImage4" />
<img src="images/Capture1.JPG" id="myImage1" style="display: none ; " alt="image1" />
<img src="images/Capture2.JPG" id="myImage2" style="display: none ; " alt="image2" />
<img src="images/Capture3.JPG" id="myImage3" style="display: none ; " alt="image3" />
<img src="images/Capture4.JPG" id="myImage4" style="display: none ; " alt="image4" />$('.img')
.on('mouseenter', function(){
$('#'+ $(this).data('target')).show();
})
.on('mouseleave', function(){
$('#'+ $(this).data('target')).hide();
});<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<img src="images/Card.png" id=I1 class="img" data-target="myImage1" />
<img src="images/Card.png" id=I2 class="img" data-target="myImage2" />
<img src="images/Card.png" id=I3 class="img" data-target="myImage3" />
<img src="images/Card.png" id=I4 class="img" data-target="myImage4" />
<img src="images/Capture1.JPG" id="myImage1" style="display: none ; " alt="image1" />
<img src="images/Capture2.JPG" id="myImage2" style="display: none ; " alt="image2" />
<img src="images/Capture3.JPG" id="myImage3" style="display: none ; " alt="image3" />
<img src="images/Capture4.JPG" id="myImage4" style="display: none ; " alt="image4" />answered Nov 21 '18 at 14:52
Taplar
15.6k21529
answered Nov 21 '18 at 14:52
Taplar
15.6k21529
answered Nov 21 '18 at 14:52
Taplar
15.6k21529
answered Nov 21 '18 at 14:52
Taplar
15.6k21529
15.6k21529
Thank you Taplar. Your code working perfectly.
– Green_Crocodile
Nov 21 '18 at 15:37
add a comment |
Thank you Taplar. Your code working perfectly.
– Green_Crocodile
Nov 21 '18 at 15:37
Thank you Taplar. Your code working perfectly.
– Green_Crocodile
Nov 21 '18 at 15:37
Thank you Taplar. Your code working perfectly.
– Green_Crocodile
Nov 21 '18 at 15:37
add a comment |
0
IF you can group the images together should do this by using pure css. If not use Taplar's answer.
Example:
html:
<!-- repeat for all your images -->
<div class="image-wrapper">
<img class="img">
<img class="image-on-hover">
</div>
CSS:
.image-on-hover{
display:none
}
.img:hover ~.image-on-hover{
display: block
}
Explanation:
The :hover Selector on CSS will be triggered when an element is hovered. With the ~ selector you can select a child element on the same level (you could call it "neighbour-element"). Since we don't want to display all neighbouring hover-images, we wrap the pairs of image and hover-images together
Different option if your grouping is exactly like the example:
.image-on-hover{
display: none;
}
/*repeat for every image pair */
#I1:hover ~ #myImage1{
display:block
}
answered Nov 21 '18 at 14:49
Heady
306318
This only works if they are grouped together, no? The OP is not grouping what is being hovered with what is being shown on hover.
– Taplar
Nov 21 '18 at 14:53
Yes, i don't know the entire use case. It really depends on how these containers are actually grouped. It can also work if the structure is different though. Would have to see the actual html
– Heady
Nov 21 '18 at 14:56
e.g. if the grouping is exactly like the example you can do this:.hover-image{ display: none}; #I1: hover ~ #myImage1{ display: block}and so on
– Heady
Nov 21 '18 at 14:56
Unfortunatly I can't group the images together. But Taplar's code it what I am looking for. but thank you for another options.
– Green_Crocodile
Nov 21 '18 at 15:39
add a comment |
0
IF you can group the images together should do this by using pure css. If not use Taplar's answer.
Example:
html:
<!-- repeat for all your images -->
<div class="image-wrapper">
<img class="img">
<img class="image-on-hover">
</div>
CSS:
.image-on-hover{
display:none
}
.img:hover ~.image-on-hover{
display: block
}
Explanation:
The :hover Selector on CSS will be triggered when an element is hovered. With the ~ selector you can select a child element on the same level (you could call it "neighbour-element"). Since we don't want to display all neighbouring hover-images, we wrap the pairs of image and hover-images together
Different option if your grouping is exactly like the example:
.image-on-hover{
display: none;
}
/*repeat for every image pair */
#I1:hover ~ #myImage1{
display:block
}
answered Nov 21 '18 at 14:49
Heady
306318
This only works if they are grouped together, no? The OP is not grouping what is being hovered with what is being shown on hover.
– Taplar
Nov 21 '18 at 14:53
Yes, i don't know the entire use case. It really depends on how these containers are actually grouped. It can also work if the structure is different though. Would have to see the actual html
– Heady
Nov 21 '18 at 14:56
e.g. if the grouping is exactly like the example you can do this:.hover-image{ display: none}; #I1: hover ~ #myImage1{ display: block}and so on
– Heady
Nov 21 '18 at 14:56
Unfortunatly I can't group the images together. But Taplar's code it what I am looking for. but thank you for another options.
– Green_Crocodile
Nov 21 '18 at 15:39
add a comment |
0
0
0
IF you can group the images together should do this by using pure css. If not use Taplar's answer.
Example:
html:
<!-- repeat for all your images -->
<div class="image-wrapper">
<img class="img">
<img class="image-on-hover">
</div>
CSS:
.image-on-hover{
display:none
}
.img:hover ~.image-on-hover{
display: block
}
Explanation:
The :hover Selector on CSS will be triggered when an element is hovered. With the ~ selector you can select a child element on the same level (you could call it "neighbour-element"). Since we don't want to display all neighbouring hover-images, we wrap the pairs of image and hover-images together
Different option if your grouping is exactly like the example:
.image-on-hover{
display: none;
}
/*repeat for every image pair */
#I1:hover ~ #myImage1{
display:block
}
answered Nov 21 '18 at 14:49
Heady
306318
IF you can group the images together should do this by using pure css. If not use Taplar's answer.
Example:
html:
<!-- repeat for all your images -->
<div class="image-wrapper">
<img class="img">
<img class="image-on-hover">
</div>
CSS:
.image-on-hover{
display:none
}
.img:hover ~.image-on-hover{
display: block
}
Explanation:
The :hover Selector on CSS will be triggered when an element is hovered. With the ~ selector you can select a child element on the same level (you could call it "neighbour-element"). Since we don't want to display all neighbouring hover-images, we wrap the pairs of image and hover-images together
Different option if your grouping is exactly like the example:
.image-on-hover{
display: none;
}
/*repeat for every image pair */
#I1:hover ~ #myImage1{
display:block
}
answered Nov 21 '18 at 14:49
Heady
306318
edited Nov 21 '18 at 15:00
answered Nov 21 '18 at 14:49
Heady
306318
answered Nov 21 '18 at 14:49
Heady
306318
answered Nov 21 '18 at 14:49
Heady
306318
306318
This only works if they are grouped together, no? The OP is not grouping what is being hovered with what is being shown on hover.
– Taplar
Nov 21 '18 at 14:53
Yes, i don't know the entire use case. It really depends on how these containers are actually grouped. It can also work if the structure is different though. Would have to see the actual html
– Heady
Nov 21 '18 at 14:56
e.g. if the grouping is exactly like the example you can do this:.hover-image{ display: none}; #I1: hover ~ #myImage1{ display: block}and so on
– Heady
Nov 21 '18 at 14:56
Unfortunatly I can't group the images together. But Taplar's code it what I am looking for. but thank you for another options.
– Green_Crocodile
Nov 21 '18 at 15:39
add a comment |
This only works if they are grouped together, no? The OP is not grouping what is being hovered with what is being shown on hover.
– Taplar
Nov 21 '18 at 14:53
Yes, i don't know the entire use case. It really depends on how these containers are actually grouped. It can also work if the structure is different though. Would have to see the actual html
– Heady
Nov 21 '18 at 14:56
e.g. if the grouping is exactly like the example you can do this:.hover-image{ display: none}; #I1: hover ~ #myImage1{ display: block}and so on
– Heady
Nov 21 '18 at 14:56
Unfortunatly I can't group the images together. But Taplar's code it what I am looking for. but thank you for another options.
– Green_Crocodile
Nov 21 '18 at 15:39
This only works if they are grouped together, no? The OP is not grouping what is being hovered with what is being shown on hover.
– Taplar
Nov 21 '18 at 14:53
This only works if they are grouped together, no? The OP is not grouping what is being hovered with what is being shown on hover.
– Taplar
Nov 21 '18 at 14:53
Yes, i don't know the entire use case. It really depends on how these containers are actually grouped. It can also work if the structure is different though. Would have to see the actual html
– Heady
Nov 21 '18 at 14:56
Yes, i don't know the entire use case. It really depends on how these containers are actually grouped. It can also work if the structure is different though. Would have to see the actual html
– Heady
Nov 21 '18 at 14:56
e.g. if the grouping is exactly like the example you can do this:
.hover-image{ display: none}; #I1: hover ~ #myImage1{ display: block} and so on– Heady
Nov 21 '18 at 14:56
e.g. if the grouping is exactly like the example you can do this:
.hover-image{ display: none}; #I1: hover ~ #myImage1{ display: block} and so on– Heady
Nov 21 '18 at 14:56
Unfortunatly I can't group the images together. But Taplar's code it what I am looking for. but thank you for another options.
– Green_Crocodile
Nov 21 '18 at 15:39
Unfortunatly I can't group the images together. But Taplar's code it what I am looking for. but thank you for another options.
– Green_Crocodile
Nov 21 '18 at 15:39
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$().hover()is an binding event listeners, just like your mouseover is binding an event listener. So of course it takes two times. Also, this logic is going to bind the hover event listener every time the image is mouseover, which is a bad idea.– Taplar
Nov 21 '18 at 14:46