Reputation: 431
sorry if question is wrong ,
i want to display
particular image
in img
tag when <input type=file>
.
$(':input[type=file]').change( function(event) {
var tmppath = URL.createObjectURL(event.target.files[0]);
$(this).closest("img").attr('src',tmppath);
});
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<input type="file" id="file1" name="file1">
<font color="#FF0000">For best view upload only <b>200*250</b></font><div class="clear"></div>
<img src="" id="img1" class="img1" height="100px" width="100px"><br>
<input type="file" id="file2" name="file2">
<font color="#FF0000">For best view upload only <b>200*250</b></font><div class="clear"></div>
<img src="" id="img2" class="img1" height="100px" width="100px"><br>
<input type="file" id="file3" name="file3">
<font color="#FF0000">For best view upload only <b>200*250</b></font><div class="clear"></div>
<img src="" id="img3" class="img1" height="100px" width="100px">
here is my jsfiddle
Upvotes: 4
Views: 6269
Reputation: 28523
Try this : You have to use .next()
instead of .closest()
because next is used to find next element and closest is used to find matching parent element. see below code
$(':input[type=file]').change( function(event) {
var tmppath = URL.createObjectURL(event.target.files[0]);
$(this).next("img").attr('src',tmppath);
});
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<input type="file" id="file1" name="file1">
<img src="" id="img1" class="img1" height="100px" width="100px"><br>
<input type="file" id="file2" name="file2">
<img src="" id="img2" class="img1" height="100px" width="100px"><br>
<input type="file" id="file3" name="file3">
<img src="" id="img3" class="img1" height="100px" width="100px">
More Information on .next() and .closest()
EDIT: As OP have changed html structure, I have made some changes in my answer as shown below. make group of input and image elements by putting it into div and make required changes in script.
$(':input[type=file]').change( function(event) {
var tmppath = URL.createObjectURL(event.target.files[0]);
//get parent using closest and then find img element
$(this).closest("div").find("img").attr('src',tmppath);
});
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div>
<input type="file" id="file1" name="file1"><br>
<font color="#FF0000">For best view upload only <b>200*250</b></font><div class="clear"></div>
<img src="" id="img1" class="img1" height="100px" width="100px"><br>
</div>
<div>
<input type="file" id="file2" name="file2"><br>
<font color="#FF0000">For best view upload only <b>200*250</b></font><div class="clear"></div>
<img src="" id="img2" class="img1" height="100px" width="100px"><br>
</div>
<div>
<input type="file" id="file3" name="file3"><br>
<font color="#FF0000">For best view upload only <b>200*250</b></font><div class="clear"></div>
<img src="" id="img3" class="img1" height="100px" width="100px">
</div>
Upvotes: 7