Display an image in a <div>, jQuery

The value of ig is being transfered fine, but I need to put this image into an image, please tell me what I am doing wrong.

var ig = '<img src='+$(this).attr('src')+' 开发者_运维百科style="z-index:1">';
$('#big').val(this)

Thanks Jean

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you need to use.html instead,

var ig = '<img src='+$(this).attr('src')+' style="z-index:1">';
$('#big').html(this)

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var ig = $('<img />').attr('src', $(this).attr('src')).css('z-index', 1);
$('#big').append(ig);

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Use .html instead of .val and pass ig. .val is used for manipulating form elements.

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You want to use html or appendTo instead of val.

var ig = '<img src='+$(this).attr('src')+' style="z-index:1">';
$('#big').html(ig);  // replaces contents

or

$(ig).appendTo('#big'); // appends to existing contents

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Use append instead of val

var ig = '<img src='+$(this).attr('src')+' style="z-index:1">';
$('#big').append(ig);

working example: http://jsbin.com/agoye3

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Incidentally, if #big is already an image, you can just change the image source:

<img id='big' src='theoldimage.jpg'>

<script>
  $('#big').attr("src", 'thenewimage.jpg');
</script>