I have this following code
$(function () {
$.post("fetch_data.php?url="+$('#url').val(), { }, function(response){
var arrValues = [ "path/to/img1", "path/to/img2", "path/to/img3", "path/to/img4" ]; //output from php
var img = new Image();
$.each(arrValues, function( intIndex, objValue ) {
img.src = objValue;
img.onload = function() {
alert('height: ' + img.height + ' width: ' + img.width);
}
});
});
});
I am ne开发者_开发技巧w in javascript/jquery, this code only return last("path/to/img4") height/width of image.
How do i make it return all image and width of the array?
I guess you have to create four image Elements... you have to modify your code a bit
try the bellow code,
$(function () {
$.post("fetch_data.php?url="+$('#url').val(), { }, function(response){
var arrValues = [ "path/to/img1", "path/to/img2", "path/to/img3", "path/to/img4" ]; //output from php
$.each(arrValues, function( intIndex, objValue ) {
var img = new Image(); //creating number of Image Elements equal to arrValues length
img.src = objValue;
img.onload = function() {
alert('height: ' + img.height + ' width: ' + img.width);
}
});
});
});
As the parameter img is defined outside your loop, it will only point to the last img u create..
try to declare the img inside the $.each loop...
and replace the img with this in the alert..
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