[英]How to display the image from ajax success function?
I have created ajax function like this...In this I will get the value from run time and i need to return the photo according to that value..In success function i need to display that image in particulat div 我已经创建了像这样的ajax函数...在这种情况下,我将从运行时获取值,并且我需要根据该值返回照片。在成功函数中,我需要在particulat div中显示该图像
var num=document.getElementById('number').value;
$.ajax({
url:"image.php?val="+num,
contentType: "image/png",
success:function(img)
{
$('#image').html('<img src="data:image/png;base64,' + img + '" />');
}
});
image.php page image.php页面
$sql_sub = select_query("select pic from photo where picnum=".$_GET['val']."");
$img = $sql_sub[0][0]->load();
header("Content-type: image/png");
ob_start();
imagepng($img);
echo "data:image/png;base64,", base64_encode(ob_get_clean());
It looks perfect..You may have an issue in tag. 看起来很完美。。您的标签可能有问题。 Check first that tag.
首先检查该标签。 However
.append
works great. 但是
.append
效果很好。
Have you tried this: 您是否尝试过:
$('body').append('<img src="https://chart.googleapis.com/chart?cht=qr&chs=200x200&chl=http%3a%2f%2fwww.facebook.com" />');
$('#div_where_you_will_sho_qr_code').append(data.toString());
or: 要么:
$('#container').html('<img src="https://chart.googleapis.com/chart?cht=qr&chs=200x200&chl=http%3a%2f%2fwww.facebook.com" />');
where #container is some DOM element to harbor your image. #container是一些DOM元素来保存您的图像。
or the way I prefer: 或我更喜欢的方式:
$('#container').html(
$('<img/>', {
src: 'https://chart.googleapis.com/chart?cht=qr&chs=200x200&chl=http%3a%2f%2fwww.facebook.com',
alt: ''
})
);
var num=document.getElementById('number').value;
$.ajax({
url:"image.php?val="+num,
type: "POST",
dataType: "html",
success:function(data)
{
$('#image').html(data));
}
});
image.php image.php
$sql_sub = select_query("select pic from photo where picnum=".$_POST'val']."");
$img = $sql_sub[0][0]->load();
$image = '<img src="data:image/png;base64,'.$img.'" />';
echo $img;
var num=document.getElementById('number').value;
$.ajax({
url:"image.php?val="+num,
type: "POST",
dataType: "html",
success:function(img)
{
$('#image').html('<img src="data:image/png;base64,' + img + '" />');
}
});
image.php image.php
$sql_sub = select_query("select pic from photo where picnum=".$_GET['val']."");
$img = $sql_sub[0][0]->load();
header("Content-type: image/png");
ob_start();
echo $img;
echo "data:image/png;base64,", base64_encode(ob_get_clean());
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