我有一个网页,将允许用户上传他们的图片,将作为他们的个人资料图片。我有这样的代码:
<form method = "post" action = "" enctype="multipart/form-data">
<input type = "hidden" name = "size" value = "1000000">
<input type = "file" name = "image">
<input type = "submit" name = "upload" value = "Upload Image">
</form>我的php:
if(isset($_POST['upload'])) {
$target = "profileimgs/".$key.basename($_FILES['image']['name']);
$image = $key.basename($_FILES['image']['name']);
$sql = "UPDATE users SET img = '$image' WHERE idnumber = $idnumber";
mysqli_query($con,$sql);
if(move_uploaded_file($_FILES['image']['tmp_name'], $target)) {
header("Location:home.php");
} else {
header("Location:error_uploading.php");
}
}通过选择图片,将显示的唯一预览是图片的文件名。我想在他们选择图像后添加他们选择的图像的“预览”。就像你在facebook上上传个人资料图片一样--你从你的电脑上选择一张照片,然后当你点击select picture时,它会自动上传你的图片到页面上,然后给你一个预览,让你看到需要点击这么多按钮的那张图片。
我能做些什么来实现这个目标吗?
发布于 2017-04-01 18:27:49
<form id="form1" runat="server">
<input type='file' id="imgInp" />
<img id="blah" src="#" alt="your image" />
</form>
<script type="text/javascript">
function readURL(input)
{
if (input.files && input.files[0]) {
var reader = new FileReader();
reader.onload = function (e) {
$('#blah').attr('src', e.target.result);
}
reader.readAsDataURL(input.files[0]);
}
}
$("#imgInp").change(function(){
readURL(this);
});
</script>
It will show instead image using javascript base encode. your php code work as well as. 发布于 2017-04-01 19:02:10
https://stackoverflow.com/questions/43160226
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