我打算在表单中预览图像或照片,但它不起作用,HTML代码如下所示:
<form action="" method="post" enctype="multipart/form-data" name="personal_image" id="newHotnessForm">
<p><label for="image">Upload Image:</label>
<input type="file" id="imageUpload"/></p>
<p><button type="submit" class="button">Save</button></p>
<div id="preview">
<img width="160px" height="120px" src="profile pic.jpg" id="thumb" />
</div>
</form>
并将JS代码/脚本包含在下面:
<script type="text/jaavascript">
$(document).ready(function(){
var thumb=$('#thumb');
new AjaxUpload('imageUpload',{
action:$('newHotnessForm').attr('action'),
name:'image',
onSubmit:function(file,extension){
$('#preview').addClass('loading');
},
onComplete:function(file,response){
thumb.load(function(){
$('#preview').removeClass('loading');
thumb.unbind();
});
thumb.attr('src',response);
}
});
});
我的表单上有两个主要问题:
为什么图像或图片的预览不起作用?如何从表单中粘贴照片当单击保存按钮时,它将转到/链接到我创建的另一个PHP或PHP页面?
发布于 2013-04-25 14:13:00
试试这个:(用于预览)
<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]);
}
}
</script>
<body>
<form id="form1" runat="server">
<input type="file" onchange="readURL(this);" />
<img id="blah" src="#" alt="your image" />
</form>
</body>
工作演示here>
发布于 2013-04-25 14:43:47
#######################
### the img page ###
#######################
<script src="https://code.jquery.com/jquery-1.9.1.min.js"></script>
<script src="https://malsup.github.com/jquery.form.js"></script>
<script type="text/javascript">
$(document).ready(function(){
$('#f').live('change' ,function(){
$('#fo').ajaxForm({target: '#d'}).submit();
});
});
</script>
<form id="fo" name="fo" action="nextimg.php" enctype="multipart/form-data" method="post">
<input type="file" name="f" id="f" value="start upload" />
<input type="submit" name="sub" value="upload" />
</form>
<div id="d"></div>
#############################
### the nextimg page ###
#############################
<?php
$name=$_FILES['f']['name'];
$tmp=$_FILES['f']['tmp_name'];
$new=time().$name;
$new="upload/".$new;
move_uploaded_file($tmp,$new);
if($_FILES['f']['error']==0)
{
?>
<h1>PREVIEW</h1><br /><img src="<?php echo $new;?>" width="100" height="100" />
<?php
}
?>
发布于 2019-06-06 23:47:27
function readURL(input) {
if (input.files && input.files[0]) {
var reader = new FileReader();
reader.onload = function(e) {
$('#ImdID').attr('src', e.target.result);
};
reader.readAsDataURL(input.files[0]);
}
}
img {
max-width: 180px;
}
<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<input type='file' onchange="readURL(this);" />
<img id="ImdID" src="" alt="Image" />
https://stackoverflow.com/questions/16207575
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