In my asp.net page I have an image upload control where users can upload there profile photo. what I want is when user browse his/her photo an immediate preview should displayed. Is there any ajax image control for that or how can I implement that?
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4 回答
0
您必须先上传文件,然后才能访问它。要进行异步文件上传,您需要多个UpdatePanel
. 此代码示例显示了在不重新加载页面的情况下进行上传的技巧。最后,要进行预览,您需要创建一个 javascript 函数来更改页面上图像的 url。
于 2010-09-27T18:23:52.223 回答
0
这是一个示例项目。
于 2010-10-30T00:33:37.053 回答
0
javascript 在上传之前不允许您访问客户端图像,因此您唯一的选择是在文件上传后显示图像。我不确定,但你可以通过在 Flash 中创建文件上传控件来做你想做的事,但我不知道 Flash,所以我无法帮助你。
于 2010-09-27T16:40:39.710 回答
0
希望这会有所帮助,我创建了一个小要点来演示如何要求用户上传图像,然后在任何服务器端进程启动之前立即显示图像:
var Image_Upload_Preview = function( file_input, image_element ){
/**
* Checks for supported features
* @returns {boolean}
*/
this.is_supported = function(){
if( ! FileReader instanceof Function ){
console.error(':( Your browser noes not support the FileReader...');
return false;
}
};
/**
* Checks the inputs provided
* @returns {boolean}
*/
this.validate_inputs = function(){
/**
* Fail if:
* 1. Not a HTML Input Element
* 2. Not a File Input Element
*
*/
if( ! $(file_input).get(0) instanceof HTMLInputElement || $(file_input).first().attr('type') != 'file' ){
console.error( 'Invalid Element provided...' );
return false;
}
/**
* Fail if:
* 1. Image Element provided is invalid
*/
if( ! $(image_element).get(0) instanceof HTMLImageElement ){
console.error( 'Invalid Image Element provided...' );
return false;
}
};
/**
* Only proceed if all the preliminary checks have passed
*/
if( this.is_supported() || this.validate_inputs() ){
return false;
}
/**
* Set the file input to only accept images
*/
$(file_input).attr('accept','image/*');
$(file_input).change(function(){
/**
* Fail if:
* 1. 'files' data is non existent
* 2. 'files' data has no data in it
*/
if( !this.files || this.files.length < 1 ){
console.error('No files data exists for this file input...');
return false;
}
var file_reader = new FileReader();
file_reader.onload = function( reader_result ) {
var image_result = null;
/**
* Legacy lookup for the result
*/
image_result = reader_result.target && reader_result.target.result ? reader_result.target.result : image_result ;
image_result = !image_result && reader_result.srcElement && reader_result.srcElement.result ? reader_result.srcElement.result : image_result ;
$( image_element ).attr( 'src', image_result );
};
file_reader.readAsDataURL( this.files[0] );
});
};
$(document).ready(function(){
/**
* Example Usages
*/
// new Image_Upload_Preview( document.getElementById('file_input_demo'), $('img') );
new Image_Upload_Preview( $('input[type=file]'), $('img') );
});
<link href="https://cdnjs.cloudflare.com/ajax/libs/uikit/2.27.2/css/uikit.min.css" rel="stylesheet"/>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script>
<div class="uk-container uk-container-center uk-text-center">
<div class="uk-thumbnail uk-margin-top">
<img class="uk-margin" src="https://placeholdit.imgix.net/~text?txtsize=56&txt=Select%20Image&w=600&h=400">
<form>
<label class="uk-button uk-button-primary uk-button-large">
<span>Select Image</span>
<input type="file" accepts="image/*" id="file_input_demo" class="uk-invisible uk-position-absolute" />
</label>
</form>
</div>
</div>
于 2017-04-10T04:50:43.920 回答