[英]how to display image to popup window
我已经使用readAsdataUrl javascript函数读取了内存中的图像。 现在,我需要在一个弹出窗口中显示此图像,当在现有页面中单击按钮时会打开该窗口。
我怎样才能做到这一点? 有什么方法可以从内存中读取图像,然后在新的弹出窗口中显示它?
进行检查-http: //www.html5rocks.com/zh-CN/tutorials/file/dndfiles/#toc-reading-files
似乎有您需要的示例:
<style>
.thumb {
height: 75px;
border: 1px solid #000;
margin: 10px 5px 0 0;
}
</style>
<input type="file" id="files" name="files[]" multiple />
<output id="list"></output>
<script>
function handleFileSelect(evt) {
var files = evt.target.files; // FileList object
// Loop through the FileList and render image files as thumbnails.
for (var i = 0, f; f = files[i]; i++) {
// Only process image files.
if (!f.type.match('image.*')) {
continue;
}
var reader = new FileReader();
// Closure to capture the file information.
reader.onload = (function(theFile) {
return function(e) {
// Render thumbnail.
var span = document.createElement('span');
span.innerHTML = ['<img class="thumb" src="', e.target.result,
'" title="', escape(theFile.name), '"/>'].join('');
document.getElementById('list').insertBefore(span, null);
};
})(f);
// Read in the image file as a data URL.
reader.readAsDataURL(f);
}
}
document.getElementById('files').addEventListener('change', handleFileSelect, false);
</script>
只需将div替换为所需的弹出窗口
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