如何使用php获取图像并在同一页面上显示
这是myphotos.html文件 我想从这个图像路径,我想在同一页上显示它我知道我会像photos.php一样保存它,但我怎么能在同一页上做,请帮助我如何使用php获取图像并在同一页面上显示,php,Php,这是myphotos.html文件 我想从这个图像路径,我想在同一页上显示它我知道我会像photos.php一样保存它,但我怎么能在同一页上做,请帮助我 <div class="container"> <div class="panel panel-default"> <div class="panel-heading"> <strong>Upload Files</strong> <sm
<div class="container">
<div class="panel panel-default">
<div class="panel-heading">
<strong>Upload Files</strong>
<small>upload any file</small>
</div>
<div class="panel-body">
<!-- Standar Form -->
<h4>Select files from your computer</h4>
<form action="" method="post" enctype="multipart/form-data" id="js-upload-form">
<div class="form-inline">
<div class="form-group">
<input type="file" name="files[]" id="js-upload-files" multiple>
</div>
<button type="submit" class="btn btn-sm btn-primary" id="js-upload-submit">Upload files</button>
</div>
</form>
</div>
</div>
我将非常感谢您,谢谢。您应该参考以下链接: 示例: HTML代码: Javascript代码:
如果我理解正确,一旦你上传你的图像,你希望你的图像在提交之前显示在表单上。对吗?提交后没有兄弟我想在同一页上显示我已经做了,图像正在存储和显示,但我想像图像滑块一样显示
<input id="imageupload" type="file" multiple />
<div id="preview-image"></div>
<script type='text/javascript' src='//code.jquery.com/jquery-1.9.1.js'></script>
<script type="text/javascript">
$("#imageupload").on('change', function () {
var countFiles = $(this)[0].files.length;
var imgPath = $(this)[0].value;
var extn = imgPath.substring(imgPath.lastIndexOf('.') + 1).toLowerCase();
var image_holder = $("#preview-image");
image_holder.empty();
if (extn == "gif" || extn == "png" || extn == "jpg" || extn == "jpeg") {
if (typeof (FileReader) != "undefined") {
for (var i = 0; i < countFiles; i++) {
var reader = new FileReader();
reader.onload = function (e) {
$("<img />", {
"src": e.target.result,
"class": "thumbimage"
}).appendTo(image_holder);
}
image_holder.show();
reader.readAsDataURL($(this)[0].files[i]);
}
} else {
alert("It doesn't supports");
}
} else {
alert("Select Only images");
}
});
</script>