所以我目前找到了这个名为cropit的照片裁剪插件。演示在这里.所以我想做的是抓取裁剪后的照片并将照片的名称上传到 mysql 数据库，并使用 php 将其保存到目录中。
到目前为止，我有这个：
.HTML：
<form method="POST">
<div class="image-editor">
<div class="cropit-image-preview-container">
<div class="cropit-image-preview"></div>
</div>
<div class="image-size-label">
Resize image
</div>
<input type="range" class="cropit-image-zoom-input">
<input type="hidden" name="image-data" class="hidden-image-data" />
<button type="submit">Submit</button>
</div>
</form>
j查询 ：
$('form').submit(function() {
// Move cropped image data to hidden input
var imageData = $('.image-editor').cropit('export');
$('.hidden-image-data').val(imageData);
// Print HTTP request params
var formValue = $(this).serialize();
$('#result-data').text(formValue);
// Prevent the form from actually submitting
return false;
});
我需要的帮助只是 php 设置代码，因为当我裁剪照片并选择提交时，jquery 返回序列化代码，并且出现所有这些我通常不熟悉的代码。下面是 jquery 返回的序列化代码的几个字符：
image-data=data%3Aimage%2Fpng%3Bbase64%2CiVBORw0KGgoAAAANSUhE...
1. 保存 base64 编码图像
<?php
//save your data into a variable - last part is the base64 encoded image
$encoded = "data%3Aimage%2Fpng%3Bbase64%2CiVBORw0KGgoAAAANSUhE";
//decode the url, because we want to use decoded characters to use explode
$decoded = urldecode($encoded);
//explode at ',' - the last part should be the encoded image now
$exp = explode(',', $decoded);
//we just get the last element with array_pop
$base64 = array_pop($exp);
//decode the image and finally save it
$data = base64_decode($base64);
$file = 'data.png';
//make sure you are the owner and have the rights to write content
file_put_contents($file, $data);
2. 获取base64编码图像的文件名
$encoded = "data%3Aimage%2Fpng%3Bbase64%2CiVBORw0KGgoAAAANSUhE";
$decoded = urldecode($encoded);
$exp = explode(';', $decoded);
$exp = explode(':', $exp[0]);
$image = array_pop($exp);
echo ($image);
我得到了Hosch Nok的答案，不解码编码的数据。不呼叫：
$decoded = urldecode($encoded);
但直接使用 $encoded 变量。