Question

所以我目前发现了这个名为cropit的照片裁剪插件。演示是here。所以我想要做的是抓取裁剪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>

jQUERY：

    $('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;
    });

所有我需要帮助的是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，所有这些我通常不熟悉的代码出现。以下是jquery返回的序列化代码的几个字符：

image-data=data%3Aimage%2Fpng%3Bbase64%2CiVBORw0KGgoAAAANSUhE...

Answer 1

<强> 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);

<强> 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);

Answer 2

通过不解码编码数据，我得到了Hosch Nok的工作答案。不打电话：

$decoded = urldecode($encoded);

但直接使用$encoded变量。

如何使用带有php的cropit jquery插件裁剪和上传照片

2 个答案: