用PHP归档的BLOB URL [英] BLOB URL to file with PHP

问题描述

我正在尝试创建文件预览系统.

I'm trying to create a file preview system.

当用户单击上传按钮并选择文件时，我使用HTML5 FILE API获取所有文件信息并为文件创建缩略图.

When the user click in a upload button and select files, I use the HTML5 FILE API to get all the file info and create a thumbnail for the file.

用户可以取消某些thoose文件的上传，也可以添加更多文件.

The user can cancel the upload of some of thoose files and add more files too.

我现在所拥有的是:

<p><input id="blob" name="blob" type="file" size="75" multiple /></p>
<form method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>">
<input id="blobtype" name="blobtype" type="text" size="20" />
<input id="bloburl" name="bloburl" type="text" size="50" />
<input id="salvar" name="salvar" type="submit" value="salvar" />
</form>
<p id="result"></p>

还有我的javascript:

And my javascript:

function cbu(file)
{
if (window.URL){ return window.URL.createObjectURL(file);}
else if (window.webkitURL){ return window.webkitURL.createObjectURL(file);}
else if (window.createObjectURL){ return window.createObjectURL(file);}
else if (window.createBlobURL){ return window.createBlobURL(file); }
}

document.getElementById("blob").onchange = function(event)
{
var result = document.getElementById("result");
var i = 0, j = event.target.files.length;
for (i = 0; i < j; i++)
{
    var url = cbu(event.target.files[i]);
    var img = document.createElement("img");
    img.onload = function(){ }
    result.appendChild(img);
    img.src = url;
    document.getElementById("blobtype").value = event.target.files[i].type;
    document.getElementById("bloburl").value = url;
    }
};

它会像这样提供缩略图:

It genarate a thumbnail like this:

<img src="blob:b9a8e310-05b3-48ba-ba24-9ca7cf287f71">

此值blob:b9a8e310-05b3-48ba-ba24-9ca7cf287f71在用户提交表单时发送到PHP文件.

This value blob:b9a8e310-05b3-48ba-ba24-9ca7cf287f71 is send to a PHP file when user submit the form.

所以我的问题是，如何将这个Blob网址转换为文件，以便可以将其保存在我的应用程序中?

So my question is, how can I transform this blob url into a file so I can save it in my app?

有更好的方法吗?

感谢您的帮助！

推荐答案

好吧，我一直在寻找解决方案，但我找到了一个解决方案:画布.

well, I keep looking for solutions and I found this one: Canvas.

Canvas元素具有toDataUrl()方法，该方法返回其包含的图像的数据URL.

Canvas element have the toDataUrl() method that returns a Data URL for the image that it contains.

所以我继续做同样的事情，但是当我提交表单时，我使用以下功能:

So I keep doing the same thing but when I submit the form I use the following function:

function putImage(){
    var url = document.getElementById("img1").src; // 'img1' is the thumbnail - I had to put an id on it
    var canvas = document.getElementById("MyCanvas");
    if (canvas.getContext) {
        var ctx = canvas.getContext("2d");

        var img = new Image();
        img.src = url;
        img.onload = function () {
            // Desenha a imagem no canvas...
            ctx.drawImage(img,0,0);

            // Grava o Data URL na imagem...
            var myImage = canvas.toDataURL("image/png");
            document.getElementById("dataurl").value = myImage;
        }
    }
};

我在画布页面上又创建了一个元素:

I create one more element on the page for canvas:

<canvas id="MyCanvas">This browser or document mode doesn't support canvas</canvas>

以及表单上的另一个用于保存数据URL的元素:

and one more element on the form to save the Data URL:

<input id="dataurl" name="dataurl" type="text" size="50" />

好吧，现在我可以在PHP中获得所有信息，并执行以下操作:

well, now I can get it all in PHP and do something like this:

if($_POST)
{
    //echo '<pre>'; print_r($_POST); echo '</pre>';

    $blob = $_POST['bloburl'];
    $type = $_POST['blobtype'];
    $data = $_POST['dataurl'];

    $contents_split = explode(',', $data);
    $encoded = $contents_split[count($contents_split)-1];
    $decoded = "";
    for ($i=0; $i < ceil(strlen($encoded)/256); $i++) {
        $decoded = $decoded . base64_decode(substr($encoded,$i*256,256)); 
    }

    $fp = fopen('imagembaixada.jpg', 'w');
        fwrite($fp, $decoded);
        fclose($fp);
    }
}

现在，我的图像已保存.

Now my image is saved.

它可以工作，但是效率很低.

It works, but looks pretty inefficient.

如果有人知道在上传图片之前创建图片预览的更好方法，请告诉我！

If someone knows a better way to create a preview for images before upload them, please tell me!

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