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我有一个 base_64(会话中)字符串。我可以在 HTML 中打印 jpeg 图像

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AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA==" />

但是当我想在 php 中显示它时它不起作用。例如http://www.freeformatter.com/base64-encoder.html将其解码为图像。我试过这种方法

`

$data = base64_decode($query);

$im = imagecreatefromstring($data);
if ($im !== false) {
    header('Content-Type: image/jpeg');
    imagepng($im);
    imagedestroy($im);
}
else {
    echo 'An error occurred.';
}

或者

header("Content-type: image/jpeg");
echo '<img src="data:image/jpeg;base64,'.$query.'" />';

或者

$imagedata=base64_decode($query);
$length = strlen($imagedata);
header('Last-Modified: '.date('r'));
header('Accept-Ranges: bytes');
header('Content-Length: '.$length);
header('Content-Type: image/jpeg');
print($imagedata);

并且没有一个工作正常。

非常感谢您的任何想法。

4

1 回答 1

0

变量 $query 不正确,你是从 $_POST 还是 $_GET 得到 $query,base_64 数据最后可能有一些“=”,当数据发送到服务器时,“=”会被编码。

于 2013-07-11T05:41:22.473 回答