如何在使用PHP上传图片之前检查/修复图片旋转 [英] How to check/fix image rotation before upload image using PHP
问题描述
上传使用iphone拍摄的图像时出现问题. 我正在尝试使用PHP和imgrotate函数自动确定是否需要将图像旋转到正确的位置,然后再将其上传到服务器.
I have a issiue when uploading a image taken using iphone. I am trying to use PHP and the imgrotate function to automaticly decide if the image needs to be rotated to the correct possition, before uploading it to the server.
我的html代码:
<form class="form-horizontal" method="post" enctype="multipart/form-data">
<div class="form-group">
<div class="col-md-9">
<div class="input-group">
<span class="input-group-btn">
<span class="btn btn-default btn-file">
Choose img<input type="file" name="file" id="imgInp">
</span>
</span>
</div>
</div>
</div>
<button type="submit">Send</button>
</form>
使用以下代码的PHP代码: 还返回错误:警告:imagerotate()期望参数1为资源,输入字符串.
The PHP code im using: Also return a error: Warning: imagerotate() expects parameter 1 to be resource, string given in.
有人针对这种情况有工作代码吗?
Anyone have a working code for this scenario?
<?php
$filename = $_FILES['file']['name'];
$exif = exif_read_data($_FILES['file']['tmp_name']);
if (!empty($exif['Orientation'])) {
switch ($exif['Orientation']) {
case 3:
$image = imagerotate($filename, -180, 0);
break;
case 6:
$image = imagerotate($filename, 90, 0);
break;
case 8:
$image = imagerotate($filename, -90, 0);
break;
}
}
imagejpeg($image, $filename, 90);
?>
推荐答案
您使用的imagerotate
错误.在传递文件名时,它期望第一个参数成为资源.检查手册
You are using imagerotate
wrong. It expect first parameter to be resource while you are passing the filename. Check manual
尝试一下:
<?php
$filename = $_FILES['file']['name'];
$filePath = $_FILES['file']['tmp_name'];
$exif = exif_read_data($_FILES['file']['tmp_name']);
if (!empty($exif['Orientation'])) {
$imageResource = imagecreatefromjpeg($filePath); // provided that the image is jpeg. Use relevant function otherwise
switch ($exif['Orientation']) {
case 3:
$image = imagerotate($imageResource, 180, 0);
break;
case 6:
$image = imagerotate($imageResource, -90, 0);
break;
case 8:
$image = imagerotate($imageResource, 90, 0);
break;
default:
$image = $imageResource;
}
}
imagejpeg($image, $filename, 90);
?>
不要忘记添加以下两行来释放内存:
Do not forget to free the memory by adding following two lines:
imagedestroy($imageResource);
imagedestroy($image);
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