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PHP上传文件并给自定义名称

转载 作者:可可西里 更新时间:2023-11-01 00:16:35 25 4
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我正在使用此功能上传我的文件:

if ((($_FILES["Artwork"]["type"] == "image/gif")
|| ($_FILES["Artwork"]["type"] == "image/jpeg")
|| ($_FILES["Artwork"]["type"] == "image/jpg")
|| ($_FILES["Artwork"]["type"] == "image/pjpeg"))
&& ($_FILES["Artwork"]["size"] < 20000000))
{
if ($_FILES["Artwork"]["error"] > 0)
{
//echo "Return Code: " . $_FILES["Artwork"]["error"] . "<br />";
}else{
$imageName = $_FILES['Artwork']['name'];
move_uploaded_file($_FILES["Artwork"]["tmp_name"],
$path_image . $imageName);
}
}else{
//echo "invalid file";
}

如何使用自定义名称更改 $imageName = $_FILES['Artwork']['name'];,但在名称中保留文件扩展名,例如:myCustomName.jpg?

谢谢!

最佳答案

您需要在代码中修改的唯一行是:

$imageName = 'CustomName.' . pathinfo($_FILES['Artwork']['name'],PATHINFO_EXTENSION);

其中“CustomName”。是您想要的图像的新名称。 pathinfo如果 PHP 函数处理路径和文件名的操作。

你的整个代码将是:

if ((($_FILES["Artwork"]["type"] == "image/gif")
|| ($_FILES["Artwork"]["type"] == "image/jpeg")
|| ($_FILES["Artwork"]["type"] == "image/jpg")
|| ($_FILES["Artwork"]["type"] == "image/pjpeg"))
&& ($_FILES["Artwork"]["size"] < 20000000))
{
if ($_FILES["Artwork"]["error"] > 0)
{
//echo "Return Code: " . $_FILES["Artwork"]["error"] . "<br />";
}else{
$imageName = 'CustomName.' . pathinfo($_FILES['Artwork']['name'],PATHINFO_EXTENSION);
move_uploaded_file($_FILES["Artwork"]["tmp_name"],
$path_image . $imageName);
}
}else{
//echo "invalid file";
}

关于PHP上传文件并给自定义名称,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/7162157/

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