Question

我有一个小问题;在PHP中，我使用curl从URL获取数据：

$url = "http://www.prelovac.com/vladimir/wp-content/uploads/2008/03/example.jpg";

我使用curl_getinfo()给了我一个数组：

Array
(
[url] => http://www.prelovac.com/vladimir/wp-content/uploads/2008/03/example.jpg
[content_type] => image/jpeg
[http_code] => 200
[header_size] => 496
[request_size] => 300
[filetime] => -1
[ssl_verify_result] => 0
[redirect_count] => 0
[total_time] => 2.735
[namelookup_time] => 0.063
[connect_time] => 0.063
[pretransfer_time] => 0.063
[size_upload] => 0
[size_download] => 34739
[speed_download] => 12701
[speed_upload] => 0
[download_content_length] => 34739
[upload_content_length] => -1
[starttransfer_time] => 1.282
[redirect_time] => 0
)

如何在链接[url] => http://www.prelovac.com/vladimir/wp-content/uploads/2008/03/example.jpg中获取图像的名称，例如

[image_name] : example
[image_ex] : jpg

感谢您的任何建议！

Answer 1

使用pathinfo

Answer 2

有时url会附加额外的参数。在这种情况下，我们可以首先删除参数部分，然后我们可以使用PHP的内置pathinfo（）函数从URL获取图像名称。

$url = 'http://images.fitnessmagazine.mdpcdn.com/sites/story/shutterstock_65560759.jpg?itok=b8HiA95H';

检查图像网址是否附加了参数。

if (strpos($url, '?') !== false) {
    $t = explode('?',$url);
    $url = $t[0];            
}

生成的url变量现在包含

http://images.fitnessmagazine.mdpcdn.com/sites/story/shutterstock_65560759.jpg

使用pathinfo()检索所需的详细信息。

$pathinfo = pathinfo($url);
echo $pathinfo['filename'].'.'.$pathinfo['extension'];

这会将shutterstock_65560759.jpg作为输出。

Answer 3

考虑以下是图像路径 $ IMAGE_URL = 'http://development/rwc/wp-content/themes/Irvine/images/attorney1.png'; 从这个url获取带扩展名的图像名，使用下面的函数basename（）; 看下面的代码

代码：

$image_url='http://development/rwc/wp-content/themes/Irvine/images/attorney1.png';
echo basename($image_url);

输出： attorney1.png

Answer 4

$url_arr = explode ('/', $arr['url']);
$ct = count($url_arr);
$name = $url_arr[$ct-1];
$name_div = explode('.', $name);
$ct_dot = count($name_div);
$img_type = $name_div[$ct_dot -1];

echo $name . "  " . $img_type;

Answer 5

$URL = urldecode('http://www.greenbiz.com/sites/default/files/imagecache/wide_large/Woman_HORIZ.jpg?sunny=20$mal+1');
$image_name = (stristr($URL,'?',true))?stristr($URL,'?',true):$URL;
$pos = strrpos($image_name,'/');
$image_name = substr($image_name,$pos+1);
$extension = stristr($image_name,'.');
if($extension == '.jpg' || $extension == '.png' || $extension == '.gif' || $extension == '.jpeg'){`enter code here`
print $image_name;
}

Answer 6

您可以使用正则表达式/(?:.+\/)(.+\.(png|jpg|jepg))[?#]?.*$/

示例：

$examples = [
    'http://images.fitnessmagazine.mdpcdn.com/sites/story/with_query.jpg?itok=b8HiA95H',
    'http://images.fitnessmagazine.mdpcdn.com/sites/story/with_hash.jpg#hghgh',
    'http://images.fitnessmagazine.mdpcdn.com/sites/story/with_query.jpg?',
    'http://images.fitnessmagazine.mdpcdn.com/sites/story/with_hash.jpg#',
    'http://images.fitnessmagazine.mdpcdn.com/sites/story/with_multydots.65560759.jpg',
    'http://images.fitnessmagazine.mdpcdn.com/sites/story/image.png',
    'http://images.fitnessmagazine.mdpcdn.com/sites/story/without_ext',   
    ];

foreach($examples as $example) {
    preg_match('/(?:.+\/)(.+\.(png|jpg|jepg))[?#]?.*$/', $example, $matches);
    
    if(isset($matches[1]) && $matches[1]) {
     echo "Url: {$example}, image name: {$matches[1]} \n";   
    } else {
        echo "Url: {$example} is not image url \n";   
    }
}

打印：

Url: http://images.fitnessmagazine.mdpcdn.com/sites/story/with_query.jpg?itok=b8HiA95H, image name: with_query.jpg 
Url: http://images.fitnessmagazine.mdpcdn.com/sites/story/with_hash.jpg#hghgh, image name: with_hash.jpg 
Url: http://images.fitnessmagazine.mdpcdn.com/sites/story/with_query.jpg?, image name: with_query.jpg 
Url: http://images.fitnessmagazine.mdpcdn.com/sites/story/with_hash.jpg#, image name: with_hash.jpg 
Url: http://images.fitnessmagazine.mdpcdn.com/sites/story/with_multydots.65560759.jpg, image name: with_multydots.65560759.jpg 
Url: http://images.fitnessmagazine.mdpcdn.com/sites/story/image.png, image name: image.png 
Url: http://images.fitnessmagazine.mdpcdn.com/sites/story/without_ext is not image url

Answer 7

如果您想忽略uri参数而从url获取文件名，则可以尝试以下操作：

$url = 'http://host/filename.ext?params';
$parsedUrl = parse_url($url);
$pathInfo = pathinfo($parsedUrl['path']);
print_r($pathInfo);
Array
(
    [dirname] => /
    [basename] => filename.ext
    [extension] => ext
    [filename] => filename
)

Answer 8

$imagePath = 'http://www.prelovac.com/vladimir/wp-content/uploads/2008/03/example.jpg';

$imageName = get_basename($imagePath);

function get_basename($filename)
{
    return preg_replace('/^.+[\\\\\\/]/', '', $imagePath);
}

Answer 9

$url ='http://hhhhhh.com/content/thees/Irvine/images/hdahdahat.png';
$img = explode(".", basename($url));

echo $img['0'];

Answer 10

使用PHP pathinfo的其他人我有另一种可能对其他人有用的方法。请参见下面的示例。

/* Image source */
$image_url = 'http://example.com/images/IMAGE_Name-Example.png';


/* Get image name with extension */
$image_name = basename($image_url); 
// returns 'IMAGE_Name-Example.png'


/* Get image name without extension, i.e. remove '.png', '.jpg', '.gif' etc */
$image_name_wo_ext = substr($image_name, 0, strripos($image_name,'.')); 
// returns 'IMAGE_Name-Example'


/* Get image name without special characters i.e. '-', '_' etc */
$image_name_wo_sc = str_replace(array('_', '-'), ' ', $image_name_wo_ext); 
// returns 'IMAGE Name Example'

如何从URL获取图像的名称？

10 个答案: