Question

当我使用文件上传时，我可以这样。

............................................... .................................................. .................................................. ...........

$image_name = $_FILES['image']['name'];
$image_size = $_FILES['image']['size'];
$image_temp = $_FILES['image']['tmp_name'];

但我想知道，如何从图像路径/链接获取图像名称，大小，温度

www.example.com/img/test.png

使用php？

感谢所有

Answer 1

    <?php
    $filename='img/1967-jaguar.jpg';//image path
    $path_parts = pathinfo($filename);
    //print_r($path_parts);
    echo 'Dir Name: '.$path_parts['dirname'].'<br>';
    echo 'Base Name: '.$path_parts['basename'].'<br>';;
    echo 'Extension: '.$path_parts['extension'].'<br>';;
    echo 'File Name: '.$path_parts['filename'].'<br>';;
    echo $file_size=filesize($filename);
    echo 'File Size: '.$file_size.'kb';
    //for image width height and mime
    $imagesize=getimagesize($filename);
    print_r($imagesize);
    ?>

Answer 2

使用pathinfo()和filesize()

关于pathinfo()，来自php手册：

<?php
$path_parts = pathinfo('/www/htdocs/img/image.jpg');

echo $path_parts['dirname'], "\n";
echo $path_parts['basename'], "\n";
echo $path_parts['extension'], "\n";
echo $path_parts['filename'], "\n"; // since PHP 5.2.0
?>

您还可以使用getimagesize()

Answer 3

您也可以使用....

＆＃34 ;; Echo $ path [＆＃39; basename＆＃39;]。＆＃34;
＆＃34 ;; Echo $ path [＆＃39; extension＆＃39;]。＆＃34;
＆＃34 ;; ？＆GT;

尝试它正常工作......

如何从图像路径/链接获取图像名称，大小，温度？

3 个答案: