PHP - 直接从 URL 将图像复制到我的服务器

PHP - Copy image to my server direct from URL

提问by Ian

Possible Duplicate:
save image from php url using php

可能的重复：
使用 php 从 php url 保存图像

I want to have PHP code for following.

我想要以下 PHP 代码。

Suppose I have one image URL, for example, http://www.google.co.in/intl/en_com/images/srpr/logo1w.png

假设我有一个图片网址，例如http://www.google.co.in/intl/en_com/images/srpr/logo1w.png

If I run one script, this image will be copied and put on my server within folder having 777 rights.

如果我运行一个脚本，这个图像将被复制并放在我的服务器上具有 777 权限的文件夹中。

Is it possible? If yea, can you please give direction for same?

是否可以？如果是的话，你能给出相同的方向吗？

Thanks,

谢谢，

Ian

伊恩

回答by dotty

Two ways, if you're using PHP5 (or higher)

两种方式，如果您使用的是 PHP5（或更高版本）

copy('http://www.google.co.in/intl/en_com/images/srpr/logo1w.png', '/tmp/file.png');

If not, use file_get_contents

如果没有，请使用file_get_contents

//Get the file
$content = file_get_contents("http://www.google.co.in/intl/en_com/images/srpr/logo1w.png");
//Store in the filesystem.
$fp = fopen("/location/to/save/image.png", "w");
fwrite($fp, $content);
fclose($fp);

From this SO post

从这个SO帖子

回答by Michael Robinson

From Copy images from url to server, delete all images after

从将图像从 url 复制到服务器，删除所有图像后

function getimg($url) {         
    $headers[] = 'Accept: image/gif, image/x-bitmap, image/jpeg, image/pjpeg';              
    $headers[] = 'Connection: Keep-Alive';         
    $headers[] = 'Content-type: application/x-www-form-urlencoded;charset=UTF-8';         
    $user_agent = 'php';         
    $process = curl_init($url);         
    curl_setopt($process, CURLOPT_HTTPHEADER, $headers);         
    curl_setopt($process, CURLOPT_HEADER, 0);         
    curl_setopt($process, CURLOPT_USERAGENT, $user_agent); //check here         
    curl_setopt($process, CURLOPT_TIMEOUT, 30);         
    curl_setopt($process, CURLOPT_RETURNTRANSFER, 1);         
    curl_setopt($process, CURLOPT_FOLLOWLOCATION, 1);         
    $return = curl_exec($process);         
    curl_close($process);         
    return $return;     
} 

$imgurl = 'http://www.foodtest.ru/images/big_img/sausage_3.jpg'; 
$imagename= basename($imgurl);
if(file_exists('./tmp/'.$imagename)){continue;} 
$image = getimg($imgurl); 
file_put_contents('tmp/'.$imagename,$image);       

回答by Shakti Singh

$url="http://www.google.co.in/intl/en_com/images/srpr/logo1w.png";
$contents=file_get_contents($url);
$save_path="/path/to/the/dir/and/image.jpg";
file_put_contents($save_path,$contents);

you must have allow_url_fopenset to on

你必须allow_url_fopen设置为on

回答by Thariama

This SO threadwill solve your problem. Solution in short:

这个SO 线程将解决您的问题。简而言之解决方案：

$url = 'http://www.google.co.in/intl/en_com/images/srpr/logo1w.png';
$img = '/my/folder/my_image.gif';
file_put_contents($img, file_get_contents($url));

回答by Shay Anderson

$url = "http://www.example/images/image.gif";
$save_name = "image.gif";
$save_directory = "/var/www/example/downloads/";

if(is_writable($save_directory)) {
    file_put_contents($save_directory . $save_name, file_get_contents($url));
} else {
    exit("Failed to write to directory "{$save_directory}");
}