php 从 URL 获取文件内容？
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/5522636/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Get file content from URL?
提问by Awan
When I use following URL in browser then it prompt me to download a text file with JSOn content.
当我在浏览器中使用以下 URL 时，它会提示我下载包含 JSOn 内容的文本文件。
https://chart.googleapis.com/chart?cht=p3&chs=250x100&chd=t:60,40&chl=Hello|World&chof=json
https://chart.googleapis.com/chart?cht=p3&chs=250x100&chd=t:60,40&chl=Hello|World&chof=json
(Click above URL see downloaded file content)
（点击上面的网址查看下载的文件内容）
Now I want to create a php page. I want that when I call this php page, it should call above URL and get content(json format) from file and show it on screen.
现在我想创建一个 php 页面。我希望当我调用这个 php 页面时，它应该调用上面的 URL 并从文件中获取内容（json 格式）并将其显示在屏幕上。
How can I do this ??
我怎样才能做到这一点 ？？
回答by John Parker
Depending on your PHP configuration, this maybe a easy as using:
根据您的 PHP 配置，这可能很容易使用：
$jsonData = json_decode(file_get_contents('https://chart.googleapis.com/chart?cht=p3&chs=250x100&chd=t:60,40&chl=Hello|World&chof=json'));
However, if allow_url_fopenisn't enabled on your system, you could read the data via CURL as follows:
但是，如果allow_url_fopen您的系统未启用，您可以通过 CURL 读取数据，如下所示：
<?php
$curlSession = curl_init();
curl_setopt($curlSession, CURLOPT_URL, 'https://chart.googleapis.com/chart?cht=p3&chs=250x100&chd=t:60,40&chl=Hello|World&chof=json');
curl_setopt($curlSession, CURLOPT_BINARYTRANSFER, true);
curl_setopt($curlSession, CURLOPT_RETURNTRANSFER, true);
$jsonData = json_decode(curl_exec($curlSession));
curl_close($curlSession);
?>
Incidentally, if you just want the raw JSON data, then simply remove the json_decode.
顺便说一句，如果您只想要原始 JSON 数据，那么只需删除json_decode.
回答by T.Todua
1) local simplest methods
1）局部最简单的方法
<?php
echo readfile("http://example.com/"); //needs "Allow_url_include" enabled
//OR
echo include("http://example.com/"); //needs "Allow_url_include" enabled
//OR
echo file_get_contents("http://example.com/");
//OR
echo stream_get_contents(fopen('http://example.com/', "rb")); //you may use "r" instead of "rb" //needs "Allow_url_fopen" enabled
?>
2) Better Way is CURL:
2）更好的方法是 CURL：
echo get_remote_data('http://example.com'); // GET request
echo get_remote_data('http://example.com', "var2=something&var3=blabla" ); // POST request
It automatically handles FOLLOWLOCATIONproblem + Remote urls: src="./imageblabla.png"turned into:src="http://example.com/path/imageblabla.png"
它自动处理FOLLOWLOCATION问题 + Remote urls:src="./imageblabla.png"变成：src="http://example.com/path/imageblabla.png"
Code : https://github.com/tazotodua/useful-php-scripts/blob/master/get-remote-url-content-data.php
代码：https: //github.com/tazotodua/useful-php-scripts/blob/master/get-remote-url-content-data.php
回答by Dr D
Don't forget: to get HTTPS contents, your OPENSSL extension should be enabled in your php.ini. (how to get contents of site use HTTPS)
不要忘记：要获取 HTTPS 内容，应该在 php.ini 中启用 OPENSSL 扩展。（如何使用HTTPS获取网站内容）
回答by cweiske
Use file_get_contentsin combination with json_decodeand echo.
使用file_get_contents结合json_decode和echo。
回答by Steve Mayne
$url = "https://chart.googleapis....";
$json = file_get_contents($url);
Now you can either echo the $json variable, if you just want to display the output, or you can decode it, and do something with it, like so:
现在你可以回显 $json 变量，如果你只想显示输出，或者你可以解码它，并用它做一些事情，像这样：
$data = json_decode($json);
var_dump($data);
