Question

我有一些代码来获取site1的json内容，但我还需要获取site2的内容。我应该为site2重新编写所有这些行吗？或者我可以在curl_setopt？

中再添加一个网址

$ch = curl_init(); 
curl_setopt($ch, CURLOPT_URL,"http://site1.com");
curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
curl_setopt($ch, CURLOPT_HEADER, 0);
$outputJson = curl_exec($ch);
if ($outputJson === FALSE) {
    echo 'Sorry, This service is currently unavailable: '. curl_error($ch);
}

Answer 1

您可以创建类似

的功能

function get_data($url)
{
     $ch = curl_init(); 
     curl_setopt($ch, CURLOPT_URL,$url);
     curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
     curl_setopt($ch, CURLOPT_HEADER, 0);
     $outputJson = curl_exec($ch);
     if ($outputJson === FALSE) {
        echo 'Sorry, This service is currently unavailable: '. curl_error($ch);
     }
     return $outputJson;
 }

并用

调用它

get_data("http://blah.com");
get_data("http://blah1.com");

这可能不是最佳解决方案，但应该适用于简单实例

Answer 2

使用多个网址卷曲可以获得更好的性能。见： http://php.net/manual/en/function.curl-multi-exec.php

并且：

http://www.rustyrazorblade.com/2008/02/curl_multi_exec/

Answer 3

您可能想尝试遍历不同的网站：

 $aSites = array("http://site1.com","http://site2.com");
 for($x=0; $x<count($aSites); $x++){
     $ch = curl_init(); 
     curl_setopt($ch, CURLOPT_URL,$aSites[$x]);
     curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
     curl_setopt($ch, CURLOPT_HEADER, 0);
     $outputJson = curl_exec($ch);
     if ($outputJson === FALSE) {
        echo 'Sorry, This service is currently unavailable: '. curl_error($ch);
    }
 }

Answer 4

<?
$url1 = "http://site1.com";  
$url2 = "http://site2.com";  

$ch = curl_init(); 
curl_setopt($ch, CURLOPT_URL, $url1);
curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
curl_setopt($ch, CURLOPT_HEADER, 0);
$outputJson = curl_exec($ch);
curl_setopt($ch, CURLOPT_URL, $url2);
$outputJson2 = curl_exec($ch);
curl_close($ch);

if ($outputJson === FALSE || $outputJson2 === FALSE) {
    echo 'Sorry, This service is currently unavailable: '. curl_error($ch);
}
?>

PHP如何使用curl打开几个源？

4 个答案: