簡單的捲曲網址存在將無法正常工作

Question

我已經寫在PHP

<?php 
error_reporting(1); 

function url_exists($url) 
{ 
    $ch = curl_init($url); 
    curl_setopt($ch, CURLOPT_HEADER, TRUE); 
    curl_setopt($ch, CURLOPT_NOBODY, TRUE); 
    curl_setopt($ch, CURLOPT_FOLLOWLOCATION, FALSE); 
    curl_setopt($ch, CURLOPT_RETURNTRANSFER, TRUE); 
    $status = array(); 
    preg_match('/HTTP\/.* ([0-9]+) .*/', curl_exec($ch) , $status); 
    return ($status[1] == 200); 
} 

echo "EX:".url_exists("http://www.google.com"); 

?> 

和

<?php 
error_reporting(1); 

$ch = curl_init("www.yahoo.com"); 
curl_setopt($ch, CURLOPT_TIMEOUT, 2); 
curl_setopt($ch, CURLOPT_CONNECTTIMEOUT, 2); 
curl_setopt($ch, CURLOPT_RETURNTRANSFER, true); 
$data = curl_exec($ch); 
$httpcode = curl_getinfo($ch, CURLINFO_HTTP_CODE); 
curl_close($ch); 
if($httpcode>=200 && $httpcode<300) 
{ 
    echo "Found@"; 
} 
else 
{ 
    echo "Not found"; 
} 

?> 

無本以上的作品2碼？爲什麼？？？昨天工作很好，爲什麼PHP不能解釋一些時間？

我從php.ini中...我想在3級不同的服務器curl擴展...

什麼，我做錯了什麼？謝謝。

Answer 1

echo "EX:".url_exists("http://www.google.com"); 

將始終回顯EX:，因爲url_exists返回布爾值。試試：

echo "EX:" . (url_exists("http://www.google.com") ? 'true' : 'false'); 

也許這樣的工作？

$ch = curl_init("http://www.example.com/favicon.ico"); 
curl_setopt($ch, CURLOPT_NOBODY, true); 
curl_exec($ch); 
$retcode = curl_getinfo($ch, CURLINFO_HTTP_CODE); 
// $retcode > 400 -> not found, $retcode = 200, found. 
curl_close($ch); 

來源https://stackoverflow.com/a/982045/956397