從網站

Question

獲取隨機地址我要搜索的鏈接或URL數量上http://public-domain-content.com 並將它們存儲在一個數組，然後就隨機選擇從陣列中的任何一個，只是顯示或回聲

我如何能做到這一點的PHP的

Answer 1

如果我明白你的要求，你可以做到這一點使用file_get_contents();

使用file_get_contents($url)，它給你一個字符串後，您可以遍歷結果字符串搜索空間，以分辨的話。計算單詞的數量，並相應地將單詞存儲在數組中。然後只需使用array_rand()

從陣列中選擇一個隨機元素然而，有時file_get_contents()存在安全問題。 您可以覆蓋這個使用下面的函數：

function get_url_contents($url) 
{ 
    $crl = curl_init(); 
    $timeout = 5; 
    curl_setopt ($crl, CURLOPT_URL,$url); 
    curl_setopt ($crl, CURLOPT_RETURNTRANSFER, 1); 
    curl_setopt ($crl, CURLOPT_CONNECTTIMEOUT, $timeout); 
    $ret = curl_exec($crl); 
    curl_close($crl); 

    return $ret; 
} 

http://php.net/manual/en/function.curl-setopt.php < ---約捲曲

示例代碼說明：

$url = "http://www.xxxxx.xxx";  //Set the website you want to get content from 
$str = file_get_contents($url); //Get the contents of the website 
$built_str = "";     //This string will hold the valid URLs 

$strarr = explode(" ", $str);  //Explode string into array(every space a new element) 

for ($i = 0; $i < count($strarr); $i++) //Start looping through the array 
{ 
    $current = @parse_url($strarr[$i]) //Attempt to parse the current element of the array 

    if ($current)      //If parse_url() returned true(URL is valid) 
    { 
     $built_str .= $current . " "; //Add the valid URL to the new string with " " 
    } 

    else 
    { 
     //URL invalid. Do something here 
    } 

} 

$built_arr = explode(" ", $built_str) //Same as we did with $str_arr. This is why we added a space to $built_str every time the URL was valid. So we could use it now to split the string into an array 

echo $built_arr[array_rand($built_arr)]; // Display a random element from our built array 

還有一個更擴展版本檢查網址，您可以在這裏探索：

http://forums.digitalpoint.com/showthread.php?t=326016

祝你好運。