检查字符串是否包含数组中的值

Question

我试图检测字符串是否包含至少一个存储在数组中的URL.

这是我的数组:

$owned_urls = array('website1.com', 'website2.com', 'website3.com');

该字符串由用户输入并通过PHP提交.在确认页面上,我想检查输入的URL是否在数组中.

我尝试过以下方法:

$string = 'my domain name is website3.com';
if (in_array($string, $owned_urls))
{
    echo "Match found"; 
    return true;
}
else
{
    echo "Match not found";
    return false;
}

无论输入什么,返回总是"未找到匹配".

这是正确的做事方式吗？

Answer 1

试试这个.

$string = 'my domain name is website3.com';
foreach ($owned_urls as $url) {
    //if (strstr($string, $url)) { // mine version
    if (strpos($string, $url) !== FALSE) { // Yoshi version
        echo "Match found"; 
        return true;
    }
}
echo "Not found!";
return false;

如果要检查不区分大小写,请使用stristr()或stripos().

Answer 2

试试这个:

$owned_urls= array('website1.com', 'website2.com', 'website3.com');

$string = 'my domain name is website3.com';

$url_string = end(explode(' ', $string));

if (in_array($url_string,$owned_urls)){
    echo "Match found"; 
    return true;
} else {
    echo "Match not found";
    return false;
}

- 谢谢

Answer 3

简单str_replace的count参数可以在这里工作:

$count = 0;
str_replace($owned_urls, '', $string, $count);
// if replace is successful means the array value is present(Match Found).
if ($count > 0) {
  echo "One of Array value is present in the string.";
}

Answer 4

如果你想要做的就是在数组中找到一个字符串,那么这样做要容易得多.

$array = ["they has mystring in it", "some", "other", "elements"];
if (stripos(json_encode($array),'mystring') !== false) {
echo "found mystring";
}

Answer 5

$string = 'my domain name is website3.com';
$a = array('website1.com','website2.com','website3.com');

$result = count(array_filter($a, create_function('$e','return strstr("'.$string.'", $e);')))>0; 
var_dump($result );

产量

bool(true)

Answer 6

这是一个小函数，用于搜索给定字符串中数组中的所有值。我在我的网站中使用它来检查访客 IP 是否在某些页面的允许列表中。

function array_in_string($str, array $arr) {
    foreach($arr as $arr_value) { //start looping the array
        if (stripos($str,$arr_value) !== false) return true; //if $arr_value is found in $str return true
    }
    return false; //else return false
}

如何使用

$owned_urls = array('website1.com', 'website2.com', 'website3.com');

//this example should return FOUND
$string = 'my domain name is website3.com';
if (array_in_string($string, $owned_urls)) {
    echo "first: Match found<br>"; 
}
else {
    echo "first: Match not found<br>";
}

//this example should return NOT FOUND
$string = 'my domain name is website4.com';
if (array_in_string($string, $owned_urls)) {
    echo "second: Match found<br>"; 
}
else {
    echo "second: Match not found<br>";
}

演示： http: //phpfiddle.org/lite/code/qf7j-8m09

• stripos功能不是很严格。它不区分大小写，或者可以匹配单词的一部分 http://php.net/manual/ro/function.stripos.php

• 如果您希望搜索区分大小写，请使用strpos http://php.net/manual/ro/function.strpos.php

• 对于精确匹配使用正则表达式（preg_match），请检查这个人的答案/sf/answers/1794371561/

Answer 7

我认为更快的方法是使用preg_match。

$user_input = 'Something website2.com or other';
$owned_urls_array = array('website1.com', 'website2.com', 'website3.com');

if ( preg_match('('.implode('|',$owned_urls_array).')', $user_input)){
    echo "Match found"; 
}else{
    echo "Match not found";
}