我试图检测一个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。
我尝试了以下方法：
$string = 'my domain name is website3.com';
if (in_array($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？
尝试这个。
$string = 'my domain name is website3.com';
foreach ($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() 。
尝试这个：
$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
如果您只想在数组中找到一个字符串，这会容易得多。
$array = ["they has mystring in it", "some", "other", "elements"];
if (stripos(json_encode($array),'mystring') !== false) {
echo "found mystring";
}
简单str_replace的计数参数将在这里工作：
$count = 0;
str_replace($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.";
}
更多信息 - https://www.techpurohit.com/extended-behaviour-explode-and-strreplace-php
$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 );
输出
bool(true)
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";
}
您可以使用 implode 和分隔符 | 连接数组值 然后使用 preg_match 搜索该值。
这是我想出的解决方案...
$emails = array('@gmail', '@hotmail', '@outlook', '@live', '@msn', '@yahoo', '@ymail', '@aol');
$emails = implode('|', $emails);
if(!preg_match("/$emails/i", $email)){
// do something
}
这是一个从给定字符串中的数组中搜索所有值的迷你函数。我在我的网站中使用它来检查访问者 IP 是否在某些页面的允许列表中。
function array_in_string($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的功能不是很严格。它不区分大小写或可以匹配单词的一部分 http://php.net/manual/ro/function.stripos.php
如果您希望该搜索区分大小写，请使用strpos http://php.net/manual/ro/function.strpos.php
对于完全匹配使用正则表达式（preg_match），检查这个人的答案https://stackoverflow.com/a/25633879/4481831
If your $string is always consistent (ie. the domain name is always at the end of the string), you can use explode() with end(), and then use in_array() to check for a match (as pointed out by @Anand Solanki in their answer).
If not, you'd be better off using a regular expression to extract the domain from the string, and then use in_array() to check for a match.
$string = 'There is a url mysite3.com in this string';
preg_match('/(?:http:\/\/)?(?:www.)?([a-z0-9-_]+\.[a-z0-9.]{2,5})/i', $string, $matches);
if (empty($matches[1])) {
// no domain name was found in $string
} else {
if (in_array($matches[1], $owned_urls)) {
// exact match found
} else {
// exact match not found
}
}
The expression above could probably be improved (I'm not particularly knowledgeable in this area)
$owned_urls= array('website1.com', 'website2.com', 'website3.com');
$string = 'my domain name is website3.com';
for($i=0; $i < count($owned_urls); $i++)
{
if(strpos($string,$owned_urls[$i]) != false)
echo 'Found';
}
您正在检查整个字符串到数组值。所以输出总是false.
在这种情况下，我同时使用array_filter和。strpos
<?php
$urls= array('website1.com', 'website2.com', 'website3.com');
$string = 'my domain name is website3.com';
$check = array_filter($urls, function($url){
global $string;
if(strpos($string, $url))
return true;
});
echo $check?"found":"not found";
我发现这快速简单，无需运行循环。
$array = array("this", "that", "there", "here", "where");
$string = "Here comes my string";
$string2 = "I like to Move it! Move it";
$newStr = str_replace($array, "", $string);
if(strcmp($string, $newStr) == 0) {
echo 'No Word Exists - Nothing got replaced in $newStr';
} else {
echo 'Word Exists - Some Word from array got replaced!';
}
$newStr = str_replace($array, "", $string2);
if(strcmp($string2, $newStr) == 0) {
echo 'No Word Exists - Nothing got replaced in $newStr';
} else {
echo 'Word Exists - Some Word from array got replaced!';
}
小解释！
$newStr用原始字符串数组中的替换值创建新变量。
进行字符串比较 - 如果值为 0，则表示字符串相等且没有任何内容被替换，因此字符串中不存在数组中的值。
如果是 2 的反之亦然，即在进行字符串比较时，原始字符串和新字符串都不匹配，这意味着有东西被替换了，因此数组中的值存在于字符串中。
我想出了这个对我有用的功能，希望这对某人有帮助
$word_list = 'word1, word2, word3, word4';
$str = 'This string contains word1 in it';
function checkStringAgainstList($str, $word_list)
{
$word_list = explode(', ', $word_list);
$str = explode(' ', $str);
foreach ($str as $word):
if (in_array(strtolower($word), $word_list)) {
return TRUE;
}
endforeach;
return false;
}
另外，请注意，如果匹配的单词是其他单词的一部分，则使用 strpos() 的答案将返回 true。例如，如果单词列表包含“st”并且您的字符串包含“street”，则 strpos() 将返回 true
$message = "This is test message that contain filter world test3";
$filterWords = array('test1', 'test2', 'test3');
$messageAfterFilter = str_replace($filterWords, '',$message);
if( strlen($messageAfterFilter) != strlen($message) )
echo 'message is filtered';
else
echo 'not filtered';
$search = "web"
$owned_urls = array('website1.com', 'website2.com', 'website3.com');
foreach ($owned_urls as $key => $value) {
if (stristr($value, $search) == '') {
//not fount
}else{
//found
}
这是搜索任何子字符串、不区分大小写且快速的最佳方法
就像我的mysql一样
前任：
select * from table where name = "%web%"