如何匹配用户的输入'没有空格数组的元素与空白在PHP - How to match user's input without white space to array which has the element with white space in PHP?

How to match user's input without white space to array which has the element with white space in PHP?

假设我的数组[Table Tennis, Basketball, Swimming, ...]中有Table Tennis，用户输入是tabletennis，我如何使它匹配?

这是我目前拥有的，$q是用户输入，$skills是我的数组:

if (preg_grep("/$q/i", $skills)){
    $skill = implode(" ", preg_grep("/$q/i", $skills));
    $searchRows[$key]['skills'] = '<a href="javascript:void(0)" class="label label-info">' . $skill . '</a>';
}else{
    $searchRows[$key]['skills'] = count($skills) . ' skills';
}

提前感谢您的帮助

也许你可以用一个更简单的解决方案?

    <?php
$testArray = array("Table Tennis", "Cooking", "Swimming"); //array to test
$testtoArray = array("tabletennis", "cooking", "swi mming");//array to test against
foreach($testArray as $arrVal)
{
    $arrValExploded = explode(" ", $arrVal);
    $var = "";
    for($i=0;$i<=(count($arrValExploded)-1);$i++)
    {
        $var = $var . lcfirst($arrValExploded[$i]);
    }
    if(in_array($var,$testtoArray))
    {
        echo $arrVal . "<br>"; //using this you can get the value from $testArray
        for($i=0;$i<count($testtoArray);$i++)
        {
            if($var == $testtoArray[$i])
            {
                echo $var . "<br>";
            }
        }
    }
    else
    {
        echo "$arrVal didn't match <br>";
    }
}
?>

这可以适用于任何数量的空格。例如:——"这是一个游戏"匹配" thisisagame "。

希望strtolower ( string $string );和str_replace(' ', '', $string);能帮到你。

例子:

$text = "Table Tennis s";    
$uinput = "tabletennis";    
$withoutwhitespace = str_replace(' ', '', $text);
$strarray = strtolower ($withoutwhitespace);
if($uinput == $strarray) {
  echo " Fine ";
} else {
  echo " Not Fine ";
}