Php大小寫不敏感的單詞從句子替換

Question

我需要從句子中替換匹配單詞。我正在使用下面的內容，但區分大小寫。我需要不區分大小寫。

$originalString = 'This is test.'; 
$findString = "is"; 
$replaceWith = "__"; 

$replacedString = (str_ireplace($findString, $replaceWith, $originalString)); 
// output : Th__ __ test. 

然後我試着

$replacedString = preg_replace('/\b('.preg_quote($findString).')\b/', $replaceWith, $originalString); 
// output : This __ test. 

它按預期工作正常，但如果我使用$findString = "Is"或"iS"或"IS"那麼它不工作。 有人可以建議我什麼是正則表達式。以獲得不區分大小寫的替換或任何其他方式來實現期望的結果。

修訂

按@nu11p01n73R答案我有以下改變，但在下面的例子是得到下降。

$originalString = 'Her (III.1) was the age of 2 years and 11 months. (shoulder a 4/5)'; 
$findStringArray = array("age", "(III.1)", "2", "months", "4"); 
foreach($findStringArray as $key => $value) { 
    $originalString = preg_replace('/\b('.preg_quote($value).')\b/i', "__", $originalString); 
} 
//Output : Her (III.1) was the __ of __ years and 11 __. (shoulder a __/5) 

//Output should be : Her __ was the __ of __ years and 11 __. (shoulder a 4/5) 

而且也是它停止工作，如果我在$findStringArray

Answer 1

所有你需要做的是一個ignore case modifier i添加到表達式的末尾作爲

$originalString = 'This Is test.'; 
$findString = "is"; 
$replaceWith = "__"; 
$replacedString = preg_replace('/\b('.preg_quote($findString).')\b/i', $replaceWith, $originalString); 
// output : This __ test. 

Answer 2

添加4/5你不能使用字邊界(III.1)或4/5，試試看：

$originalString = 'Her (III.1) was the age of 2 years and 11 months. (shoulder a 4/5)'; 
$findStringArray = array("age", "(III.1)", "2", "months", "4/5"); 
foreach($findStringArray as $key => $value) { 
    $originalString = preg_replace('~(?<=^|[. (])'.preg_quote($value).'(?=[.)]|$)~i', "__", $originalString); 
    //       __^  __^       __^ __^ 
} 
echo $originalString,"\n"; 

編輯：我已將/的分隔符更改爲~，並在查看中添加了括號。

輸出：

Her __ was the __ of __ years and 11 __. (shoulder a __)