Question

我需要从句子中替换匹配的单词。我正在使用下面但它区分大小写。我需要不区分大小写。

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

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

然后我试过

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

它按预期正常工作，但如果我使用$findString = "Is"或"iS"或"IS"，那么它无效。任何人都可以告诉我什么是正则表达式。获得不区分大小写的替换或任何其他方式来实现期望的结果。

已更新

根据@nu11p01n73R回答，我已在下方进行了更改，但在下面的示例中，它会下降。

$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)

如果我在4/5上添加$findStringArray

，它也会停止工作

Answer 1

您需要做的就是在正则表达式的末尾添加ignore case modifier i

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

Answer 2

您不能在(III.1)或4/5上使用字边界，请尝试使用：

$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";

修改：我已将分隔符从/更改为~，并在外观中添加了括号。

<强>输出：

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

从句子中取代Php案例不敏感的单词

2 个答案: