Question

我想替换字符串（或说一个段落）中所有出现的特定术语。

这些术语可以是Blue Restaurant Ocean / Blue Restaurant，Blue's Restaurant。

$inputext = "The name is Ocean/Blue Restaurant "; 
$term = "Ocean/Blue Restaurant";
$rule = "/". $term."/i";
$replacetext = "[X]";
$outputext = preg_replace($rule, $replacetext, $inputext);
echo $outputext,"\n";

所需的输出

 The name is [X]

当输入字符“ /”时，它不起作用。

我的主要目标是替换段落中所有出现的短语。是否有使用preg_replace的通用方法？

谢谢您的帮助。

Answer 1

使用preg-quote

进行逃脱示例

<a href="#'.$updateid.'" attr="'.$_SESSION['id'].'"  type="'.$updateid.'" id="'.$f_uid.'"  class="report_offence glyphicon glyphicon-warning-sign" title="Edit this status" >Report</a>

使用$inputext = "Those terms can be Blue Restaurant Ocean/Blue Restaurant, Blue's Restaurant."; // you put your phrases here, remember to escape // double quote if you want to use it as phrase $termList = [ "Ocean Restaurant", "Ocean/Blue Restaurant", "Blue Restaurant", "Blue's Restaurant", "so called \"double quote\" probably", ]; $pattern = ''; $termsCount = count($termList); for ($i=0; $i<$termsCount; $i++) { $item = '(' . preg_quote($termList[$i], '/') . ')'; if ($i < $termsCount -1) { $item .= '|'; } $pattern .= $item; } $rule = '/' . $pattern . '/i'; $replacetext = "[X]"; $outputext = preg_replace($rule, $replacetext, $inputext); echo $outputext,"\n";和array_map的示例：

implode

Answer 2

您的代码存在的问题是，您应转义正则表达式中用作定界符的字符。此外，您不需要preg_*函数。您只需要strtr()：

$str = "The name is Ocean/Blue Restaurant Blue Restaurant Blue's Restaurant"; 
echo strtr($str, ["Blue Restaurant" => "[x]", "Ocean/Blue Restaurant" => "[x]", "Blue's Restaurant" => "[x]"]);

输出：

The name is [x] [x] [x]

处理字符/ preg_replace中-PHP

2 个答案: