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[英]Regular Expression to check if string ends with one underscore and two letters with php
[英]PHP Regular Expression: String starts with and ends with
$ptn = "/^Response.+?[:] /";
$str = "Response from Moore Auto: Thanks for your feedback";
$rpltxt = "";
echo preg_replace($ptn, $rpltxt, $str);
“Moore Auto”是一個變量名,所以我只需要冒號和空格后面的文字。 在這種情況下,期望的最終結果將是字符串“感謝您的反饋”。 非常感激!
使用substr()
簡單,如下所示:
$str = 'Response from Moore Auto: Thanks for your feedback';
echo substr($str, strpos($str,':')+2); //echoes "Thanks for your feedback"
如果有多個冒號,Damiens解決方案不起作用。 如果第一部分不包含冒號,這應該始終有效:
<?php
$ptn = "/^Response[^:]+:\s*(.*)$/";
$str = "Response from Moore Auto: Thanks for your feedback";
if (preg_match($ptn, $str, $match)) {
$text = $match[1];
echo $text; //Thanks for your feedback
}
?>
嘗試
<?php
$ptn = "/^(Response.+[:])(.*?)/";
$str = "Response from Moore Auto: Thanks for your feedback";
$rpltxt = "$2";
echo preg_replace($ptn, $rpltxt, $str);
?>
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