Question

$ptn = "/^Response.+?[:] /";
$str = "Response from Moore Auto: Thanks for your feedback";
$rpltxt = "";
echo preg_replace($ptn, $rpltxt, $str);

“Moore Auto”是一个变量名，所以我只需要冒号和空格后面的文字。在这种情况下，期望的最终结果将是字符串“感谢您的反馈”。非常感谢！

Answer 1

简单地使用substr()，就像这样：

$str = 'Response from Moore Auto: Thanks for your feedback';
echo substr($str, strpos($str,':')+2);  //echoes "Thanks for your feedback"

Answer 2

如果有多个冒号，Damiens解决方案不起作用。如果第一部分不包含冒号，这应该始终有效：

<?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
}
?>

Answer 3

尝试

<?php 
$ptn = "/^(Response.+[:])(.*?)/";
$str = "Response from Moore Auto: Thanks for your feedback";
$rpltxt = "$2";
echo preg_replace($ptn, $rpltxt, $str);
?>

PHP正则表达式：字符串以及以...结尾

3 个答案: