解析“ *” -量词{x，y}不跟随 [英] parsing "*" - Quantifier {x,y} following nothing

问题描述

失败。
我该如何解决？

fails when I try Regex.Replace() method. how can i fix it?

Replace.Method (String, String, MatchEvaluator, RegexOptions)

我尝试输入代码

<%# Regex.Replace( (Model.Text ?? "").ToString(), patternText, "<b>" + patternText + "</b>", RegexOptions.IgnoreCase | RegexOptions.Multiline)%>

推荐答案

您是否尝试仅使用字符串 * 作为正则表达式？至少这就是导致您在此处出现错误的原因：

Did you try using only the string "*" as a regular expression? At least that's what causes your error here:

PS Home:\> "a" -match "*"
The '-match' operator failed: parsing "*" - Quantifier {x,y} following nothing..
At line:1 char:11
+ "a" -match  <<<< "*"

字符 * 是特殊字符在正则表达式中，因为它允许前置标记出现零次或多次。

The character * is special in regular expressions as it allows the preceding token to appear zero or more times. But there actually has to be something preceding it.

如果要匹配文字星号，请使用 \ * 作为正则表达式。否则，您需要指定可能重复的内容。例如，正则表达式 a * 不匹配任何内容，或者匹配任意多个 a s。

If you want to match a literal asterisk, then use \* as regular expression. Otherwise you need to specify what may get repeated. For example the regex a* matches either nothing or arbitrary many as in a row.

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