问题
I am trying below code but getting error
String x = "aaa XXX bbb";
String replace = "XXX";
String y = "xy$z";
String z=y.replaceAll("$", "\\$");
x = x.replaceFirst(replace, z);
System.out.println(x);
Error
Exception in thread "main" java.lang.IllegalArgumentException: Illegal group reference
at java.util.regex.Matcher.appendReplacement(Unknown Source)
at java.util.regex.Matcher.replaceFirst(Unknown Source)
at java.lang.String.replaceFirst(Unknown Source)
at Test.main(Test.java:10)
I want result as
aaa xy$z bbb
回答1:
Use replace() instead, which doesn't use regular expressions, since you don't need them at all:
String x = "aaa XXX bbb";
String replace = "XXX";
String y = "xy$z";
x = x.replace(replace, y);
System.out.println(x);
This will print aaa xy$z bbb
, as expected.
回答2:
If the replacement string includes a dollar sign or a backslash character, you should use
Matcher.quoteReplacement()
So change
String z=y.replaceAll("$", "\\$");`
to
String z = Matcher.quoteReplacement(y);
回答3:
The reason for the error is that after the line:
String z=y.replaceAll("$", "\\$");
The value of z
is: xy$z$
what you really want to do is:
String x = "aaa XXX bbb";
String replace = "XXX";
String y = "xy\\$z";
x = x.replaceFirst(replace, y);
System.out.println(x);
which will output:
aaa xy$z bbb
回答4:
The problem id due to replaceFirst
The value of String z=y.replaceAll("$", "\\$");
is xy$z$
Replaces the first substring of this string that matches the given regular expression with the given replacement.
An invocation of this method of the form str.replaceFirst(regex, repl)
yields exactly the same result as the expression
Pattern.compile(regex).matcher(str).replaceFirst(repl)
Note that backslashes (\
) and dollar signs ($
) in the replacement string may cause the results to be different than if it were being treated as a literal replacement string;
来源:https://stackoverflow.com/questions/12208834/how-to-escape-in-java