我想匹配以同一报价开始和结尾的字符串,但仅以开始和结尾的引号:
"foo bar"
'foo bar'
"the quick brown fox"
,但我不希望这些匹配或剥离:
foo "bar"
foo 'bar'
'foo bar"
"the lazy" dogs
我尝试了此Java Regexp,但对所有情况都没有用:
Pattern.compile("^"|^'(.+)"$|'$").matcher(quotedString).replaceAll("");
我认为有一种方法可以做lookahead,但在这种情况下我不知道如何使用它。
或设置单独检查它们的IF语句会更有效?
Pattern startPattern = Pattern.compile("^"|^');
Pattern endPattern = Pattern.compile("$|'$");
if (startPattern.matcher(s).find() && endPattern.matcher(s).find()) {
...
}
(当然,这与我不想要的'foo bar"
匹配(
您正在寻找的正则是
^(["'])(.*)1$
替换字符串为 "$2"
:
Pattern pattern = Pattern.compile("^(["'])(.*)\1$");
String output = pattern.matcher(input).replaceAll("$2");
演示:https://ideone.com/3a5pet
这是一种可以检查您所有要求的方法:
public static boolean matches (final String str)
{
boolean matches = false;
// check for null string
if (str == null) return false;
if ((str.startsWith(""") && str.endsWith(""")) ||
(str.startsWith("'") && str.endsWith("'")))
{
String strip = str.substring(1, str.length() - 1);
// make sure the stripped string does not have a quote
if (strip.indexOf(""") == -1 && strip.indexOf("'") == -1)
{
matches = true;
}
}
return matches;
}
测试:
public static void main(String[] args)
{
System.out.println("Should Passn-----------");
System.out.println(matches(""foo bar""));
System.out.println(matches("'foo bar'"));
System.out.println(matches(""the quick brown fox""));
System.out.println("nShould Failn-----------");
System.out.println(matches("foo "bar""));
System.out.println(matches("foo 'bar'"));
System.out.println(matches("'foo bar""));
System.out.println(matches(""the lazy" dogs"));
}
输出:
Should Pass
-----------
true
true
true
Should Fail
-----------
false
false
false
false