我需要提取看起来像这样的字符串的值:
nameClass (val1)(val2)
拥有:
nameClass
val1
val2
问题在于它也必须适用于此:
nameClass
和
nameClass (val1)(val2)(val1)...(valn)
我尝试创建正则表达式,但它只适用于
nameClass (val1)(val2)
变体,看起来像这样(经过 Viorel Moraru 改进后):
String pattern = "((?:[a-z]+[A-Z][a-z]+))(([ |(]+)([-|+]?\\d+)([ |(|)]+)([-|+]?\\d+)([ |)]+))*";
如何使模式适用于所有
nameClass
和
nameClass (val1)(val2)(val1)...(valn)
Java代码:
String txt = "inputTestdata(12)(-13)";
String patern = "((?:[a-z]+[A-Z][a-z]+))([ |(]+)([-|+]?\\d+)([ |(|)]+)([-|+]?\\d+)([ |)]+)";
Pattern p = Pattern.compile(patern);
Matcher m = p.matcher(txt);
if (m.find())
{
for (int i = 1; i < m.groupCount(); i ++)
{
System.out.print(m.group(i) + "\n");
}
}
答案 0 :(得分:4)
您可以使用此代码:
String s = "nameClass(val1)(val2)(val3)";
Pattern p = Pattern.compile("^(\\w+) *(.*)$");
Matcher m = p.matcher(s);
String ps = "";
if (m.matches())
{
ps = m.group(2);
System.out.printf("Outside parantheses:<%s>\n", m.group(1));
}
Pattern p1 = Pattern.compile("\\(([^)]*)\\)");
Matcher m1 = p1.matcher(ps);
while (m1.find())
{
System.out.printf("Inside parentheses:<%s>%n", m1.group(1));
}
<强>输出:强>
Outside parantheses:<nameClass>
Inside parentheses:<val1>
Inside parentheses:<val2>
Inside parentheses:<val3>
答案 1 :(得分:3)
假设:
nameClass
...为什么不用括号替换括号内的所有内容?
例如:
Pattern p = Pattern.compile("\\((.+?)\\)");
String[] inputs = {"nameClass", "nameClass (var1)", "nameClass (var1) (var2)"};
Matcher m;
for (String input: inputs) {
m = p.matcher(input);
System.out.println("Input: " + input + " --> replacement: " + m.replaceAll("$1"));
// resetting matcher after "replaceAll" and accessing values directly by group 1 reference
m.reset();
while (m.find()) {
System.out.println("\tFound value: " + m.group(1));
}
}
输出:
Input: nameClass --> replacement: nameClass
Input: nameClass (var1) --> replacement: nameClass var1
Found value: var1
Input: nameClass (var1) (var2) --> replacement: nameClass var1 var2
Found value: var1
Found value: var2
答案 2 :(得分:0)
我不是正则表达式专家,但这有用吗?
\s*\w+\s*(\(\w+\))*