英文:
String matching in math expression x <= 900 (int, int)
问题
以下是翻译好的部分：
我有一个数字，比如说900。
我提取一个带有格式的字符串 "x <= int (int, int)"，例如： "x <= 900 (4, 7)"。
我想要验证它是否符合格式 "x <= 900 (int, int)"。
我如何使用字符串模式匹配器来实现呢？
我目前尝试了以下方法，但结果是false。
1) Pattern pattern = Pattern.compile("x <= 900 (\\d, \\d)");
2) Pattern pattern = Pattern.compile("x <= 900 ([0-9], [0-9])");
String expectedString = "x <= 900 (4, 7)";
Matcher m = pattern.matcher(expectedString);
System.out.println(m.matches());
英文:
I have the number, say 900.
I extract a string with the format "x <= int (int, int)" for example. "x <= 900 (4, 7)"
I want to verify that it matches the format "x <= 900 (int, int)"
How can I do it using String pattern matchers?
I have tried the following so far, the result is false.
1) Pattern pattern = Pattern.compile("x <= 900 (\\d, \\d)");
2) Pattern pattern = Pattern.compile("x <= 900 ([0-9], [0-9])");
String expectedString = "x <= 900 (4, 7)";
Matcher m = pattern.matcher(expectedString);
System.out.println(m.matches());
答案1
得分: 0
你需要使用反斜杠（\）来转义括号，因为它们是正则表达式元字符。要匹配一个或多个数字，使用\d+。
Pattern pattern = Pattern.compile("x <= \\d+ \\(\\d+, \\d+\\)");
<kbd>演示</kbd>
英文:
You need to escape the parentheses using backslashes (\) as they are regular expression metacharacters. To match one or more digits, use \d+.
Pattern pattern = Pattern.compile("x <= \\d+ \\(\\d+, \\d+\\)");
<kbd>Demo</kbd>
答案2
得分: 0
尝试转义括号 - 否则，它会将括号内的所有内容解释为一个组（而不是像您想要的那样将括号视为文字字符）。以下是我做的一个示例（诚然，我使用的是 .NET 语法，但对于 Java，我认为类似）：
^x <= [0-9]+ \\([0-9]+, [0-9]+\\)$
注意：在字符串中，您可能还需要转义 \\，所以字符串应该是：
String regex = "^x <= [0-9]+ \\([0-9]+, [0-9]+\\)$";
英文:
Try escaping the parenthesis - otherwise, it'll interpret everything inside the parenthesis as a group (instead of treating the parenthesis as a literal character like you want). Here's one I did (admittedly using .NET syntax, but it'll be similar for Java I think):
^x <= [0-9]+ \([0-9]+, [0-9]+\)$
NOTE: in the string, you'll likely need to escape the \ too, so the string would be:
String regex = "^x <= [0-9]+ \\([0-9]+, [0-9]+\\)$:
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