如何解决我可以在某些字符之间显示字符串吗？
我有一个巨大的日志文件，里面有数百个异常，每个异常也有发生的时间，但它们不一定在每个异常的同一个地方（在堆栈跟踪中）。我能看到的唯一共同点是它们总是在方括号之间。
我知道日志文件中的日期总是采用 [25/05/21 10:28:41:231 BST] 的形式。有没有一种方法可以只在 List<String> 上显示“[]”之间的字符。或者只是在字符串上，如果我可以将其解析为字符串。我当时正在考虑编写一些逻辑来测试字符串 25/05/21 10:28:41:231 BST 是否是一个日期，以避免在括号之间有更多数据时得到不需要的结果。
到目前为止，我尝试用“[”拆分它，但遇到了一些问题。数据文件中有一些“[”，这意味着时间和日期不一定是 String 数组中的第一个输入，所以我不能全部选择它们。
解决方法
某种形式
"\[(\d\d/\d\d/\d\d \d\d:\d\d:\d\d:\d\d [A-Z]{3})\]"
例如在斯卡拉
scala> val pattern = "\\[(\\d\\d/\\d\\d/\\d\\d \\d\\d:\\d\\d:\\d\\d:\\d\\d\\d [A-Z]{3})\\]".r
val pattern: scala.util.matching.Regex = \[(\d\d/\d\d/\d\d \d\d:\d\d:\d\d:\d\d\d [A-Z]{3})\]
scala> pattern.matches("[25/05/21 10:28:41:231 BST]")
val res5: Boolean = true
您可以试试这个 getBetween() 方法。您可能会发现它对其他方面也很有用：
/**
* Retrieves any string data located between the supplied string leftString
* parameter and the supplied string rightString parameter.<br><br>
* <p>
* This method will return all instances of a substring located between the
* supplied Left String and the supplied Right String which may be found
* within the supplied Input String.<br>
*
* @param inputString (String) The string to look for substring(s) in.<br>
*
* @param leftString (String) What may be to the Left side of the substring
* we want within the main input string. Sometimes the
* substring you want may be contained at the very
* beginning of a string and therefore there is no
* Left-String available. In this case you would simply
* pass a Null String ("") to this parameter which
* basically informs the method of this fact. Null can
* not be supplied and will ultimately generate a
* NullPointerException.<br>
*
* @param rightString (String) What may be to the Right side of the
* substring we want within the main input string.
* Sometimes the substring you want may be contained at
* the very end of a string and therefore there is no
* Right-String available. In this case you would simply
* pass a Null String ("") to this parameter which
* basically informs the method of this fact. Null can
* not be supplied and will ultimately generate a
* NullPointerException.<br>
*
* @param options (Optional - Boolean - 2 Parameters):<pre>
*
* ignoreLetterCase - Default is false. This option works against the
* string supplied within the leftString parameter
* and the string supplied within the rightString
* parameter. If set to true then letter case is
* ignored when searching for strings supplied in
* these two parameters. If left at default false
* then letter case is not ignored.
*
* trimFound - Default is true. By default this method will trim
* off leading and trailing white-spaces from found
* sub-string items. General sentences which obviously
* contain spaces will almost always give you a white-
* space within an extracted sub-string. By setting
* this parameter to false,leading and trailing white-
* spaces are not trimmed off before they are placed
* into the returned Array.</pre>
*
* @return (String[] Array) Returns a Single Dimensional String Array of all
* the sub-strings found within the supplied Input String which are
* between the supplied Left-String and supplied Right-String.
*/
public static String[] getBetween(String inputString,String leftString,String rightString,boolean... options) {
// Return null if nothing was supplied.
if (inputString.isEmpty() || (leftString.isEmpty() && rightString.isEmpty())) {
return null;
}
// Prepare optional parameters if any supplied.
// If none supplied then use Defaults...
boolean ignoreCase = false; // Default.
boolean trimFound = true; // Default.
if (options.length > 0) {
if (options.length >= 1) {
ignoreCase = options[0];
if (options.length >= 2) {
trimFound = options[1];
}
}
}
// Remove any control characters from the
// supplied string (if they exist).
String modString = inputString.replaceAll("\\p{Cntrl}","");
// Establish a List String Array Object to hold
// our found substrings between the supplied Left
// String and supplied Right String.
List<String> list = new ArrayList<>();
// Use Pattern Matching to locate our possible
// substrings within the supplied Input String.
String regEx = java.util.regex.Pattern.quote(leftString) + "{1,}"
+ (!rightString.isEmpty() ? "(.*?)" : "(.*)?")
+ java.util.regex.Pattern.quote(rightString);
if (ignoreCase) {
regEx = "(?i)" + regEx;
}
java.util.regex.Pattern pattern = java.util.regex.Pattern.compile(regEx);
java.util.regex.Matcher matcher = pattern.matcher(modString);
while (matcher.find()) {
// Add the found substrings into the List.
String found = matcher.group(1);
if (trimFound) {
found = found.trim();
}
list.add(found);
}
return list.toArray(new String[list.size()]);
}
要使用它，您可以执行以下操作：
String[] res = getBetween(logString,"[","]");
System.out.println(res[0]);
String[] parts = res[0].split("\\s+");
String date = parts[0];
String time = parts[1] + " " + parts[2];
System.out.println("Date: --> " + date);
System.out.println("Time: --> " + time);
版权声明：本文内容由互联网用户自发贡献，该文观点与技术仅代表作者本人。本站仅提供信息存储空间服务，不拥有所有权，不承担相关法律责任。如发现本站有涉嫌侵权/违法违规的内容， 请发送邮件至 dio@foxmail.com 举报，一经查实，本站将立刻删除。
