How do I remove all the ending caret characters from end of a string?
如何从字符串末尾删除所有结束插入符？
For example: I have a string with value "t^e^st ^^^^^^^^" My output should be "t^e^st". How can I do this using Java regex?
例如：我有一个值为“t ^ e ^ st ^^^^^^^^”的字符串我的输出应为“t ^ e ^ st”。如何使用Java正则表达式执行此操作？
2 个解决方案
#1
2
"t^e^st ^^^^^^^^".replaceAll("\\^*$", '')
The caret also needs to be escaped. \\s* takes care of any trailing whitespace also.
插入符也需要逃脱。 \\ s *也会处理任何尾随空格。
#2
0
Try anchoring a repeated ^ to the end of the String, like this:
尝试将重复的^锚定到String的末尾，如下所示：
"t^e^st ^^^^^^^^".replaceAll("(.*)\\^*$", "$1").trim();
“t ^ e ^ st ^^^^^^^^”。replaceAll（“（。*）\\ ^ * $”，“$ 1”）。trim（）;
This looks for and captures any characters ((.*)) that are followed by 0 or more carats (\\^*) at the end of the string ($). Then it replaces all of that with the first capture group ($1) and trims the string to remove any leading/trailing spaces. The result should be just "t^e^st".
这会查找并捕获字符串末尾（0）后面跟着0或更多克拉（\\ ^ *）的任何字符（（。*））。然后它用第一个捕获组（$ 1）替换所有这些并修剪字符串以删除任何前导/尾随空格。结果应该只是“t ^ e ^ st”。
#1
2
"t^e^st ^^^^^^^^".replaceAll("\\^*$", '')
The caret also needs to be escaped. \\s* takes care of any trailing whitespace also.
插入符也需要逃脱。 \\ s *也会处理任何尾随空格。
#2
0
Try anchoring a repeated ^ to the end of the String, like this:
尝试将重复的^锚定到String的末尾，如下所示：
"t^e^st ^^^^^^^^".replaceAll("(.*)\\^*$", "$1").trim();
“t ^ e ^ st ^^^^^^^^”。replaceAll（“（。*）\\ ^ * $”，“$ 1”）。trim（）;
This looks for and captures any characters ((.*)) that are followed by 0 or more carats (\\^*) at the end of the string ($). Then it replaces all of that with the first capture group ($1) and trims the string to remove any leading/trailing spaces. The result should be just "t^e^st".
这会查找并捕获字符串末尾（0）后面跟着0或更多克拉（\\ ^ *）的任何字符（（。*））。然后它用第一个捕获组（$ 1）替换所有这些并修剪字符串以删除任何前导/尾随空格。结果应该只是“t ^ e ^ st”。