删除字符串中的某些字符，Java
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/13229157/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Removing Certain Characters inside a String, Java
提问by Christian Eric Paran
My problem here is that i want a Character remove in some parts of a String but I do not know how to restrict the removing.
我的问题是我想在字符串的某些部分删除一个字符，但我不知道如何限制删除。
Example:
例子：
A computer is a general purpose device that can be\n
programmed to carry out a finite set of\n
millions to billions of times more capable.\n
\n
In this era mechanical analog computers were used\n
for military applications.\n
1.1 Limited-function early computers\n
1.2 First general-purpose computers\n
1.3 Stored-program architecture\n
1.4 Semiconductors and\n
this here example is the content of my string, what i want to happen is to remove the \nof lines 1 and 2 above but not to remove the \nin line 5 onwards. How do i remove the \nwithout removing the other \n?. My Goal here is to make the string a paragraph without \nafter line. like the example the first 3 lines can be a paragraph and the next lines are in bullet form(example). what i am saying is that I do not want to remove \nin bulleted characters.
这里的示例是我的字符串的内容，我想要发生的是删除\n上面的第 1 行和第 2 行，但不删除第\n5 行以后的内容。我如何删除\n而不删除另一个\n？. 我的目标是使字符串成为没有\n行后的段落。像示例一样，前 3 行可以是一个段落，下一行是项目符号形式（示例）。我的意思是我不想删除\n项目符号字符。
The real contents of the string is dynamic.
字符串的真实内容是动态的。
I have tried using String.replaceAll("\n", " ")well clearly that would not work it will remove all the \ni have thought of using Regexto determine what is Alphanumberic but it would remove some letters after \n
我已经尝试过String.replaceAll("\n", " ")清楚地使用它是行不通的，它会删除\n我想Regex用来确定什么是字母数字的所有内容，但它会在之后删除一些字母\n
回答by Rohit Jain
Try using this regex: -
尝试使用此正则表达式：-
str = str.replaceAll("(.+)(?<!\.)\n(?!\d)", " ");
System.out.println(str);
This will replace your \nif it is not precededby a dot - termination of a paragraph, and it is not followedby a digit, for when it is followed by a bulleted point. (like, your \nin first bullet point is followed by a 1.2. So, it will not be replaced.).
这将替换你的，\n如果它是not preceded由 a代替的dot - termination of a paragraph，它是not followed由 a digit，因为它后面是一个项目符号。（例如，您\n的第一个项目符号后跟一个1.2。因此，它不会被替换。）。
(.+)at the start, ensures that you are not replacing ablank line.
(.+)在开始时，确保您不会替换blank line.
This will work for the string you have shown.
这将适用于您显示的字符串。
Explanation: -
解释： -
(.+)-> A capture group, capturing anything, occurring at least once.(?<!\\.)-> This is callednegative-look-behind. It matches thestringfollowing it, only if that string is not preceded by adot(.)given in thenegative-look-behindpattern.
For e.g.: -You don't need to replace\nafter the line: -millions to billions of times more capable.\n.(?!\\d)-> This is callednegative -look-ahead. It matches string behind it, only if that string is not followed by adigit (\\d)given in thenegative-look-aheadpattern.
For e.g.: -In your bulleted points,computers\nis followed by1.2. where1is a digit. So, you don't want to replace that\n.Now,
$1and$2represent the groups captured in the pattern match. Since you just want to replace"\n". So, we took the remaining pattern match as it is, while replacing"\n"with aspace.
(.+)-> 一个捕获组，捕获任何东西，至少发生一次。(?<!\\.)-> 这叫做negative-look-behind. 它匹配string后面的它，仅当该字符串之前没有在模式中dot(.)给定negative-look-behind。
例如： -您不需要\n在行之后替换： -millions to billions of times more capable.\n。(?!\\d)-> 这叫做negative -look-ahead. 它匹配后面的字符串，仅当该字符串后面没有模式中的digit (\\d)给定negative-look-ahead。
例如： -在您的项目符号中，computers\n后跟1.2. 哪里1是数字。所以，你不想替换那个\n。现在，
$1并$2表示在模式匹配捕获的组。由于您只想替换"\n". 因此，我们按原样采用剩余的模式匹配，同时替换"\n"为space.
So, $1is representation for 1st group- (.+)
所以，$1是代表1st group-(.+)
Note, look-aheadand look-behindregexes are non-capturinggroups.
注意，look-ahead和look-behind正则表达式的non-capturing群体。
For More Details, follow these links: -
有关更多详细信息，请访问以下链接：-
回答by Vincent Jia
I suspect your requirement is to remove the \n of lines 1 and 2 .
What you can do is as below:
我怀疑您的要求是删除第 1 行和第 2 行的 \n 。
你可以做的是如下：
split your string into segments,
String[] array = yourString.split("\n");
concat every segments by adding \n tag, except line 1,2
array[1] + array[2] + array[3] + '\n' + array[4] + '\n' ...// and so forth
将你的字符串分成几段，
String[] array = yourString.split("\n");
通过添加 \n 标记连接每个段，第 1,2 行除外
数组[1] + 数组[2] + 数组[3] + '\n' + 数组[4] + '\n' ...// 等等
