java - 包含开头有 4 个空格的行组

Question

我想将所有行包含在 <pre> 中开头有 4 个空格的标记。

我试过什么？

^[ ]{4}(.*)(?!=^[ ]{4})

DEMO

输入:

Here is the code:
    String name = "Jon";
    System.out.println("Hello "+name);
output:
    Hello Jon

实际输出:

Here is the code: 
    <pre>String name = "Jon";</pre> 
    <pre>System.out.println("Hello "+name);</pre>
output: 
    <pre>Hello Jon</pre> 

预期输出:

Here is the code:
<pre>
    String name = "Jon";
    System.out.println("Hello "+name);
</pre>
output:
<pre>
    Hello Jon
</pre>

Java 示例代码:

text.replaceAll(regex, "<pre>$1</pre>");

Answer 1

您可以使用:

String out = input.replaceAll("(?m)((?:^ {4}\\S.*$\\r?\\n)*^ {4}\\S.*$)", 
                              "<pre>\\n$1\\n</pre>");

RegEx Demo

解释:

(?m)                    # enable multilie mode
^ {4}\\S.*$             # match a line with exact 4 spaces at start
\\r?\\n                 # followed by a line feed character
(?:^ {4}\\S.*$\\r?\\n)* # match 0 or more of such lines
^ {4}\\S.*$             # followed by a line with exact 4 spaces at start
<pre>\\n$1\\n</pre>     # replace by <pre> newline matched block newline </pre>