Question

我搜索过，找不到我的问题。我用linux输出ls -l保存了文件，内容是：

drwxr-xr-x  2 usr usr 4096 Jan 20 17:49 file1
drwxrwxr-x  4 usr usr 4096 Jan 20 18:00 file2
drwx------  2 usr usr 4096 Feb  3 08:48 catalog1

我想以小时为例离开第八列，并切断其余部分。我该怎么办？我是java和编程的初学者。

Answer 1

您可以使用正则表达式来匹配时间戳（因为它可以保证类似时间的值不会出现在任何其他字段中）。类似的东西：

// Populate this with the output of the ls -l command
String input;

// Create a regular expression pattern to match.
Pattern pattern = Pattern.compile("\\d{2}:\\d{2}");

// Create a matcher for this pattern on the input string.
Matcher matcher = pattern.matcher(input);

// Try to find instances of the given regular expression in the input string.    
while (matcher.find()){
  System.out.println(matcher.group());
}

要检索任意列，您可以选择为要检索的任何列编写正则表达式，或者您可能希望仅拆分空格字符上的每一行，然后按索引选择。例如，要获取所有文件大小：

String input;

String[] inputLines = input.split("\n");
for (String inputLine : inputLines) {
  String[] columns = inputLine.split(" ");
  System.out.println(columns[4]); // Where 4 indicates the filesize column
}

Answer 2

您需要使用StringTokenizer来提取您要查找的确切信息。请尝试以下代码：

String value =  "drwxr-xr-x  2 usr usr 4096 Jan 20 17:49 file1\n"+
                "drwxrwxr-x  4 usr usr 4096 Jan 20 18:00 file2\n"+
                "drwx------  2 usr usr 4096 Feb  3 08:48 catalog1";
StringBuffer sBuffer = new StringBuffer(10);
StringTokenizer sTokenizer = new StringTokenizer(value,"\n");
while (sTokenizer.hasMoreTokens())
{
    String sValue = sTokenizer.nextToken();
    StringTokenizer sToken = new StringTokenizer(sValue," ");
    int counter = 0;
    while (sToken.hasMoreTokens())
    {
        String token = sToken.nextToken();
        counter++;
        if (counter == 8)//8 is the column that you want to leave.
        {
            sBuffer.append(token+"\n");
            break;
        }
    }           
}
System.out.println(sBuffer.toString());

在java中从文件中删除列

2 个答案: