svn - 根据 svn 状态为文件名着色

Question

调用 ls 时，我希望文件名具有不同的颜色，具体取决于它们的颠覆状态。例如，添加的文件将为青色，修改的文件为红色等。 bash 的强大功能是否有可能？在这方面有什么准备吗？

Answer 1

据我所知，用纯 bash (脚本除外)是不可能实现的。

你可以很容易地使用脚本(bash、python、perl，无论你的毒药)获得彩色文件列表。这是一个用 python 编写的相当粗略的概念验证实现:https://gist.github.com/776093

#!/usr/bin/env python
import re
from subprocess import Popen, PIPE

colormap = {
    "M" : "31", # red
    "?" : "37;41", # grey
    "A" : "32", # green
    "X" : "33", # yellow
    "C" : "30;41", # black on red
    "-" : "31", # red
    "D" : "31;1", # bold red
    "+" : "32", # green
}
re_svnout = re.compile(r'(.)\s+(.+)$')
file_status = {}


def colorise(line, key):
    if key in colormap.keys():
        return "\001\033[%sm%s\033[m\002" % (colormap[key], line)
    else:
        return line

def get_svn_status():
    cmd = "svn status"
    output = Popen(cmd, shell=True, stdout=PIPE)
    for line in output.stdout:
        match = re_svnout.match(line)
        if match:
            status, f = match.group(1), match.group(2)

            # if sub directory has changes, mark it as modified
            if "/" in f:
                f = f.split("/")[0]
                status = "M"

            file_status[f] = status

if __name__ == "__main__":
    get_svn_status()
    for L in Popen("ls", shell=True, stdout=PIPE).stdout:
        line = L.strip()
        status = file_status.get(line, False)
        print colorise(line, status)