使用bash /剪切/分割提取字符串的一部分 [英] extract part of a string using bash/cut/split

问题描述

我有一个这样的字符串：

I have a string like this:

/var/cpanel/users/joebloggs:DNS9=domain.com

我需要提取用户名：此字符串从joebloggs并将其存储在一个变量

I need to extract the username: joebloggs from this string and store it in a variable

该字符串的格式将永远与joebloggs的例外，domain.com所以我想字符串可以用一刀切分裂两次？

The format of the string will always be the same with exception of joebloggs and domain.com so I am thinking the string can be split twice using "cut"?

首先将拆分使用分割字符串了。我们将在第一部分存储在varibale传递给第二分割功能

First split would split the string up using : and we would store the first part in a varibale to pass to the second split function.

二分裂将使用分割字符串/和最后一个字（joebloggs）存储到一个变量

Second split would split the string using / and store the last word (joebloggs) into a variable

我知道如何使用数组和分裂做到这一点在PHP，但在bash我有点失去了。

I know how to do this in php using arrays and splits but in bash I am a bit lost.

推荐答案

要使用参数扩展而无需任何额外的进程... <从这个字符串在bash提取物 joebloggs / p>

To extract joebloggs from this string in bash using parameter expansion without any extra processes...

MYVAR="/var/cpanel/users/joebloggs:DNS9=domain.com" 

NAME=${MYVAR%:*}  # get the part before the colon
NAME=${NAME##*/}  # get the part after the last slash
echo $NAME

不依赖于 joebloggs 是在路径特定深度。

如果你想要去一​​个有点过分使用的grep ：

If you want to go a bit overboard using grep:

echo MYVAR | grep -oE '/[^/]+:' | cut -c2- | rev | cut -c2- | rev

---

摘要

几个参数扩展模式的概述，以供参考...

An overview of a few parameter expansion modes, for reference...

${MYVAR#pattern}       # delete shortest match of pattern from the beginning
${MYVAR##pattern}      # delete longest match of pattern from the beginning
${MYVAR%pattern}       # delete shortest match of pattern from the end
${MYVAR%%pattern}      # delete longest match of pattern from the end

所以＃表示从一开始比赛（想到一个注释行）和％从最终手段。一个实例是指最短的两个实例是指最长的。

So # means match from the beginning (think of a comment line) and % means from the end. One instance means shortest and two instances means longest.

您也可以替换使用特定的字符串或模式：

You can also replace particular strings or patterns using:

${MYVAR/search/replace}

的模式是相同的格式，文件名匹配，因此 * （任何字符）是很常见，常接一个特定符号如 / 或。

The pattern is in the same format as file-name matching, so * (any characters) is common, often followed by a particular symbol like / or .

例子：

给定一个变量像

MYVAR="users/joebloggs/domain.com" 

删除离开路径文件名（所有字符以斜线）：

Remove the path leaving file name (all characters up to a slash):

echo ${MYVAR##*/}
domain.com

删除文件名，留下的路径（后删除最短的匹配最后 / ）

echo ${MYVAR%/*}
users/joebloggs

得到的只是文件扩展名（最后期限前删除所有）：

Get just the file extension (remove all before last period):

echo ${MYVAR##*.}
com

的注：的做两次手术，你不能将它们合并，但分配给一个中间变量。因此，要获得该文件的名称不带路径或扩展名：

NOTE: To do two operations, you can't combine them, but have to assign to an intermediate variable. So to get the file name without path or extension:

NAME=${MYVAR##*/}      # remove part before last slash
echo ${NAME%.*}        # from the new var remove the part after the last period
domain

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