在 bash 中连接两个没有换行符的命令的输出 [英] Concatenate in bash the output of two commands without newline character

问题描述

我需要什么:

假设我有两个命令，A 和 B，每个命令都返回一个单行字符串(即，一个没有换行符的字符串，除了在最后).我需要一个命令(或管道命令序列)C 将命令 A 和 B 的输出连接到同一行并插入 1 个空格他们之间的性格.

Suppose I have two commands, A and B, each of which returns a single-line string (i.e., a string with no newline character, except possibly 1 at the very end). I need a command (or sequence of piped commands) C that concatenates the output of commands A and B on the same line and inserts 1 space character between them.

工作原理示例:

例如，假设命令 A 的输出是这里的引号 之间的字符串:

For example, suppose the output of command A is the string between the quotation marks here:

"The quick"

假设命令 B 的输出是这里的引号 之间的字符串:

And suppose the output of command B is the string between the quotation marks here:

"brown fox"

然后我希望命令C的输出是在引号之间的字符串:

Then I want the output of command(s) C to be the string between the quotation marks here:

"The quick brown fox"

我尝试过的最佳解决方案:

在尝试自己找出 C 时，似乎以下管道命令序列应该起作用:

In trying to figure out C by myself, it seemed that the follow sequence of piped commands should work:

{ echo "The quick" ; echo "brown fox" ; } | xargs -I{} echo {} | sed 's/
//'

不幸的是，这个命令的输出是

Unfortunately, the output of this command is

The quick
brown fox

推荐答案

你可以使用tr:

{ echo "The quick"; echo "brown fox"; } | tr "
" " "

或使用 sed:

{ echo "The quick"; echo "brown fox"; } | sed ':a;N;s/
/ /;ba'

输出:

The quick brown fox 

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