在Bash脚本中使用awk [英] using awk in Bash script

问题描述

我正在学习Bash脚本.我的 data.txt 包含:

I am learning Bash scripts. my data.txt contains:

A Orange
B Apple

我尝试过

echo "enter"
read fruit
awk -F'[: ]' '$2 == "Apple"' data.txt
a=`awk -F'[: ]' '\$2 == "$fruit"' data.txt` # returns null
echo "$a"

当我将变量'a'与它一起使用时，为什么我的awk命令不起作用?

Why isn't my awk command is not working when I am using my variable 'a' with it?

推荐答案

问题的症结在于您试图使用单引号字符串，就好像它是双引号一个.

The crux of the problem is that you're trying to use a single-quoted string as if it were a double-quoted one.

您想要使用时要做的事情:

'\$2 == "$fruit"' # DOES NOT WORK - nothing is escaped or expanded

是:

"\$2 == \"$fruit\""  # keep literal $2, expand $fruit

用单引号括起来的外壳程序字符串中的任何内容都是 ，因此不会扩展变量引用，也不需要转义任何内容-甚至也无法转义 ( \ 也将成为字符串的文字部分.

Anything inside a single-quoted shell string is taken literally, so variable references aren't expanded, and nothing needs escaping - nor indeed can be escaped (the \ would become a literal part of the string too).

也就是说，最好将shell的世界和Awk脚本分开，以防止混淆，这意味着:

That said, it's better to keep the worlds of the shell and the Awk script separate, so as to prevent confusion, which means:

• 使用单引号字符串作为Awk脚本.
• 通过Awk的 -v 选项将任何shell变量值传递给脚本，以创建 Awk 变量.

• Use a single-quoted string as the Awk script.
• Pass any shell-variable values to the script via Awk's -v option to create Awk variables.

如果我们将它们放在一起:

If we put it all together:

a=$(awk -F'[: ]' -v fruit="$fruit" '$2 == fruit' data.txt)

请注意，我使用了现代的命令替换语法 $(...)，即优于传统语法`...` .

Note that I've used modern command-substitution syntax $(...), which is preferable to legacy syntax `...`.

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