如何使用环境变量在bash脚本引用字符串内 [英] How to use environment variable inside a quoted string in BASH script

问题描述

我已经试过各种形式在bash脚本如下：

I've tried various forms of the following in a bash script:

#!/bin/bash
svn diff $@ --diff-cmd /usr/bin/diff -x "-y -w -p -W $COLUMNS"

但我不能让语法正确展开列环境变量。

我已经试过各种形式如下：

I've tried various forms of the following:

svn diff $@ --diff-cmd /usr/bin/diff -x '-y -w -p -W $COLUMNS'

和

svn diff $@ --diff-cmd /usr/bin/diff -x '-y -w -p -W ${COLUMNS}'

和

eval svn diff $@ --diff-cmd /usr/bin/diff -x "-y -w -p -W $COLUMNS"

建议？

推荐答案

如果不确定，可以使用终端上的COLS的要求，而忘记列：

If unsure, you might use the 'cols' request on the terminal, and forget COLUMNS:

COLS=$(tput cols)

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