bash 中如何比较两个字符串

JBug

我按照Learning the bash shell中写的脚本, 运行不正确:

#!/bin/bash
# file name cmp.sh

if [ "$1" = "$2" ]; then
    echo "$1 is equal $2"
fi
if [ "$1" > "$2" ]; then
    echo "$1 large than $2"
fi
if [ "$1" < "$2" ]; then
    echo "$1 less than $2"
fi

输入:
$> source cmp.sh hello helo
输出:
hello large than helo
hello less than helo
好象只有(=, !=, -n, -z) 才能用于字符串操作.

那位dx指点一下怎么比较字符串.

kiron

据我所知，只有的确是只有(=, !=, -n, -z)才能用于字符串操作

moke33

if [ "-x$1" -eq "-x$2" ]; then
echo "$1 is equal $2"
elif [ "-x$1" > "-x$2" ]; then
echo "$1 large than $2"
else
echo "$1 less than $2"
fi

JBug

用-x也不行呀, 运行结果如下
$ ./cmp.sh hello helo
./cmp.sh: line 17: [: -xhello: integer expression expected
hello large than helo

tigerpower

#!/bin/bash
if [[ "$1" = "$2" ]]; then
echo "$1 is equal $2"
fi
if [[ "$1" > "$2" ]]; then
echo "$1 large than $2"
fi
if [[ "$1" < "$2" ]]; then
echo "$1 less than $2"
fi

《学习bash》，这本书我也有，不怎么灵的。。。

ebok

不小心挖坟了，在[..]中，要用到<或>比较字符串，需要用 \ 转义
如
[ "$a" \< "$b" ]