Question

在定义了此功能的脚本中：

#!/bin/bash --

outtrue  (){ printf "%15s is T|%s" "$*" "$?"; }
outfalse (){ printf "%15s is F|%s" "$*" "$?"; }

tval1(){  [ "$@" ]  && outtrue "$@" || outfalse "$@"; }
tval2(){  [ "$*" ]  && outtrue "$@" || outfalse "$@"; }
tval3(){ [[ "$@" ]] && outtrue "$@" || outfalse "$@"; }

tval(){
    printf "test %s\t" "$1"; shift
    case $1 in
        (1) shift; tval1 "$@" ;;
        (2) shift; tval2 "$@" ;;
        (3) shift; tval3 "$@" ;;
    esac
    printf "\n"
}

此测试工作完美（如预期）：

xyz="str";  tval "-n var"       1   "-n" "$xyz"
xyz="str";  tval "-z var"       1   "-z" "$xyz"
one=1;      tval "1 -eq \$one"  1   "1" "-eq" "$one"
one=2;      tval "1 -eq \$one"  1   "1" "-eq" "$one"

结果：

test -n var              -n str is T|0
test -z var              -z str is F|1
test 1 -eq $one         1 -eq 1 is T|0
test 1 -eq $one         1 -eq 2 is F|1

然而，改变＆＃34; $ @＆＃34;到＆＃34; $ *＆＃34;或使用bash测试[[...]]制作测试失败（请注意，此处使用了函数2或3）：

one=1;  tval "1 -eq \$one"  2   "1" "-eq" "$one"
one=2;  tval "1 -eq \$one"  2   "1" "-eq" "$one"

one=1;  tval "1 -eq \$one"  3   "1" "-eq" "$one"
one=2;  tval "1 -eq \$one"  3   "1" "-eq" "$one"

结果：

test 1 -eq $one         1 -eq 1 is T|0
test 1 -eq $one         1 -eq 2 is T|0

test 1 -eq $one         1 -eq 1 is T|0
test 1 -eq $one         1 -eq 2 is T|0

你看到了什么问题吗？

Answer 1

报价!!

使用“$ *”和bash [[...]]命令行值不会被拆分。

如果单词不为空，则测试等同于[ word ]，因此为真。

这个shell测试有什么问题？

1 个答案: