Question

我有一个var，它应该包含一个或其他字符串值或退出程序。

第一个似乎对我有用：

[ !"$JOB_FILE" == "all_jobs.pl" ] || [ !"$JOB_FILE" == "incomplete.out" ] && { echo -e "\n Only all_jobs.pl or incomplete.out are valid job files\n"; exit 127 ; }

然而，第二个不是：

[ "$JOB_FILE" != "all_jobs.pl" ] || [ "$JOB_FILE" != "incomplete.out" ] && { echo -e "\n Only all_jobs.pl or incomplete.out are valid job files\n"; exit 127 ; }

两个不相等的定义之间有什么区别。

Answer 1

这些是正确的形式：

[ ! "$JOB_FILE" == "all_jobs.pl" ] && [ ! "$JOB_FILE" == "incomplete.out" ] && {
    echo -e "\n Only all_jobs.pl or incomplete.out are valid job files\n"
    exit 127
}

[ "$JOB_FILE" != "all_jobs.pl" ] && [ "$JOB_FILE" != "incomplete.out" ] && {
    echo -e "\n Only all_jobs.pl or incomplete.out are valid job files\n"
    exit 127
}

在Bash中，这更好：

[[ $JOB_FILE != "all_jobs.pl" && $JOB_FILE != "incomplete.out" ]] && {
    echo -e "\n Only all_jobs.pl or incomplete.out are valid job files\n"
    exit 127
}

或者这个：

shopt -s extglob  ## Not necessary on Bash 4.1 or newer.

[[ $JOB_FILE != @(all_jobs.pl|incomplete.out) ]] && {
    echo -e "\n Only all_jobs.pl or incomplete.out are valid job files\n"
    exit 127
}

使用regex也是可能的，但它更难逃脱。

Answer 2

好像你有一个逻辑错误。如果两个语句都为真，但您使用||，则需要执行回显。你真的想要使用&&。

if [[ "$JOB_FILE" != "all_jobs.pl" && "$JOB_FILE" != "incomplete.out" ]]; then 
   echo -e "\n Only all_jobs.pl or incomplete.out are valid job files\n"
   exit 127
fi

Answer 3

一种方法是检查文件名是否是预期的文件名，并使用逻辑OR来链条：

[ "$JOB_FILE" = "all_jobs.pl" ] || [ "$JOB_FILE" = "incomplete.out" ] || { echo -e "\n Only all_jobs.pl or incomplete.out are valid job files\n"; exit 127 ; }

Answer 4

逻辑是相反的。

您想要检查JOB_FILE是“A”还是“B”。在其他情况下，请输出错误消息。

JOB_FILE  status
A         ok
B         ok
rest      error

因此

A   B   (A or B)  !(A or B)    !A or !B         !A and !B
1   0       1         0       0 or 1 => 1      0 and 1 => 0
0   1       1         0       1 or 0 => 1      1 and 0 => 0
0   0       0         1       1 or 1 => 1      1 and 1 => 1
--------------------------------------------------------------
                              always true      just true if not A and not B

1=True，0=False。

在这种逻辑味道之后，你需要的是：

[ ! "$JOB_FILE" == "all_jobs.pl" ] && [ ! "$JOB_FILE" == "incomplete.out" ] && { echo ... }
                                   ^^

比较bash中不匹配的字符串

4 个答案: