比較bash中不匹配的字符串

Question

我有一個變量，應包含一個或其他字符串值或退出程序。

第一個似乎對我有用：

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

但是，第二個沒有：

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

這兩個不相等的定義之間有什么區別？

Answer 1

這些是正確的形式：

[ ! "$JOB_FILE" == "all_jobs.pl" ] && [ ! "$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
}

而在Bash中，這樣做更好：

[[ $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
}

或這個：

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
}

也可以使用regex但更難逃脫。

Answer 2

看來您有邏輯錯誤。 如果兩個語句都為真，則要執行回顯，但是您正在使用|| 。 您的真正意思是使用&& 。

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

一種方法是寧願檢查文件名是否為預期的文件名，並使用邏輯或鏈接條件：

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

Answer 4

邏輯是相反的。

您要檢查JOB_FILE是“ A”還是“ B”。 在其他情況下，請打印錯誤消息。

JOB_FILE  status
A         ok
B         ok
rest      error

於是

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

其中1=True ， 0=False 。

了解了邏輯之后，您需要的是：

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