如果使用bash的if语句和调用函数 [英] if statement and calling function in if using bash

问题描述

我写了一个函数:

check_log(){
    if [ -f "/usr/apps/appcheck.log" ]
    then
         return 1
    else
         return 0
    fi
}

然后我在"if"条件下调用此函数:

Then I call this function in an "if" condition:

if [ check_log ];
then
    ........statements....
fi

这项工作有效吗?我在这里感到困惑，因为bash成功返回0，失败返回1，但是我的函数返回1并且条件正在检查1/0，它得到1并且应该给出失败，但是在我的shell脚本条件通过了.

Will this work? I am confused here because bash returns 0 on success and 1 on failure, but my function is returning 1 and the condition is checking for 1/0, it gets 1 and it should give failures, but in my shell script the condition is passing.

有人可以阐明这个问题吗?

Can anyone shed light on this issue?

推荐答案

if [ check_log ];

使用方括号时，将调用test命令.它等效于if test check_log，它是if test -n check_log的简写，反过来意味着如果"check_log"不是空字符串".它根本不会调用您的check_log函数.

When you use square brackets you're invoking the test command. It's equivalent to if test check_log which is shorthand for if test -n check_log, which in turn means "if "check_log" is not an empty string". It doesn't call your check_log function at all.

将其更改为此:

if check_log;

顺便说一句，该函数可以更简单地写为:

By the way, the function could be more simply written as:

check_log() {
    ! [ -f "/usr/apps/appcheck.log" ]
}

函数的返回值是最后一条命令的退出状态，因此不需要显式的return语句.

The return value from a function is the exit status of the last command, so no need for explicit return statements.

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