命令没有bash函数调用发现 [英] Command not found on bash function call

问题描述

您好我是新来bash脚本编程，但不明白为什么我得到的未找到命令的错误，当我试图给一个局部变量参数20120920 5.这个函数调用的结果

Hi i'm new to bash scripting but cannot understand why i get the command not found error when i try to assign to a local variable the result of this function call with parameters 20120920 5.

    #!/bin/bash

function nDaysAgo () #date # daysago
{
 date -d "${1} - ${2} days" +%Y%m%d;
}

这样的脚本名称是ndaysago，我是第一个调用的脚本。 nday​​sago然后像这样分配值：

so the script name is ndaysago, i'm first invoking the script with . ndaysago and then assigning the value like this:

newdate= nDaysAgo 20120910 5

它打印：20120905：找不到命令

it prints: 20120905: command not found

含义的日期执行制作，但随后试图使用输出作为命令，那不是我所希望的。

Meaning that the date execution is made but then tries to use the output as command, thats not what i would expect.

我也尝试过像这样的函数内分配新的值给一个变量：

i have also tried assigning the new value to a variable within the function like so:

#!/bin/bash

function nDaysAgo () #date # daysago
{
 var=$(date -d "${1} - ${2} days" +%Y%m%d)
}

但仍然一无所获，MMMMMM

but still nothing, mmmmmm

推荐答案

空格不在身边允许为变量赋值时， = 。要调用你应该使用这就是所谓的 $（...）语法功能=htt​​p://tldp.org/LDP/abs/html/ commandsub.html相对=nofollow>命令替换。

Spaces are not allowed around the = when assigning a variable. To invoke a function you should use the $(...) syntax which is called command substitution.

更改为：

newdate=$(nDaysAgo 20120910 5)

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