为什么这个bash expr命令不能工作?(复制)

This question already has an answer here:

这个问题已经有了答案:

• Command not found error in Bash variable assignment 4 answers
• 命令未在Bash变量赋值4中找到错误

I am trying to increment a variable in a bash script and it is not working. Here is my code:

我试图在bash脚本中增加一个变量，但它不工作。这是我的代码:

#! /bin/bash

COUNTER=0
while [ $COUNTER -lt 5 ]
do
    echo "i will add this line to file mycreation">>./myfile
    COUNTER = `expr $COUNTER + 1`
done

The quotes around the COUNTER assignment are backticks.

反任务的引号是背景音。

I tried replacing COUNTER with $COUNTER like this:

我试着用$COUNTER来代替它:

$COUNTER = `expr $COUNTER + 1`

But that did not solve the problem and gave me the following error:

但这并没有解决问题，给了我以下的错误:

line7: 0: command not found. 

4 个解决方案

#1

As @Cory rightly pointed out, there should not be spaces around the equal sign or else bash will confuse COUNTER for a command.

正如@Cory正确地指出的，等号周围不应该有空格，否则bash会混淆COUNTER作为命令。

COUNTER=$(expr $COUNTER + 1)

going off-topic ...

That said, you could avoid having bash fork a subprocess by using the following alternatives:

也就是说，您可以通过使用以下替代方法来避免使用bash fork子进程:

• Using the bash builtin 'let' command:

  使用bash内置的“让”命令:

  let COUNTER="COUNTER + 1"
  

• or, using bash c-style expression:

  或者，使用bash c-style表达式:

  (( COUNTER++ ))
  

In fact, your while loop can be written as:

实际上，您的while循环可以写成:

for ((COUNTER=0; COUNTER <= 5 ; COUNTER++))
do
    echo "i will add this line to file mycreation">>./myfile
done

Breaking down the error message

When you were met with the error:

当你遇到错误时:

line 7:   0:    command not found.
'-----'  '--'  '------------------'
   |       |                 |
location   |            Description of error.
          culprit 

my guess is what you had on line 7 was

我猜你在第7行是什么

$COUNTER = `expr $COUNTER + 1`
--------   --------------------
    |                 |
Evaluated to 0        |
                  Evaluated to 1

What bash ends up see is 0 = 1 and since bash statements are generally in the form command arg1 arg1 ..., bash interprets it as run the command 0 with arguments = 1. Thus the error message : 0: command not found.

bash的最终结果是0 = 1，由于bash语句通常使用的是命令arg1 arg1……， bash将其解释为运行参数= 1的命令0。因此，错误消息:0:未找到命令。

When you removed the spaces around the equal sign, what bash ends up interpreting is:

当你移除等号周围的空格时，bash最终解释的是:

0=1

which means run command 0=1 with no arguments, hence the error 0=1: command not found.

这意味着在没有参数的情况下运行命令0=1，因此错误0=1:没有找到命令。

Variable assignments should be in the form VAR_NAME=VALUE (without the $), so the syntax you should be using is:

变量赋值的形式应该是VAR_NAME=VALUE(没有$)，所以应该使用的语法是:

COUNTER=`expr $COUNTER + 1` # or any of the variants above

which bash evaluates and eventually interpret as:

bash评估并最终解释为:

COUNTER=2

#2

Remove the spaces around the equals sign:

删除等号附近的空格:

COUNTER=`expr $COUNTER + 1`

#3

Another way.

另一种方式。

COUNTER=$(($COUNTER + 1))

#4

for i in {0..4}; do 
    echo "i will add this line to file mycreation" >> ./myfile
done