如果语句不按预期工作，则Shell [英] Shell if statements not working as intended

问题描述

我试图在shell中创建一个搜索程序作为练习，但是当我尝试使用if语句处理空行时，我收到一条消息，说shell遇到了意外的操作符。

I am attempting to create a search program in shell as an exercise, but when I try to handle empty lines with an if-statement I get a message saying that the shell encountered an unexpected operator.

    #!/bin/sh

file=$1
token=$2

while read line
do
    if [ ! -z $line ]
    then
      set $line
      if [ $1 = $token ] 
      then 
        echo $line
      fi
    fi
done < $file

当我使用 match_token animals_blanks dog 我得到

./match_token: 8: [: cat: unexpected operator
./match_token: 8: [: dog: unexpected operator
./match_token: 8: [: dog: unexpected operator
./match_token: 8: [: cow: unexpected operator
./match_token: 8: [: lion: unexpected operator
./match_token: 8: [: bear: unexpected operator
./match_token: 8: [: wolf: unexpected operator

animals_blanks文件包含：

The animals_blanks files contains:

cat meow kitten

dog ruff pup
dog bark

cow moo calf

lion roar cub

bear roar cub
wolf howl pup

推荐答案

引用变量：

 if [ ! -z "$line" ]

但通常会有人写：

 if [ -n "$line" ]

当您保持变量不加引号时， [命令会看到如下内容： [-n cat dog] ，这是一个错误，因为它只需要在 -n 之后只有一个参数。通过引用变量，表达式变为 [-ncat dog] ，它只有一个参数，正如 [。请注意，确实没有理由进行该测试，或使用 set ;当读取时，读取可以为您分割线：

When you leave the variable unquoted, the [ command sees something like: [ -n cat dog ], which is an error because it expects only one argument after -n. By quoting the variable, the expression becomes [ -n "cat dog" ] which has only one argument, as expected by [. Note that there's really no reason to do that test, or to use set; read can split the line for you when it reads:

while read animal sound offspring; do
    test "$animal" = "$token" && echo $animal $sound $offspring
done < $file

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