Shell脚本错误

Question

谁能说出这个脚本出了什么问题吗？因为我得到的错误如下

./timer: line 9: [13: command not found
./timer: line 12: [13: command not found

我的脚本如下所示

#!/bin/bash

   while :
     do
    HOUR=$(date +%H)
    MINUTE=$(date +%M)
    SECOND=$(date +%S)

        if [$HOUR == 13] && [$MINUTE == 12] && [$SECOND == 1]
     then ./binary

    elif [$HOUR == 18] && [$MINUTE == 30] && [$SECOND == 1] 
     then ./binary
    fi

    done 

Answer 1

在…之间留个空格。示例：

if [$HOUR == 13] && [$MINUTE == 12] && [$SECOND == 1]

应该变成

if [ $HOUR == 13 ] && [ $MINUTE == 12 ] && [ $SECOND == 1 ]

Answer 2

我认为您必须使用"${VARIABLE}"，并注意方括号中的空格

这将提供：

if [ "${HOUR}" == 13 ] && [ "${HOUR}" == 12 ] && [ "${HOUR}" == 1 ]

希望这能有所帮助！

Answer 3

bash中的测试操作符需要在左方括号和右方括号之间留出空格，请尝试

[ $HOUR == 13 ] && [ $MINUTE == 12 ] && [ $SECOND == 1 ]

和

[ $HOUR == 18 ] && [ $MINUTE == 30 ] && [ $SECOND == 1 ]

这里有一个您可能会发现有用的http://tldp.org/LDP/abs/html/testconstructs.html链接