我想像这样表示多个条件:
if [ ( $g -eq 1 -a "$c" = "123" ) -o ( $g -eq 2 -a "$c" = "456" ) ]
then
echo abc;
else
echo efg;
fi
但当我执行脚本时，它会显示
syntax error at line 15: `[' unexpected,
第15行显示if ....在哪里
这种情况有什么问题?我猜是()有问题。
当前回答
在Bash中，可以使用以下技术进行字符串比较
if [ $var OP "val" ]; then
echo "statements"
fi
例子:
var="something"
if [ $var != "otherthing" ] && [ $var != "everything" ] && [ $var != "allthings" ]; then
echo "this will be printed"
else
echo "this will not be printed"
fi
其他回答
使用/bin/bash可以工作:
if [ "$option" = "Y" ] || [ "$option" = "y" ]; then
echo "Entered $option"
fi
你也可以链两个以上的条件:
if [ \( "$1" = '--usage' \) -o \( "$1" = '' \) -o \( "$1" = '--help' \) ]
then
printf "\033[2J";printf "\033[0;0H"
cat << EOF_PRINT_USAGE
$0 - Purpose: upsert qto http json data to postgres db
USAGE EXAMPLE:
$0 -a foo -a bar
EOF_PRINT_USAGE
exit 1
fi
#!/bin/bash
current_usage=$( df -h | grep 'gfsvg-gfslv' | awk {'print $5'} )
echo $current_usage
critical_usage=6%
warning_usage=3%
if [[ ${current_usage%?} -lt ${warning_usage%?} ]]; then
echo OK current usage is $current_usage
elif [[ ${current_usage%?} -ge ${warning_usage%?} ]] && [[ ${current_usage%?} -lt ${critical_usage%?} ]]; then
echo Warning $current_usage
else
echo Critical $current_usage
fi
在Bash中，可以使用以下技术进行字符串比较
if [ $var OP "val" ]; then
echo "statements"
fi
例子:
var="something"
if [ $var != "otherthing" ] && [ $var != "everything" ] && [ $var != "allthings" ]; then
echo "this will be printed"
else
echo "this will not be printed"
fi
如果字符串变量中有空格并检查是否存在，请小心。一定要正确引用它们。
if [ ! "${somepath}" ] || [ ! "${otherstring}" ] || [ ! "${barstring}" ] ; then
