我有这个Bash脚本，在第16行有一个问题。 如何获取第15行之前的结果并添加 变成第16行中的变量?
#!/bin/bash
num=0
metab=0
for ((i=1; i<=2; i++)); do
for j in `ls output-$i-*`; do
echo "$j"
metab=$(cat $j|grep EndBuffer|awk '{sum+=$2} END { print sum/120}') (line15)
num= $num + $metab (line16)
done
echo "$num"
done
当前回答
在Bash中,
num=5
x=6
(( num += x ))
echo $num # ==> 11
请注意，Bash只能处理整数算术，因此如果您的AWK命令返回一个分数，那么您将需要重新设计:以下是在AWK中重写的代码，以完成所有数学运算。
num=0
for ((i=1; i<=2; i++)); do
for j in output-$i-*; do
echo "$j"
num=$(
awk -v n="$num" '
/EndBuffer/ {sum += $2}
END {print n + (sum/120)}
' "$j"
)
done
echo "$num"
done
其他回答
#!/bin/bash
read X
read Y
echo "$(($X+$Y))"
#!/usr/bin/bash
#integer numbers
#===============#
num1=30
num2=5
echo $(( num1 + num2 ))
echo $(( num1-num2 ))
echo $(( num1*num2 ))
echo $(( num1/num2 ))
echo $(( num1%num2 ))
read -p "Enter first number : " a
read -p "Enter second number : " b
# we can store the result
result=$(( a+b ))
echo sum of $a \& $b is $result # \ is used to espace &
#decimal numbers
#bash only support integers so we have to delegate to a tool such as bc
#==============#
num2=3.4
num1=534.3
echo $num1+$num2 | bc
echo $num1-$num2 | bc
echo $num1*$num2 |bc
echo "scale=20;$num1/$num2" | bc
echo $num1%$num2 | bc
# we can store the result
#result=$( ( echo $num1+$num2 ) | bc )
result=$( echo $num1+$num2 | bc )
echo result is $result
##Bonus##
#Calling built in methods of bc
num=27
echo "scale=2;sqrt($num)" | bc -l # bc provides support for calculating square root
echo "scale=2;$num^3" | bc -l # calculate power
我也很喜欢这个方法。有更少的杂乱:
count=$[count+1]
适用于MacOS
#!/bin/bash
sum=0
for (( i=1; i<=5; i++ )); do
sum=$(echo "$sum + 1.1" | bc) # bc: if you want to use decimal
done
echo "Total: $sum"
您应该将metab声明为整数，然后使用算术求值
declare -i metab num
...
num+=metab
...
更多信息请参见6.5 Shell算术。
