bash 在unix中用echo打印两个没有空格的变量
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Print two variables without space with echo in unix
提问by Piyush Baijal
VAR1=ABC;
VAR2=XYZ;
VAR3=PQR;
I want to print value of VAR1,VAR2 and VAR3 in a for loop. Please help!
我想在 for 循环中打印 VAR1、VAR2 和 VAR3 的值。请帮忙！
In a for loopi am trying to do echo $VAR$i
在for loop我试图做的echo $VAR$i
Expected O/P:
预期 O/P：
ABC
XYZ
PQR
Actual Output:
实际输出：
1
2
3
回答by choroba
Use an array:
使用数组：
VAR[1]=ABC
VAR[2]=XYZ
VAR[3]=PQR
for i in 1 2 3 ; do
echo ${VAR[i]}
done
Or, use the variable indirection:
或者，使用变量间接：
VAR1=ABC
VAR2=XYZ
VAR3=PQR
for i in 1 2 3 ; do
name=VAR$i
echo ${!name}
done
回答by fedorqui 'SO stop harming'
You can do it like this:
你可以这样做：
for i in 1 2 3
do
eval "echo $VAR${i}"
done
The key point is the evalcommand.
关键是eval命令。
Test
测试
$ VAR1=ABC; VAR2=XYZ; VAR3=PQR;
$ for i in 1 2 3; do eval "echo $VAR${i}"; done
ABC
XYZ
PQR
回答by Alfishe
echo "${VAR1}${VAR2}${VAR3}"
Simple as that. I don't think it make sense to create useless cycles just to concatenate several values into a string
就那么简单。我认为仅仅为了将几个值连接成一个字符串而创建无用的循环是没有意义的
回答by tusharmakkar08
Instead of
代替
echo $VAR$i在 for 循环中做
varx=$varx$var$i
and initialize varx as empty string and at end do
并将 varx 初始化为空字符串，最后做
echo $varx
This is for printing without space as the question suggests.
正如问题所暗示的那样，这是用于没有空间的打印。
