bash 从一个 awk 命令设置多个变量？
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set multiple variables from one awk command?
提问by X.M.
This is a very common script:
这是一个非常常见的脚本：
#!/bin/bash
teststr="col1 col2"
var1=`echo ${teststr} | awk '{print }'`
var2=`echo ${teststr} | awk '{print }'`
echo var1=${var1}
echo var2=${var2}
However I dont like this, especially when there are more fields to parse. I guess there should be a better way like:
但是我不喜欢这个，尤其是当有更多的字段需要解析时。我想应该有更好的方法，例如：
(var1,var2)=`echo ${teststr} | awk '{print $1 $2}' (in my imagination)
(var1,var2)=`echo ${teststr} | awk '{print $1 $2}'（在我的想象中）
Is that so? Thanks for help to improve effeciency and save some CPU power.
是这样吗？感谢您帮助提高效率并节省一些 CPU 功率。
回答by potong
This might work for you:
这可能对你有用：
var=(col0 col1 col2)
echo "${var[1]}"
col1
回答by Hemant Patel
Bash has Array Support, We just need to supply values dynamically :)
Bash 有数组支持，我们只需要动态提供值:)
function test_set_array_from_awk(){
# Note : -a is required as declaring array
let -a myArr;
# Hard Coded Valeus
# myArr=( "Foo" "Bar" "other" );
# echo "${myArr[1]}" # Print Bar
# Dynamic values
myArr=( $(echo "" | awk '{print "Foo"; print "Bar"; print "Fooo-And-Bar"; }') );
# Value @index 0
echo "${myArr[0]}" # Print Foo
# Value @index 1
echo "${myArr[1]}" # Print Bar
# Array Length
echo ${#myArr[@]} # Print 3 as array length
# Safe Reading with Default value
echo "${myArr[10]-"Some-Default-Value"}" # Print Some-Default-Value
echo "${myArr[10]-0}" # Print 0
echo "${myArr[10]-''}" # Print ''
echo "${myArr[10]-}" # Print nothing
# With Dynamic Index
local n=2
echo "${myArr["${n}"]-}" # Print Fooo-And-Bar
}
# calling test function
test_set_array_from_awk
Bash Array Documentation : http://tldp.org/LDP/abs/html/arrays.html
Bash 阵列文档：http: //tldp.org/LDP/abs/html/arrays.html
