傳遞與空格的字符串作爲參數傳給擊功能

Question

鑑於下面的字符串變量

VAR="foo   bar" 

我需要它被傳遞給一個bash功能，並訪問它，按照慣例，通過$1。到目前爲止，我一直無法弄清楚如何做到這一點：

#!/bin/bash 
function testfn(){ 
    echo "in function: $1" 
} 
VAR="foo   bar" 
echo "desired output is:" 
echo "$(testfn 'foo   bar')" 
echo "Now, what about a version with \$VAR?" 
echo "Note, by the way, that the following doesn't do the right thing:" 
echo $(testfn "foo   bar") #prints: "in function: foo bar" 

Answer 1

Bash是聰明和對雙引號匹配的內部或一個$(...)結構之外。

因此，echo "$(testfn "foo bar")"是有效的，並且testfn函數的結果將僅被視爲echo內部命令的單個參數。