bash：將多行字符串讀入多個變量

Question

如何用一個換行符分隔字符串三行到三個變量？

# test string 
s='line 01 
line 02 
line 03' 

# this doesn't seem to make any difference at all 
IFS=$'\n' 

# first naive attempt 
read a b c <<< "${s}" 

# this prints 'line 01||': 
# everything after the first newline is dropped 
echo "${a}|${b}|${c}" 

# second attempt, remove quotes 
read a b c <<< ${s} 

# this prints 'line 01 line 02 line 03||': 
# everything is assigned to the first variable 
echo "${a}|${b}|${c}" 

# third attempt, add -r 
read -r a b c <<< ${s} 

# this prints 'line 01 line 02 line 03||': 
# -r switch doesn't seem to make a difference 
echo "${a}|${b}|${c}" 

# fourth attempt, re-add quotes 
read -r a b c <<< "${s}" 

# this prints 'line 01||': 
# -r switch doesn't seem to make a difference 
echo "${a}|${b}|${c}" 

我也嘗試過使用echo ${s} | read a b c代替<<<，但不能得到這工作的。

這可以在bash中完成嗎？

Answer 1

讀取默認輸入定界符爲\ n

{ read a; read b; read c;} <<< "${s}" 

-d炭：允許指定另一個輸入定界符

例如是有在輸入字符串中沒有字符SOH（1個ASCII）

IFS=$'\n' read -r -d$'\1' a b c <<< "${s}" 

編輯：-d可以接受一個空參數，該空間在-d和null參數之間是強制的：

IFS=$'\n' read -r -d '' a b c <<< "${s}" 

編輯：關於任意數量的行

function read_n { 
    local i s n line 
    n=$1 
    s=$2 
    arr=() 
    for ((i=0;i<n;i+=1)); do 
     IFS= read -r line 
     arr[i]=$line 
    done <<< "${s}" 
} 

nl=$'\n' 
read_n 10 "a${nl}b${nl}c${nl}d${nl}e${nl}f${nl}g${nl}h${nl}i${nl}j${nl}k${nl}l" 

printf "'%s'\n" "${arr[@]}" 

Answer 2

您正在尋找readarray命令，不read溶液後評論。

readarray -t lines <<< "$s" 

（理論上，$s不需要在這裏引用，除非你正在使用bash 4.4或更高版本，我無論如何都會引用它，因爲在bash以前版本的一些bug。）

一旦該行處於數組，你可以，如果你需要

a=${lines[0]} 
b=${lines[1]} 
c=${lines[2]}