拆分包含 '&&' 的 Bash 字符串

Question

I have a bash script like this:

#!/bin/bash

params="path=/me & you/folder/file.json&&user=Bob"
splitParams=(${params//&&/ })

for i in "${!splitParams[@]}"
do
    echo "$i: ${splitParams[i]}"
done

It produces:

0: path=/me
1: &
2: you/folder/file.json
3: user=Bob

But when I type em in shell:

➜  params="path=/me & you/folder/file.json&&user=Bob"
➜  splitParams=(${params//&&/ })
➜  echo $splitParams
path=/me & you/folder/file.json user=Bob

I expect the output from script is like when I typed in shell:

0: path=/me & you/folder/file.json
1: user=Bob

Answer 1

echo $splitParams only prints the first element. Use declare -p to see exactly what's in the array:

$ declare -p splitParams
declare -a splitParams=([0]="path=/me" [1]="&" [2]="you/folder/file.json" [3]="user=Bob")

So now how can you split the string at && but not have whitespace mess things up? Well, relying on the shell to do it is error-prone. Changing && into spaces and splitting on spaces runs into trouble when the input string already contains spaces.

A better way to do it is to change the delimiters into newlines. Then instead of using array=($string) to break $string apart, use readarray so spaces don't trigger splitting, only newlines.

$ readarray -t splitParams < <(sed 's/&&/\n/g' <<< "$params")
$ declare -p splitParams
declare -a splitParams=([0]="path=/me & you/folder/file.json" [1]="user=Bob")

Answer 2

grep -oP '(?<=^|&&).*?(?=&&|$)' <<< "path=/me & you/folder/file.json&&user=Bob"

prints:

path=/me & you/folder/file.json
user=Bob

Answer 3

You may use something like this:

params="path=/me & you/folder/file.json&&user=Bob"
while IFS=  read -r -d $'\0'
do
  splitParams+=("$REPLY")
done < <(echo "${params}" | awk -F'&&' '{ for (i=1;i<=NF;i++) printf("%s\0",$i); }')

for i in "${!splitParams[@]}"
do
    echo "$i: ${splitParams[i]}"
done

Output:

0: path=/me & you/folder/file.json
1: user=Bob

Answer 4

You can use awk for that purpose, but remember that splitParams can not be accessible outside of the awk. We will use awk's split function:

split([String to be proccessed], [Array for results], [Delimeter])

#!/bin/bash
params="path=/me & you/folder/file.json && user=Bob"
awk -F: '{split($0, splitParams, "&&"); print splitParams[1] "\n"splitParams[2]}' <<< $params

Output:

 path=/me & you/folder/file.json
  user=Bob