如何在Bash中将字符串分割成数组？

由于有很多种方法可以解决这个问题，让我们首先定义一下我们想在解决方案中看到的内容。

1. Bash提供了一个内置的readarray命令来实现这个目的。让我们使用它。
2. 避免使用丑陋和不必要的技巧，比如更改IFS、循环、使用eval或添加额外的元素然后将其删除。
3. 找到一个简单、易读的方法，可以轻松地适应类似的问题。

readarray命令最容易使用换行符作为分隔符。对于其他分隔符，它可能会向数组中添加一个额外的元素。最清晰的方法是在传入之前先将输入调整为适合readarray的形式。

在这个例子中，输入没有多字符分隔符。如果我们运用一点常识，最好将其理解为逗号分隔的输入，每个元素可能需要修剪。我的解决方案是将输入按逗号拆分成多行，修剪每个元素，然后将所有内容传递给readarray。

string='  Paris,France  ,   All of Europe  '
readarray -t foo < <(tr ',' '\n' <<< "$string" |sed 's/^ *//' |sed 's/ *$//')

# Result:
declare -p foo
# declare -a foo='([0]="Paris" [1]="France" [2]="All of Europe")'

编辑：我的解决方案允许逗号分隔符周围的不一致空格，同时允许元素包含空格。很少有其他解决方案能处理这些特殊情况。

我还避免了看起来像是hack的方法，比如创建一个额外的数组元素然后再删除它。如果您不认为这是最佳答案，请留下评论解释。

如果您想尝试纯Bash实现相同的方法，并且使用更少的子shell，那是可能的。但结果更难阅读，而且这种优化可能是不必要的。

string='     Paris,France  ,   All of Europe    '
foo="${string#"${string%%[![:space:]]*}"}"
foo="${foo%"${foo##*[![:space:]]}"}"
foo="${foo//+([[:space:]]),/,}"
foo="${foo//,+([[:space:]])/,}"
readarray -t foo < <(echo "$foo")

这是我的技巧！

使用bash拆分字符串是一件相当无聊的事情。我们只有有限的方法适用于少数情况（按“;”，“/”，“.”等拆分），或者输出会有各种副作用。

下面的方法需要进行一些操作，但我相信它将适用于我们大多数的需求！

#!/bin/bash

# --------------------------------------
# SPLIT FUNCTION
# ----------------

F_SPLIT_R=()
f_split() {
    : 'It does a "split" into a given string and returns an array.

    Args:
        TARGET_P (str): Target string to "split".
        DELIMITER_P (Optional[str]): Delimiter used to "split". If not 
    informed the split will be done by spaces.

    Returns:
        F_SPLIT_R (array): Array with the provided string separated by the 
    informed delimiter.
    '

    F_SPLIT_R=()
    TARGET_P=$1
    DELIMITER_P=$2
    if [ -z "$DELIMITER_P" ] ; then
        DELIMITER_P=" "
    fi

    REMOVE_N=1
    if [ "$DELIMITER_P" == "\n" ] ; then
        REMOVE_N=0
    fi

    # NOTE: This was the only parameter that has been a problem so far! 
    # By Questor
    # [Ref.: https://unix.stackexchange.com/a/390732/61742]
    if [ "$DELIMITER_P" == "./" ] ; then
        DELIMITER_P="[.]/"
    fi

    if [ ${REMOVE_N} -eq 1 ] ; then

        # NOTE: Due to bash limitations we have some problems getting the 
        # output of a split by awk inside an array and so we need to use 
        # "line break" (\n) to succeed. Seen this, we remove the line breaks 
        # momentarily afterwards we reintegrate them. The problem is that if 
        # there is a line break in the "string" informed, this line break will 
        # be lost, that is, it is erroneously removed in the output! 
        # By Questor
        TARGET_P=$(awk 'BEGIN {RS="dn"} {gsub("\n", "3F2C417D448C46918289218B7337FCAF"); printf $0}' <<< "${TARGET_P}")

    fi

    # NOTE: The replace of "\n" by "3F2C417D448C46918289218B7337FCAF" results 
    # in more occurrences of "3F2C417D448C46918289218B7337FCAF" than the 
    # amount of "\n" that there was originally in the string (one more 
    # occurrence at the end of the string)! We can not explain the reason for 
    # this side effect. The line below corrects this problem! By Questor
    TARGET_P=${TARGET_P%????????????????????????????????}

    SPLIT_NOW=$(awk -F"$DELIMITER_P" '{for(i=1; i<=NF; i++){printf "%s\n", $i}}' <<< "${TARGET_P}")

    while IFS= read -r LINE_NOW ; do
        if [ ${REMOVE_N} -eq 1 ] ; then

            # NOTE: We use "'" to prevent blank lines with no other characters 
            # in the sequence being erroneously removed! We do not know the 
            # reason for this side effect! By Questor
            LN_NOW_WITH_N=$(awk 'BEGIN {RS="dn"} {gsub("3F2C417D448C46918289218B7337FCAF", "\n"); printf $0}' <<< "'${LINE_NOW}'")

            # NOTE: We use the commands below to revert the intervention made 
            # immediately above! By Questor
            LN_NOW_WITH_N=${LN_NOW_WITH_N%?}
            LN_NOW_WITH_N=${LN_NOW_WITH_N#?}

            F_SPLIT_R+=("$LN_NOW_WITH_N")
        else
            F_SPLIT_R+=("$LINE_NOW")
        fi
    done <<< "$SPLIT_NOW"
}

# --------------------------------------
# HOW TO USE
# ----------------

STRING_TO_SPLIT="
 * How do I list all databases and tables using psql?

\"
sudo -u postgres /usr/pgsql-9.4/bin/psql -c \"\l\"
sudo -u postgres /usr/pgsql-9.4/bin/psql <DB_NAME> -c \"\dt\"
\"

\"
\list or \l: list all databases
\dt: list all tables in the current database
\"

[Ref.: https://dba.stackexchange.com/questions/1285/how-do-i-list-all-databases-and-tables-using-psql]


"

f_split "$STRING_TO_SPLIT" "bin/psql -c"

# --------------------------------------
# OUTPUT AND TEST
# ----------------

ARR_LENGTH=${#F_SPLIT_R[*]}
for (( i=0; i<=$(( $ARR_LENGTH -1 )); i++ )) ; do
    echo " > -----------------------------------------"
    echo "${F_SPLIT_R[$i]}"
    echo " < -----------------------------------------"
done

if [ "$STRING_TO_SPLIT" == "${F_SPLIT_R[0]}bin/psql -c${F_SPLIT_R[1]}" ] ; then
    echo " > -----------------------------------------"
    echo "The strings are the same!"
    echo " < -----------------------------------------"
fi

对于多行元素，为什么不考虑使用类似于

$ array=($(echo -e $'a a\nb b' | tr ' ' '§')) && array=("${array[@]//§/ }") && echo "${array[@]/%/ INTERELEMENT}"

a a INTERELEMENT b b INTERELEMENT

另一种方法是：

string="Paris, France, Europe"
IFS=', ' arr=(${string})

现在你的元素存储在“arr”数组中。 要遍历这些元素：

for i in ${arr[@]}; do echo $i; done

另一种方法可以是：

str="a, b, c, d"  # assuming there is a space after ',' as in Q
arr=(${str//,/})  # delete all occurrences of ','

执行完这行代码后，“arr”就是一个包含四个字符串的数组。 这不需要处理IFS、read或任何其他特殊的东西，因此更简单和直接。