在Tab中将制表符分隔的字符串拆分为数组

Question

line="\t\t\t1\t2\t\t3"

What I do:

IFS=$'\t'  DIRS=($line);

What I want to get:

DIRS[0]=NULL; DIRS[1]=NULL; DIRS[2]=NULL; DIRS[3]=1;DIRS[4]=2;DIRS[5]=NULL;DIRS[6]=3;

What I actually get:

DIRS[0]=1; DIRS[1]=2; DIRS[2]=3

Is that possible to get what I want to get?

Answer 1

Bash treats whitespace specially in IFS:

If IFS has a value other than the default, then sequences of the whitespace characters space and tab are ignored at the beginning and end of the word, as long as the whitespace character is in the value of IFS (an IFS whitespace character). Any character in IFS that is not IFS whitespace, along with any adjacent IFS whitespace characters, delimits a field. A sequence of IFS whitespace characters is also treated as a delimiter.

So, use a non-whitespace character. (BTW, I use $'\\t' for a tab, search man bash for Quoting )

#!/bin/bash
line=$'\t\t\t1\t2\t\t3'

IFS=$':' DIRS=(${line//$'\t'/:})       # Replace tabs with colons.

for (( i=0 ; i<${#DIRS[@]} ; i++ )); do
    echo "$i: [${DIRS[i]}]"
done