regex - 带有正则表达式的 Bash 脚本在 Ubuntu 上不起作用

Question

我有一个在我的 OpenSuSE 机器上运行的 Bash 脚本，但是当复制到我的 Ubuntu 机器时，它不起作用。脚本从文件中读取。该文件具有由空格（制表符和空格）分隔的字段。

#!/bin/bash
function test1()
{
    while read LINE
    do
        if [[ $LINE =~ "^$" || $LINE =~ "^#.*" ]] ; then
            continue;
        fi
        set -- $LINE
        local field1=$1
        local field2=$2
    done < test.file
}

test1

与 test.file 包含：

# Field1Header    Field2Header
abcdef            A-2
ghijkl            B-3

似乎有两个问题：

(1) $field2，带连字符的那个，是空白的

(2) 去除以# 开头的空行和行的正则表达式不起作用

有谁知道怎么了？正如我所说，它在 OpenSuSE 上运行良好。

谢谢，保罗

Answer 1

1. Quoting is wrong, that probably accounts for the regex failing.
2. No need to use bashisms.
3. No need to use set

Try

while read field1 field2 dummy
do
    if ! test "${field1%%#*}"
    then
        continue
    fi
    # do stuff here
done

EDIT: The obvious version using set

while read -r line
do
    if ! test "${line%%#*}"
    then
        continue
    fi
    set -- $line
    do_stuff_with "$@"
done

Answer 2

Apparently, as of bash 3.2 the regular expression should not be quoted. So this should work:

#!/bin/bash
while read LINE
do
    if [[ $LINE =~ ^$ || $LINE =~ ^#.* ]] ; then
        continue;
    fi
    set -- $LINE
    local field1=$1
    local field2=$2
done < test.file

Edit: you should probably use Jo So's answer as it's definitely cleaner. But I was explaining why the regex fails and the reason behind the different behavior between OpenSuse and Ubuntu(different version of bash, very probably)

Answer 3

On my ubuntu there is no expresion like "=~" for test command. Just use this one:

if [[ $LINE = "" || ${LINE:0:1} = "#" ]] ; then
    continue;
fi