在 Bash 中使用 2 个不同的分隔符拆分字符串
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/25163486/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Split string using 2 different delimiters in Bash
提问by geekyjazzy
I am trying to split a string in BASH based on 2 delimiters - Space and the \. This is the string:-
我正在尝试根据 2 个分隔符 - 空格和 \ 在 BASH 中拆分字符串。这是字符串：-
var_result="Pass results_ADV__001__FUNC__IND\ADV__001__FUNC__IND_08_06_14_10_04_34.tslog"
I want it to split in 3 parts as follows:-
我希望它分成 3 部分，如下所示：-
part_1=Pass
part_2=results_ADV__001__FUNC__IND
part_3=ADV__001__FUNC__IND_08_06_14_10_04_34.tslog
I have tried using IFS and it splits the first one well. But the second split some how removes the "\" and sticks the entire part and I get the split as :-
我试过使用 IFS，它很好地分割了第一个。但是第二次拆分一些如何删除“\”并粘贴整个部分，我将拆分为：-
test_res= Pass
log_file_info=results_ADV__001__FUNC__INDADV__001__FUNC__IND_08_06_14_10_04_34.tslog
The IFS I used is as follows:-
我使用的 IFS 如下：-
echo "$var_result"
IFS=' ' read -a array_1 <<< "$var_result"
echo "test_res=${array_1[0]}, log_file_info=${array_1[1]}"
Thanks in advance.
提前致谢。
回答by Mark Setchell
I think you need this:
我认为你需要这个：
IFS=' |\' read -ra array_1 <<< "$var_result"
回答by cryptobionic
This should do it
这应该做
#!/bin/bash
var_result="Pass results_ADV__001__FUNC__IND\ADV__001__FUNC__IND_08_06_14_10_04_34.tslog"
field1=$(echo "$var_result" | awk -F ' ' '{print $(NF-1)}');
field2=$(echo "$var_result" | awk -F \ '{print $(NF-1)}');
field1=$(echo "$field1" | awk -F ' ' '{print $(NF-1)}');
field3=$(echo "$var_result" | awk -F \ '{print }');
echo $field1;
echo $field2;
echo $field3;
