解析此字符串以获取 bash 中的特定信息答案

【问题标题】：Parsing this string for particular information in bash解析此字符串以获取 bash 中的特定信息
【发布时间】：2019-12-11 21:24:29
【问题描述】：

我正在编写一个 bash 脚本，其中一部分需要我通过解析字符串中的有用信息并丢弃其余部分来检索作业的标题和截止日期。

我尝试使用 sed 一点点，但似乎无法让它按我想要的方式工作。

我尝试过的脚本中的一件事是 LABS=$(sed 's/<a.*/a>//' $LABS)

["<a href=\"https://classroom.github.com/a/WOWerwCz\">lab01</a>",
    "Lab 1", bblearn_content_base + "/resources/Labs/1.html",
    "7/3/2019", 1,
    "<a href=\"https://classroom.github.com/a/k3dVwTMy\">lab02</a>",
    "Lab 2", bblearn_content_base + "/resources/Labs/2.html",
    "7/12/2019", 1,
    "<a href=\"https://classroom.github.com/a/z1chUDd4\">lab03</a>",
    "Lab 3", bblearn_content_base + "/resources/Labs/3.html",
    "7/20/2019", 1,
    "<a href=\"https://classroom.github.com/a/iHbdXqs4\">lab04</a>",
    "Lab 4", bblearn_content_base + "/resources/Labs/4.html",
    "7/31/2019", 1,
    "<a href=\"https://classroom.github.com/a/WgyMWn68\">lab05</a>",
    "Lab 5", bblearn_content_base + "/resources/Labs/5.html",
    "8/5/2019", 1,
    "<a href=\"https://classroom.github.com/a/4anRjuDB\">lab06</a>",
    "Lab 6", bblearn_content_base + "/resources/Labs/6.html",
    "8/10/2019", 1,
    "<a href=\"https://classroom.github.com/a/qTyBR1R8\">lab07</a>",
    "Lab 7", bblearn_content_base + "/resources/Labs/7.html",
    "8/16/2019", 1,
    "<a href=\"https://classroom.github.com/a/UIJsxfA5\">lab08</a>",
    "Lab 8", bblearn_content_base + "/resources/Labs/8.html",
    "8/22/2019", 1,
    "<a href=\"https://classroom.github.com/a/XYUPMOiC\">lab09</a>",
    "Lab 9", bblearn_content_base + "/resources/Labs/9.html",
    "8/25/2019", 1,
    "<a href=\"https://classroom.github.com/a/ZJQ70-dy\">lab10</a>",
    "Lab 10", bblearn_content_base + "/resources/Labs/10.html",
    "9/1/2019", 1
]

基本上我需要做的是删除所有不是“Lab x”和随后日期的内容，但我正在努力弄清楚如何。

【问题讨论】：

@adam 我认为您的问题缺乏澄清。您能否举例说明您正在编写的脚本所期望的结果？

标签： bash parsing awk grep

【解决方案1】：

我假设您可以在一个 awk 脚本中编写所有处理任务。但这里有一个小awk 脚本来完成这项任务：

script.awk

BEGIN {FPAT="\"[^\"]+";}  # extract the first field between " into variable $1
NR % 3 == 0 { # on each 3rd line
    print labName, substr($1,2); # print labName and $1 without initial "
}
{             # on each line 
    labName = substr($1,2); # read the labName (or whatever there) from $1 (without initial ")
}

运行：

awk -f script.awk input.txt

或一个班轮：

awk 'BEGIN {FPAT="\"[^\"]+";}NR % 3 == 0 {print labName, substr($1,2);}{labName = substr($1,2);}' input.txt

输出：

Lab 1 7/3/2019
Lab 2 7/12/2019
Lab 3 7/20/2019
Lab 4 7/31/2019
Lab 5 8/5/2019
Lab 6 8/10/2019
Lab 7 8/16/2019
Lab 8 8/22/2019
Lab 9 8/25/2019
Lab 10 9/1/2019

【讨论】：

【解决方案2】：

你可以试试这个：

LABS=$(sed '/<a href/d;/^[][]/d;s/\s*\("[^"]*"\).*/\1/' $LABS)

输出

"Lab 1"
"7/3/2019"
"Lab 2"
"7/12/2019"
"Lab 3"
"7/20/2019"
"Lab 4"
"7/31/2019"
"Lab 5"
"8/5/2019"
"Lab 6"
"8/10/2019"
"Lab 7"
"8/16/2019"
"Lab 8"
"8/22/2019"
"Lab 9"
"8/25/2019"
"Lab 10"
"9/1/2019"

解释

/<a href/d                # remove all lines with hrefs
;
/^[][]/d                  # remove all lines starting with [ or ] 
;
s/\s*\("[^"]*"\).*/\1/    # remove leading spaces, and extract content between "..."

【讨论】：