测试模式 A 和 B，并返回介于两者之间的文本行，包括模式 A 和 B

Question

你好，再次堆栈溢出！

我有一个非常大的平面文件，我希望导出与 2 种不同模式匹配的所有记录。 问题是每条记录中的行数不同，并且记录相互渗透。 记录的最后一行是门 ID，第一行是用户：。

我正在测试电子邮件地址中的@，最后一次登录包含"登录时间：2013-08"。我需要导出所有行，包括电子邮件地址行和上次登录行。以下是 2 个样本。 我试过像这样使用 awk：

awk '/login time: 2013-08/{e=0}/@ /{gsub("^.*@ ","",$0);e=1}{if(e==1){print}}'  filename

当然，那失败了。

所以这是示例数据

User: [e-mail]
First Name: Afshi
Last Name: Noghami
Is Delegated Admin: False
IP Whitelisted: False
Account Suspended: False
Must Change Password: False
Unique ID: 102209840259208897543
ID TPYE: Cx4
Creation Time: 2013-06-07T04:14:42.000Z
Last login time: Never
Path: /Members/Inactive
IMs:
Addresses:
Organizations:
Phones:
Relations:
Door IDs:
User: [e-mail]
First Name: JISS
Last Name: NALLIKUZHY
Is a Super Admin: False
Is Delegated Admin: False
Has Agreed to Terms: True
IP Whitelisted: False
Account Suspended: False
Must Change Password: False
Unique ID: 109765147242431344122
ID TYPE: Cx4
Mailbox setup: True
Included: False
Creation Time: 2013-06-07T03:32:52.000Z
Last login time: 2013-08-02T07:13:02.000Z
Path: /Members/Inactive
IMs:
Addresses:
Organizations:
Phones:
Relations:
Door IDs:

对于具有上次登录日期的每条记录，所需的输出如下所示：

User: [e-mail]  
First Name: JISS  
Last Name: NALLIKUZHY  
Is a Super Admin: False  
Is Delegated Admin: False  
Has Agreed to Terms: True  
IP Whitelisted: False  
Account Suspended: False  
Must Change Password: False  
Unique ID: 109765147242431344122  
ID TYPE: Cx4  
Mailbox setup: True  
Included: False  
Creation Time: 2013-06-07T03:32:52.000Z  
Last login time: 2013-08-02T07:13:02.000Z 

Answer 1

也许这样的东西可以为你工作：

awk '$1=="User:",/login time: 2013-08/' file

Answer 2

awk '/User:/{if(NR!=1){for(i=0;i<j;i++)print a[i]>"file"k;j=0;k++;}a[j++]=$0;next}{a[j++]=$0;}END{for(i=0;i<j;i++)print a[i]>"file"k}' i=0 k=1  grepper.txt

其中 grepper.txt 包含输入数据

这会将文件拆分为多个文件，每个文件有一条记录（当然有多行）。

然后 grep 并丢弃不需要的文件。

循环内

grep "login time: 2013-08" fileN && grep "User:" fileN | grep "@" || rm -f fileN

Answer 3

首先，将每条记录读入一个字段数组：

BEGIN { FS = ": " }   # each line has fieldname and value
/^$/ { next }         # skip blank records
$1 == "User" {        # first field of new record
    delete fields     # delete current array
    fields[$1] = $2 } # store field value in array
$1 == "Door IDs" {    # last field of current record
    fields[$1] = $2   # store field value in array
    do_process() }    # process current record
$1 != "User" &&       # fields between first ...
$2 != "Door IDs" {    #             ... and last
    fields[$1] = $2 } # store field value in array

然后，对记录执行任何需要执行的操作。在这里，我打印了"用户"和"上次登录时间"字段，但您可以执行所需的任何处理：

function do_process() {
    print fields["User"], fields["Last login time"] }

请注意，我还没有测试过这段代码...

编辑：根据下面的评论进行了修改。我假设用户字段总是标记新记录的开始。下面是用于读取和存储记录的代码的修订版本：

BEGIN { FS = ": "       # each line has fieldname and value
        first = 1 }     # flag for first record
/^$/ { next }           # skip blank records
$1 == "User" {          # first field of new record
    if (first > 1)      # no data the first time; skip
        do_process()    # process current record
    delete fields       # reset fields for new record
    fields[$1] = $2 }   # store field value in array
$1 == "Door IDs" {      # last field of current record
    fields[$1] = $2     # store field value in array
    do_process() }      # process current record
/./ { fields[$1] = $2 } # store field value in array
END { if (first > 1)    # last record not processed
        do_process() }  # process last record

然后，您可以根据需要处理数据。

Answer 4

将行从 ^User 分组到Door ID，然后仅在匹配时才打印 @.*login time: 20[0-9] ...

我想我终于明白了你的需要：

试试这个：

sed -ne '/^Door ID/!H;/^User:/h;/^Door ID/{x;G;/@.*login time: 20[0-9]/p}' file

这将符合您的要求。

合并每个数据包后，您甚至可以删除与 2013-08 匹配的所有条目：

sed -ne '/^Door ID/!H;/^User:/h;/^Door ID/{x;G;/@.*login time: 20[0-9]/{/login time: 2013-08/!p}}' file