测试模式A和B并返回其间的文本行，包括模式A和amp;乙

时间：2013-09-18 13:50:44

标签： bash awk grep pattern-matching

再次问好stackoverflow！

我有一个非常大的平面文件，我希望导出所有匹配2种不同模式的记录。问题是每条记录中的行数不同，记录相互渗透。记录的最后一行是Door ID，第一行是User :.

我正在电子邮件地址中对@进行测试，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

当然失败了......

所以这是样本数据

User: afshin@runners.org
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: jjnalli@runners.org
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: jjnalli@runners.org  
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

4 个答案:

答案 0 :(得分：1)

也许这样的事情对你有用：

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

答案 1 :(得分：1)

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，每个文件只有一条记录（当然有多行）。

然后grep并丢弃不需要的文件。

在循环中

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

答案 2 :(得分：1)

将行从^User分组到Door ID，而不是仅在匹配@.*login time: 20[0-9]时打印...

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

答案 3 :(得分：1)

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，做你需要做的任何事情。在这里，我打印用户和上次登录时间字段，但您可以进行所需的任何处理：

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

请注意我没有测试过这段代码......

编辑：根据以下评论修改。我假设User字段始终标记新记录的开头。这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

然后您可以随意处理数据。

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