Question

我有一个文件：

$ cat dd.txt
*************************** 1. row ***************************
            File: log-bin.00005
        Position: 120614936
    Binlog_Do_DB:
Binlog_Ignore_DB:

我只想要来自文件log-bin.000055, 120614936

的上述值

现在我想在我的奴隶表中更新file,position。

update db.position set binlog_file='log-bin.000055',position='120614936' where master_host='master.tb.com';

如何为它编写简单的脚本？

Answer 1

使用GNU Grep和positive lookbehind：

$ grep -Po "(?<=File: )[^ ]+|(?<=Position: )[0-9]+" dd.txt
log-bin.00005
120614936

# Store file in variable 
$ file=$(grep -Po "(?<=File: )[^ ]+" dd.txt)

# Store position in variable
$ pos=$(grep -Po "(?<=Position: )[0-9]+" dd.txt)

# Generate query using variable and store in variable
$ query="update db.position set binlog_file='"$file"',position='$pos' where master_host='master.tb.com';"

$ echo $query    

update db.position set binlog_file='log-bin.00005',position='120614936' where master_host='master.tb.com';

# Pipe query to mysql
$ echo $query | mysql -u user -ppassword

作为剧本：

#!/bin/bash

file=$(grep -Po "(?<=File: )[^ ]+" dd.txt)
pos=$(grep -Po "(?<=Position: )[0-9]+" dd.txt)

$ query="update db.position set binlog_file='"$file"',position='$pos' where master_host='master.tb.com';"
$ echo $query | mysql -u user -ppassword

Answer 2

要获取指定的数据，您可以执行以下操作：

cat dd.txt | egrep "File|Position" | cut -d: -f2

如何从文件中获取特定单词以及如何更新表中的单词

2 个答案: