查询osd上的pg数
本文中的命令的第一版来源于国外的一个博客,后面的版本为我自己修改的版本
查询的命令如下:
ceph pg dump | awk '
/^pg_stat/ { col=1; while($col!="up") {col++}; col++ }
/^[0-9a-f]+\.[0-9a-f]+/ { match($0,/^[0-9a-f]+/); pool=substr($0, RSTART, RLENGTH); poollist[pool]=0;
up=$col; i=0; RSTART=0; RLENGTH=0; delete osds; while(match(up,/[0-9]+/)>0) { osds[++i]=substr(up,RSTART,RLENGTH); up = substr(up, RSTART+RLENGTH) }
for(i in osds) {array[osds[i],pool]++; osdlist[osds[i]];}
}
END {
printf("\n");
printf("pool :\t"); for (i in poollist) printf("%s\t",i); printf("| SUM \n");
for (i in poollist) printf("--------"); printf("----------------\n");
for (i in osdlist) { printf("osd.%i\t", i); sum=0;
for (j in poollist) { printf("%i\t", array[i,j]); sum+=array[i,j]; poollist[j]+=array[i,j] }; printf("| %i\n",sum) }
for (i in poollist) printf("--------"); printf("----------------\n");
printf("SUM :\t"); for (i in poollist) printf("%s\t",poollist[i]); printf("|\n");
}'
默认的输出如下:
pool : 0 1 2 | SUM
----------------------------------------
osd.4 54 133 79 | 266
osd.5 57 104 88 | 249
osd.6 61 132 86 | 279
osd.7 54 114 85 | 253
osd.8 63 123 85 | 271
osd.0 62 120 87 | 269
osd.1 52 126 81 | 259
osd.2 52 103 88 | 243
osd.3 57 125 89 | 271
----------------------------------------
SUM : 512 1080 768 |
这个有个问题就是osd是乱序的,并且对于一个存储池来说不清楚哪个osd的pg是最多的
重构第一版:
跟上面的相比按顺序来排列
ceph pg dump | awk '
/^pg_stat/ { col=1; while($col!="up") {col++}; col++ }
/^[0-9a-f]+\.[0-9a-f]+/ { match($0,/^[0-9a-f]+/); pool=substr($0, RSTART, RLENGTH); poollist[pool]=0;
up=$col; i=0; RSTART=0; RLENGTH=0; delete osds; while(match(up,/[0-9]+/)>0) { osds[++i]=substr(up,RSTART,RLENGTH); up = substr(up, RSTART+RLENGTH) }
for(i in osds) {array[osds[i],pool]++; osdlist[osds[i]];}
}
END {
printf("\n");
slen=asorti(poollist,newpoollist);
printf("pool :\t");for (i=1;i<=slen;i++) {printf("%s\t", newpoollist[i])}; printf("| SUM \n");
for (i in poollist) printf("--------"); printf("----------------\n");
slen1=asorti(osdlist,newosdlist)
delete poollist;
for (i=1;i<=slen1;i++) { printf("osd.%i\t", newosdlist[i]); sum=0;
for (j=1;j<=slen;j++) { printf("%i\t", array[newosdlist[i],newpoollist[j]]); sum+=array[newosdlist[i],newpoollist[j]]; poollist[j]+=array[newosdlist[i],newpoollist[j]] }; printf("| %i\n",sum)
}
for (i in poollist) printf("--------"); printf("----------------\n");
printf("SUM :\t"); for (i=1;i<=slen;i++) printf("%s\t",poollist[i]); printf("|\n");
}'
输出结果为下面的,可以看到现在是按顺序来的,存储池是顺序的,osd编号也是顺序的
pool : 0 1 2 | SUM
----------------------------------------
osd.0 62 120 87 | 269
osd.1 52 126 81 | 259
osd.2 52 103 88 | 243
osd.3 57 125 89 | 271
osd.4 54 133 79 | 266
osd.5 57 104 88 | 249
osd.6 61 132 86 | 279
osd.7 54 114 85 | 253
osd.8 63 123 85 | 271
----------------------------------------
SUM : 512 1080 768 |
重构第二版:
包含osd pool的排序,包含osd的排序,输出平均pg数目,输出最大的osd编号,输出超过平均值的百分比
ceph pg dump | awk '
/^pg_stat/ { col=1; while($col!="up") {col++}; col++ }
/^[0-9a-f]+\.[0-9a-f]+/ { match($0,/^[0-9a-f]+/); pool=substr($0, RSTART, RLENGTH); poollist[pool]=0;
up=$col; i=0; RSTART=0; RLENGTH=0; delete osds; while(match(up,/[0-9]+/)>0) { osds[++i]=substr(up,RSTART,RLENGTH); up = substr(up, RSTART+RLENGTH) }
for(i in osds) {array[osds[i],pool]++; osdlist[osds[i]];}
}
END {
printf("\n");
slen=asorti(poollist,newpoollist);
printf("pool :\t");for (i=1;i<=slen;i++) {printf("%s\t", newpoollist[i])}; printf("| SUM \n");
for (i in poollist) printf("--------"); printf("----------------\n");
slen1=asorti(osdlist,newosdlist)
delete poollist;
for (i=1;i<=slen1;i++) { printf("osd.%i\t", newosdlist[i]); sum=0;
for (j=1;j<=slen;j++) { printf("%i\t", array[newosdlist[i],newpoollist[j]]); sum+=array[newosdlist[i],newpoollist[j]]; poollist[j]+=array[newosdlist[i],newpoollist[j]];if(array[newosdlist[i],newpoollist[j]] != 0){poolhasid[j]+=1 };if(array[newosdlist[i],newpoollist[j]]>maxpoolosd[j]){maxpoolosd[j]=array[newosdlist[i],newpoollist[j]];maxosdid[j]=newosdlist[i]}}; printf("| %i\n",sum)}
for (i in poollist) printf("--------"); printf("----------------\n");
printf("SUM :\t"); for (i=1;i<=slen;i++) printf("%s\t",poollist[i]); printf("|\n");
printf("AVE :\t"); for (i=1;i<=slen;i++) printf("%d\t",poollist[i]/poolhasid[i]); printf("|\n");
printf("max :\t"); for (i=1;i<=slen;i++) printf("%s\t",maxpoolosd[i]); printf("|\n");
printf("osdid :\t"); for (i=1;i<=slen;i++) printf("osd.%s\t",maxosdid[i]); printf("|\n");
printf("per:\t"); for (i=1;i<=slen;i++) printf("%.1f%\t",100*(maxpoolosd[i]-poollist[i]/poolhasid[i])/(poollist[i]/poolhasid[i])); printf("|\n");
}'
输出如下:
pool : 0 1 2 | SUM
----------------------------------------
osd.0 62 120 87 | 269
osd.1 52 126 81 | 259
osd.2 52 103 88 | 243
osd.3 57 125 89 | 271
osd.4 54 133 79 | 266
osd.5 57 104 88 | 249
osd.6 61 132 86 | 279
osd.7 54 114 85 | 253
osd.8 63 123 85 | 271
----------------------------------------
SUM : 512 1080 768 |
AVE : 56 120 85 |
max : 63 133 89 |
osdid : osd.8 osd.4 osd.3 |
per: 10.7% 10.8% 4.3% |
重构第三版:
包含osd pool的排序,包含osd的排序,输出平均pg数目,输出最大的osd编号,输出最大超过平均值的百分比,输出最少pg的osd编号,输出最小低于平均值的百分比
ceph pg dump | awk '
/^pg_stat/ { col=1; while($col!="up") {col++}; col++ }
/^[0-9a-f]+\.[0-9a-f]+/ { match($0,/^[0-9a-f]+/); pool=substr($0, RSTART, RLENGTH); poollist[pool]=0;
up=$col; i=0; RSTART=0; RLENGTH=0; delete osds; while(match(up,/[0-9]+/)>0) { osds[++i]=substr(up,RSTART,RLENGTH); up = substr(up, RSTART+RLENGTH) }
for(i in osds) {array[osds[i],pool]++; osdlist[osds[i]];}
}
END {
printf("\n");
slen=asorti(poollist,newpoollist);
printf("pool :\t");for (i=1;i<=slen;i++) {printf("%s\t", newpoollist[i])}; printf("| SUM \n");
for (i in poollist) printf("--------"); printf("----------------\n");
slen1=asorti(osdlist,newosdlist)
delete poollist;
for (j=1;j<=slen;j++) {maxpoolosd[j]=0};
for (j=1;j<=slen;j++) {for (i=1;i<=slen1;i++){if (array[newosdlist[i],newpoollist[j]] >0 ){minpoolosd[j]=array[newosdlist[i],newpoollist[j]] ;break } }};
for (i=1;i<=slen1;i++) { printf("osd.%i\t", newosdlist[i]); sum=0;
for (j=1;j<=slen;j++) { printf("%i\t", array[newosdlist[i],newpoollist[j]]); sum+=array[newosdlist[i],newpoollist[j]]; poollist[j]+=array[newosdlist[i],newpoollist[j]];if(array[newosdlist[i],newpoollist[j]] != 0){poolhasid[j]+=1 };if(array[newosdlist[i],newpoollist[j]]>maxpoolosd[j]){maxpoolosd[j]=array[newosdlist[i],newpoollist[j]];maxosdid[j]=newosdlist[i]};if(array[newosdlist[i],newpoollist[j]] != 0){if(array[newosdlist[i],newpoollist[j]]<=minpoolosd[j]){minpoolosd[j]=array[newosdlist[i],newpoollist[j]];minosdid[j]=newosdlist[i]}}}; printf("| %i\n",sum)} for (i in poollist) printf("--------"); printf("----------------\n");
slen2=asorti(poollist,newpoollist);
printf("SUM :\t"); for (i=1;i<=slen;i++) printf("%s\t",poollist[i]); printf("|\n");
printf("Osd :\t"); for (i=1;i<=slen;i++) printf("%s\t",poolhasid[i]); printf("|\n");
printf("AVE :\t"); for (i=1;i<=slen;i++) printf("%.2f\t",poollist[i]/poolhasid[i]); printf("|\n");
printf("Max :\t"); for (i=1;i<=slen;i++) printf("%s\t",maxpoolosd[i]); printf("|\n");
printf("Osdid :\t"); for (i=1;i<=slen;i++) printf("osd.%s\t",maxosdid[i]); printf("|\n");
printf("per:\t"); for (i=1;i<=slen;i++) printf("%.1f%\t",100*(maxpoolosd[i]-poollist[i]/poolhasid[i])/(poollist[i]/poolhasid[i])); printf("|\n");
for (i=1;i<=slen2;i++) printf("--------");printf("----------------\n");
printf("min :\t"); for (i=1;i<=slen;i++) printf("%s\t",minpoolosd[i]); printf("|\n");
printf("osdid :\t"); for (i=1;i<=slen;i++) printf("osd.%s\t",minosdid[i]); printf("|\n");
printf("per:\t"); for (i=1;i<=slen;i++) printf("%.1f%\t",100*(minpoolosd[i]-poollist[i]/poolhasid[i])/(poollist[i]/poolhasid[i])); printf("|\n");
}'
输出如下:
dumped all in format plain
pool : 0 1 2 | SUM
----------------------------------------
osd.0 206 206 53 | 465
osd.1 22 19 5 | 46
osd.2 202 196 49 | 447
osd.3 19 25 6 | 50
osd.4 29 35 9 | 73
osd.5 34 31 6 | 71
----------------------------------------
SUM : 512 512 128 |
AVE : 85 85 21 |
max : 206 206 53 |
osdid : osd.0 osd.0 osd.0 |
per: 141.4% 141.4% 148.4% |
---------------------------------------
min : 19 19 5 |
osdid : osd.3 osd.1 osd.1 |
per: -77.7% -77.7% -76.6% |
Luminous以及之后的版本
ceph pg dump | awk '
/^PG_STAT/ { col=1; while($col!="UP") {col++}; col++ }
/^[0-9a-f]+\.[0-9a-f]+/ { match($0,/^[0-9a-f]+/); pool=substr($0, RSTART, RLENGTH); poollist[pool]=0;
up=$col; i=0; RSTART=0; RLENGTH=0; delete osds; while(match(up,/[0-9]+/)>0) { osds[++i]=substr(up,RSTART,RLENGTH); up = substr(up, RSTART+RLENGTH) }
for(i in osds) {array[osds[i],pool]++; osdlist[osds[i]];}
}
END {
printf("\n");
slen=asorti(poollist,newpoollist);
printf("pool :\t");for (i=1;i<=slen;i++) {printf("%s\t", newpoollist[i])}; printf("| SUM \n");
for (i in poollist) printf("--------"); printf("----------------\n");
slen1=asorti(osdlist,newosdlist)
delete poollist;
for (j=1;j<=slen;j++) {maxpoolosd[j]=0};
for (j=1;j<=slen;j++) {for (i=1;i<=slen1;i++){if (array[newosdlist[i],newpoollist[j]] >0 ){minpoolosd[j]=array[newosdlist[i],newpoollist[j]] ;break } }};
for (i=1;i<=slen1;i++) { printf("osd.%i\t", newosdlist[i]); sum=0;
for (j=1;j<=slen;j++) { printf("%i\t", array[newosdlist[i],newpoollist[j]]); sum+=array[newosdlist[i],newpoollist[j]]; poollist[j]+=array[newosdlist[i],newpoollist[j]];if(array[newosdlist[i],newpoollist[j]] != 0){poolhasid[j]+=1 };if(array[newosdlist[i],newpoollist[j]]>maxpoolosd[j]){maxpoolosd[j]=array[newosdlist[i],newpoollist[j]];maxosdid[j]=newosdlist[i]};if(array[newosdlist[i],newpoollist[j]] != 0){if(array[newosdlist[i],newpoollist[j]]<=minpoolosd[j]){minpoolosd[j]=array[newosdlist[i],newpoollist[j]];minosdid[j]=newosdlist[i]}}}; printf("| %i\n",sum)} for (i in poollist) printf("--------"); printf("----------------\n");
slen2=asorti(poollist,newpoollist);
printf("SUM :\t"); for (i=1;i<=slen;i++) printf("%s\t",poollist[i]); printf("|\n");
printf("Osd :\t"); for (i=1;i<=slen;i++) printf("%s\t",poolhasid[i]); printf("|\n");
printf("AVE :\t"); for (i=1;i<=slen;i++) printf("%.2f\t",poollist[i]/poolhasid[i]); printf("|\n");
printf("Max :\t"); for (i=1;i<=slen;i++) printf("%s\t",maxpoolosd[i]); printf("|\n");
printf("Osdid :\t"); for (i=1;i<=slen;i++) printf("osd.%s\t",maxosdid[i]); printf("|\n");
printf("per:\t"); for (i=1;i<=slen;i++) printf("%.1f%\t",100*(maxpoolosd[i]-poollist[i]/poolhasid[i])/(poollist[i]/poolhasid[i])); printf("|\n");
for (i=1;i<=slen2;i++) printf("--------");printf("----------------\n");
printf("min :\t"); for (i=1;i<=slen;i++) printf("%s\t",minpoolosd[i]); printf("|\n");
printf("osdid :\t"); for (i=1;i<=slen;i++) printf("osd.%s\t",minosdid[i]); printf("|\n");
printf("per:\t"); for (i=1;i<=slen;i++) printf("%.1f%\t",100*(minpoolosd[i]-poollist[i]/poolhasid[i])/(poollist[i]/poolhasid[i])); printf("|\n");
}'
luminous之后的版本json的输出小写改成大写了,需要替换几个字段,上面的已经增加了
上面的处理使用的是awk处理,开始的时候看不懂什么意思,然后就去看了这本书《The AWK Programming Language》
语法的解释
/^pg_stat/ { col=1; while($col!="up") {col++}; col++ }
这个是匹配pg dump 的输出结果里面pg_stat那个字段,开始计数为1,不是up值就将col的值加1,这个匹配到的就是我们经常看到的[1,10]这个值最后的col++是将col值+1,因为字段里面有up,up_primary,我们需要的是up_primary
/[1]+.[0-9a-f]+/ { match($0,/[2]+/); pool=substr($0, RSTART, RLENGTH); poollist[pool]=0;
这个是匹配前面的 1.17a pg号 ,使用自带的match函数 做字符串的过滤统计匹配.号前面的存储池ID, 并得到 RSTART, RLENGTH 值,这个是取到前面的存储池ID,使用substr 函数,就可以得到pool的值了,poollist[pool]=0,是将数组的值置为0
up=$col; i=0; RSTART=0; RLENGTH=0; delete osds; while(match(up,/[0-9]+/)>0) { osds[++i]=substr(up,RSTART,RLENGTH); up = substr(up, RSTART+RLENGTH) }
先将变量置0,然后将osd编号一个个输入到osds[i]的数组当中去
for(i in osds) {array[osds[i],pool]++; osdlist[osds[i]];}
将osds数组中的值输入到数组当中去,并且记录成osdlist,和数组array[osd[i],pool]
printf("\n");
printf("pool :\t"); for (i in poollist) printf("%s\t",i); printf("| SUM \n");
打印osd pool的编号
for (i in poollist) printf("--------"); printf("----------------\n");
根据osd pool的长度打印----
for (i in osdlist) { printf("osd.%i\t", i); sum=0;
打印osd的编号
for (j in poollist) { printf("%i\t", array[i,j]); sum+=array[i,j]; poollist[j]+=array[i,j] }; printf("| %i\n",sum) }
打印对应的osd的pg数目,并做求和的统计
for (i in poollist) printf("--------"); printf("----------------\n");
printf("SUM :\t"); for (i in poollist) printf("%s\t",poollist[i]); printf("|\n");
打印新的poollist里面的求和的值
修改版本里面用到的函数
slen1=asorti(osdlist,newosdlist)
这个是将数组里面的下标进行排序,这里是对osd和poollist的编号进行排序 slen1是拿到数组的长度,使用for进行遍历输出
脚本的逻辑
- 匹配到pg的id和pg对应的osd,
- 使用数组的方式,将统计到的osd id存储起来,
- 然后打印数组
其他资源:
pg设置的计算器:
http://ceph.com/pgcalc/
pg的查询和设置:
http://ceph.com/docs/master/rados/operations/placement-groups/
变更记录
| Why | Who | When |
|---|---|---|
| 创建 | 武汉-运维-磨渣 | 2015-10-04 |
| 修改 | 武汉-运维-磨渣 | 2016-08-24 |
| 修改有0的统计BUG | 武汉-运维-磨渣 | 2016-09-08 |
| 增加luminous版本脚本 | 武汉-运维-磨渣 | 2020-06-17 |
引用博客地址如下:
http://cephnotes.ksperis.com/blog/2015/02/23/get-the-number-of-placement-groups-per-osd/
查询osd上的pg数的更多相关文章
- Ceph之PG数调整
1. PG介绍 PG, Placement Groups.CRUSH先将数据分解成一组对象,然后根据对象名称.复制级别和系统中的PG数等信息执行散列操作,再将结果生成PG ID.可以将PG看做一个逻辑 ...
- ceph之查看osd上pg的分布
一.概述 osd上pg的分布决定了数据分布的均匀与否,所以能直观的看到pg到osd的上分布是很有必要的: ceph只身提供了相关的命令: #ceph pg ls-by-osd.{osd_id} #fo ...
- PG数计算
PG数计算 原地址:http://xiaqunfeng.cc/2017/09/15/too-many-PGs-per-OSD/ ceph告警问题:”too many PGs per OSD” 的解决方 ...
- 调整ceph的pg数(pg_num, pgp_num)
https://www.jianshu.com/p/ae96ee24ef6c 调整ceph的pg数 PG全称是placement groups,它是ceph的逻辑存储单元.在数据存储到cesh时,先打 ...
- 每个pool pg数计算
ceph PGs per Pool Calculator 原文档:http://xiaqunfeng.cc/2017/09/18/ceph-PGs-per-Pool-Calculator/ 2017- ...
- 李呈祥:bilibili在湖仓一体查询加速上的实践与探索
导读: 本文主要介绍哔哩哔哩在数据湖与数据仓库一体架构下,探索查询加速以及索引增强的一些实践.主要内容包括: 什么是湖仓一体架构 哔哩哔哩目前的湖仓一体架构 湖仓一体架构下,数据的排序组织优化 湖仓一 ...
- CDOJ 1104 求两个数列的子列的交集 查询区间小于A的数有多少个 主席树
求两个数列的子列的交集 Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://acm.uestc.edu.cn/#/problem/show/1104 ...
- mysql 存储过程:提供查询语句并返回查询执行影响的行数
mysql 存储过程:提供查询语句并返回查询执行影响的行数DELIMITER $$ DROP PROCEDURE IF EXISTS `p_get_select_row_number`$$ CREAT ...
- sqlserver查询所有表的行数的sql语句
原文:sqlserver查询所有表的行数的sql语句 select object_name(id),rowcnt from sysindexes where indid<2 and object ...
随机推荐
- GA001-181-21
Composite State with History The Composite State with History Pattern describes an entity (e.g. Cl ...
- spring boot:使mybatis访问多个druid数据源(spring boot 2.3.2)
一,为什么要使用多个数据源? 1,什么情况下需要使用多个数据源? 当我们需要访问不同的数据库时,则需要配置配置多个数据源, 例如:电商的业务数据库(包括用户/商品/订单等) 和统 ...
- linux(centos8):用grep命令查找文件内容
一,grep的用途: linux平台有最常用的三大文本处理工具:awk/sed/grep grep的功能:搜索指定文件的内容,按照指定的模式匹配,并输出匹配内容所在的行. 需要注意的地方:grep只支 ...
- 3.QOpenGLWidget-通过着色器来渲染渐变三角形
在上章2.通过QOpenGLWidget绘制三角形,我们学习绘制三角形还是单色的,本章将为三角形每个顶点着色. 1.着色器描述 着色器的开头总是要声明版本,接着是输入和输出变量.uniform和m ...
- C# 获取页面Post过来的数据
/// <summary> /// 获取post过来的数据 /// </summary> /// <param name="page">< ...
- MySQL数据库基础-3-SQL 基本概念
SQL 基本概念 约束:constraint,表中的数据要遵守的限制 主键:一个或多个字段的组合,填入的数据必须能在本表中唯一标识本行:必须提供数据,即NOT NULL,一个表只能有一个 惟一键:一个 ...
- Typora图片一键传——picgo使用兰空图床
Typora picgo使用兰空图床 最近smms图床免费版的速度眼看着越来越慢,特别是到晚上,图片几乎是在原地打转,于是我想替换掉Typora中picgo默认使用的smms图床,网上的教程清一色地只 ...
- 手写webpack核心原理,再也不怕面试官问我webpack原理
手写webpack核心原理 目录 手写webpack核心原理 一.核心打包原理 1.1 打包的主要流程如下 1.2 具体细节 二.基本准备工作 三.获取模块内容 四.分析模块 五.收集依赖 六.ES6 ...
- ActiveMQ第一个示例
首先先安装ActiveMQ:https://www.cnblogs.com/hejianliang/p/9149590.html 创建Java项目,把 activemq-all-5.15.4.jar ...
- 国云数据:中国版的Snowflake,国内数据中台领导者
[股神巴菲特加持,今年最受关注美股IPO ] 这段时间, 由股神巴菲特54年来首次打新的美股IPO公司Snowflake迅速得到业界重点关注.Snowflake已于2020年9月16日正式上市,发行价 ...