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
- [转]CSS学习笔记
原文:http://www.fx114.net/qa-266-93710.aspx 01.什么是CSS. CSS指层叠样式表(Cascading Style Sheets). ·样式定义如 ...
- 【转】Python3 正则表达式特殊符号及用法（详细列表）
转载自鱼c论坛:https://fishc.com.cn/forum.php?mod=viewthread&tid=57691&extra=page%3D1%26filter%3Dty ...
- 使用 Azure静态web应用+Github全自动部署VUE站点
什么事Azure静态web应用 Azure 静态 Web 应用是一种服务,可从 GitHub 存储库自动构建完整的堆栈 Web 应用,并将其部署到 Azure,目前它还是预览版. Azure 静态 W ...
- tomcat 启动失败
1.tomcat单独通过脚本可以启动,但是http://localhost:8080加载页面失败 tomcat 启动失败: a.检查JAVA_HOME和CATALINA_HOME是否配置正确:或者直接 ...
- oracle 1day
1.主流数据库: 2.项目选择数据库的原则: 3.oracle 常用用户sys (sysdba系统管理员),system(sysoper系统操作员),scott(密码tiger) sys login: ...
- non-local denoising methods
NL-Means算法 在噪声先验为高斯噪声的基础上, 进行non-local的平均,在2005年由Baudes提出,该算法使用自然图像中普遍存在的冗余信息来去噪声.与常用的双线性滤波.中值滤波等利用图 ...
- D. Concatenated Multiples 解析(思維)
Codeforce 1029 D. Concatenated Multiples 解析(思維) 今天我們來看看CF1029D 題目連結 題目 給你一個序列\(a\)和一個數字\(k\),求有幾種ind ...
- 面试时说Redis是单线程的，被喷惨了！
Redis是单线程的,这话搁以前,是横着走的,谁都知道的真理.现在不一样,Redis 变了.再说这句话,多少得有质疑的语气来跟你辩驳一番.意志不坚定的,可能就缴械投降,顺着别人走了. 到底是什么样的, ...
- Java 中的 反射机制
概念明确 什么是类的对象? 类的对象就是:基于某个类 new 出来的对象,也称为实例对象.这个很容易理解,就不过多阐述了. 什么是类对象? 类对象就是:类加载的产物,封装了一个类的所有信息(类名.父类 ...
- 【Luogu】 P6274 [eJOI2017]六 题解
首先,题目说了最多\(6\)个质因数. 如此小的数据范围,不是状压还是啥? 然后,我们可以发现一个性质:只要两个因数有相同的质因数(不管次数是多少),两者就不互质. 这启示我们用一个二进制数来表示一类 ...