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% |


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 ,网上说这个是awk的圣经,这本书在京东卖1000RMB+,可见这本书的价值,下载地址为: http://pan.baidu.com/s/1gdwbF71,关于原始脚本的意思在这里做一个简单的语法的解释,以及作者脚本的逻辑。


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

/^[0-9a-f]+.[0-9a-f]+/ { match($0,/^[0-9a-f]+/); 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

引用博客地址如下:

http://cephnotes.ksperis.com/blog/2015/02/23/get-the-number-of-placement-groups-per-osd/