MYSQL INNODB 组合索引分支节点数据解析

1、本文证明组合索引的所有键值在分支节点(非叶子结点也进行了存储)。
2、本文给出B+ 索引如何进行验证其B+树结构

关于B树结构(不是B+树)可以参考：
http://blog.itpub.net/7728585/viewspace-2126929/

脚本：
mysql> create table testzh(id int  primary key auto_increment ,id2 int,id3 int,name varchar(20), key(id2,id3));
Query OK, 0 rows affected (0.07 sec)


 delimiter //
 create procedure ins()
     begin
    declare i int;
     set i=0;
     while i<100000 do
         insert into testzh(id2,id3,name) values(FLOOR((RAND()*100000)),FLOOR((RAND()*100000)),'gaopeng');
         set i=i+1;
     end while;
  end;
//
delimiter ;


写一个程序 主要读取下面几位每个块的：
64-66字节：B+层次，0是叶子结点，向上分别是1,2.... 如果是三层结构的则根结点为层次为2，分支为1，叶子结点为0
66-74字节：索引ID，对应INNODB_SYS_INDEXES 的INDEX_ID字段
24-26字节：块类型，我们主要查看0X45BF的类型是数据类型    

程序附在最后，下面是跑出来的结果我的：
key(id2,id3)的INDEX_ID是
mysql> select * from information_schema.INNODB_SYS_INDEXES where  TABLE_ID=132;
+----------+---------+----------+------+----------+---------+-------+-----------------+
| INDEX_ID | NAME    | TABLE_ID | TYPE | N_FIELDS | PAGE_NO | SPACE | MERGE_THRESHOLD |
+----------+---------+----------+------+----------+---------+-------+-----------------+
|      160 | PRIMARY |      132 |    3 |        1 |       3 |   207 |              50 |
|      161 | id2     |      132 |    0 |        2 |       4 |   207 |              50 |
+----------+---------+----------+------+----------+---------+-------+-----------------+




注意程序计数块从0开始，为贴近C/C++数组规定
[root@testmy test]# ./b_level testzh.ibd|grep 161
Block id is 4:Index page no is  161 : B+ Tree Level is  1
Block id is 8:Index page no is  161 : B+ Tree Level is  0
Block id is 9:Index page no is  161 : B+ Tree Level is  0
Block id is 12:Index page no is  161 : B+ Tree Level is  0
Block id is 14:Index page no is  161 : B+ Tree Level is  0
Block id is 19:Index page no is  161 : B+ Tree Level is  0
Block id is 20:Index page no is  161 : B+ Tree Level is  0
Block id is 21:Index page no is  161 : B+ Tree Level is  0
Block id is 22:Index page no is  161 : B+ Tree Level is  0
Block id is 31:Index page no is  161 : B+ Tree Level is  0
Block id is 32:Index page no is  161 : B+ Tree Level is  0
Block id is 34:Index page no is  161 : B+ Tree Level is  0
Block id is 35:Index page no is  161 : B+ Tree Level is  0
Block id is 36:Index page no is  161 : B+ Tree Level is  0
Block id is 37:Index page no is  161 : B+ Tree Level is  0
Block id is 38:Index page no is  161 : B+ Tree Level is  0
Block id is 41:Index page no is  161 : B+ Tree Level is  0
Block id is 53:Index page no is  161 : B+ Tree Level is  0
Block id is 54:Index page no is  161 : B+ Tree Level is  0
Block id is 55:Index page no is  161 : B+ Tree Level is  0
...........
Block id is 348:Index page no is  161 : B+ Tree Level is  0
Block id is 349:Index page no is  161 : B+ Tree Level is  0
Block id is 350:Index page no is  161 : B+ Tree Level is  0
Block id is 351:Index page no is  161 : B+ Tree Level is  0
Block id is 352:Index page no is  161 : B+ Tree Level is  0
Block id is 353:Index page no is  161 : B+ Tree Level is  0
Block id is 354:Index page no is  161 : B+ Tree Level is  0
Block id is 355:Index page no is  161 : B+ Tree Level is  0
Block id is 356:Index page no is  161 : B+ Tree Level is  0
Block id is 357:Index page no is  161 : B+ Tree Level is  0
Block id is 358:Index page no is  161 : B+ Tree Level is  0
Block id is 359:Index page no is  161 : B+ Tree Level is  0
Block id is 360:Index page no is  161 : B+ Tree Level is  0
Block id is 361:Index page no is  161 : B+ Tree Level is  0
Block id is 362:Index page no is  161 : B+ Tree Level is  0
Block id is 363:Index page no is  161 : B+ Tree Level is  0
Block id is 364:Index page no is  161 : B+ Tree Level is  0
这就是我161索引key(id2,id3)全部块，
这里我们看到我们的这个组合索引是 2层的 B+树结构
Block id is 4:Index page no is  161 : B+ Tree Level is  1
这个块就是根结点，那么我们需要读取它需要用到我写过的另外一个小程序
，用于读取其数据，但是这个程序写死了读取2个INT类型数据如下，按照顺序给出
并且给出偏移量,并且给出指向的块(注意这个指向的块是我猜测的,随后验证)：

那我们就读取BLOCK 4(dataview 程序放到最后)
[root@testmy test]# ./dataview testzh.ibd 4
Index_no is:161
find first one record!
B:23,A:59613,offset:126,leaf block:8
B:736,A:31951,offset:2106,leaf block:249
B:1591,A:58218,offset:1072,leaf block:203
B:2390,A:34725,offset:2326,leaf block:324
B:3231,A:16448,offset:500,leaf block:54
B:3647,A:95182,offset:3118,leaf block:360
B:4050,A:42271,offset:1776,leaf block:235
B:4751,A:62810,offset:1028,leaf block:201
B:5614,A:53731,offset:1886,leaf block:240
B:6482,A:29372,offset:346,leaf block:34
B:7216,A:26606,offset:2524,leaf block:333
B:8028,A:78263,offset:1138,leaf block:206
B:8777,A:61401,offset:2238,leaf block:255
B:9562,A:34379,offset:698,leaf block:63
B:10353,A:93823,offset:2150,leaf block:251
B:11114,A:50825,offset:1358,leaf block:216
B:11859,A:57029,offset:2634,leaf block:338
B:12611,A:67208,offset:214,leaf block:19
B:13378,A:43201,offset:2062,leaf block:247
B:14179,A:72734,offset:940,leaf block:197
B:14972,A:39942,offset:2458,leaf block:330
B:15743,A:9319,offset:632,leaf block:60
B:16488,A:33875,offset:2392,leaf block:327
B:17320,A:32881,offset:1204,leaf block:209
B:18117,A:6361,offset:2084,leaf block:248
B:18966,A:70177,offset:368,leaf block:35
B:19722,A:16497,offset:1710,leaf block:232
B:20563,A:60667,offset:896,leaf block:195
B:21357,A:38077,offset:2040,leaf block:246
B:22151,A:15974,offset:522,leaf block:55
B:22612,A:89791,offset:3008,leaf block:355
B:23070,A:74154,offset:1534,leaf block:224
B:23935,A:42535,offset:852,leaf block:193
B:24814,A:51733,offset:1644,leaf block:229
B:25714,A:232,offset:170,leaf block:12
B:26468,A:119,offset:2678,leaf block:340
B:27177,A:3573,offset:1402,leaf block:218
B:27870,A:22983,offset:2700,leaf block:341
B:28679,A:71426,offset:676,leaf block:62
B:29439,A:77570,offset:2216,leaf block:254
B:30160,A:66775,offset:1424,leaf block:219
B:30865,A:17274,offset:2722,leaf block:342
B:31611,A:25970,offset:390,leaf block:36
B:32377,A:56419,offset:1930,leaf block:242
B:33117,A:8818,offset:1292,leaf block:213
B:33838,A:19538,offset:2898,leaf block:350
B:34489,A:92156,offset:742,leaf block:129
B:35286,A:6546,offset:2194,leaf block:253
B:36076,A:40723,offset:1314,leaf block:214
B:36763,A:20769,offset:2876,leaf block:349
B:37450,A:46558,offset:236,leaf block:20
B:38214,A:7960,offset:2568,leaf block:335
B:38945,A:498,offset:984,leaf block:199
B:39726,A:40552,offset:2260,leaf block:321
B:40462,A:1439,offset:588,leaf block:58
B:41208,A:19781,offset:2612,leaf block:337
B:41939,A:99119,offset:1248,leaf block:211
B:42637,A:26247,offset:2436,leaf block:329
B:43441,A:1592,offset:434,leaf block:38
B:44198,A:74037,offset:2480,leaf block:331
B:44889,A:50243,offset:1226,leaf block:210
B:45599,A:7689,offset:2788,leaf block:345
B:46378,A:44374,offset:786,leaf block:131
B:47200,A:3361,offset:2370,leaf block:326
B:47932,A:52288,offset:1336,leaf block:215
B:48736,A:82841,offset:2128,leaf block:250
B:49492,A:68781,offset:148,leaf block:9
B:50257,A:38011,offset:2348,leaf block:325
B:51074,A:96951,offset:1446,leaf block:220
B:51847,A:62231,offset:2744,leaf block:343
B:52618,A:31192,offset:654,leaf block:61
B:53362,A:72348,offset:2590,leaf block:336
B:54029,A:82275,offset:1380,leaf block:217
B:54788,A:28840,offset:2546,leaf block:334
B:55538,A:98163,offset:324,leaf block:32
B:56313,A:20704,offset:2282,leaf block:322
B:57065,A:90078,offset:1116,leaf block:205
B:57525,A:17975,offset:3096,leaf block:359
B:57973,A:52757,offset:1798,leaf block:236
B:58427,A:8888,offset:3140,leaf block:361
B:58890,A:65370,offset:764,leaf block:130
B:59627,A:28852,offset:1996,leaf block:320
B:60454,A:26779,offset:1160,leaf block:207
B:61261,A:35533,offset:2304,leaf block:323
B:62009,A:21826,offset:280,leaf block:22
B:62517,A:26392,offset:2986,leaf block:354
B:62950,A:3417,offset:1556,leaf block:225
B:63882,A:92888,offset:874,leaf block:194
B:64651,A:5358,offset:1732,leaf block:233
B:65511,A:32339,offset:544,leaf block:56
B:66290,A:26183,offset:1952,leaf block:243
B:67063,A:60386,offset:1182,leaf block:208
B:67767,A:48347,offset:2502,leaf block:332
B:68527,A:49114,offset:412,leaf block:37
B:69290,A:16760,offset:1864,leaf block:239
B:70112,A:32543,offset:1050,leaf block:202
B:70941,A:92527,offset:1908,leaf block:241
B:71758,A:23571,offset:610,leaf block:59
B:72210,A:32172,offset:3030,leaf block:356
B:72689,A:9869,offset:1666,leaf block:230
B:73155,A:27492,offset:3162,leaf block:362
B:73580,A:84669,offset:918,leaf block:196
B:74380,A:61469,offset:1754,leaf block:234
B:75228,A:21920,offset:192,leaf block:14
B:75625,A:16519,offset:3052,leaf block:357
B:76082,A:45066,offset:1600,leaf block:227
B:76901,A:3530,offset:962,leaf block:198
B:77370,A:45042,offset:3184,leaf block:363
B:77786,A:63188,offset:1688,leaf block:231
B:78604,A:5492,offset:566,leaf block:57
B:79500,A:6597,offset:1842,leaf block:238
B:80265,A:85811,offset:1094,leaf block:204
B:81035,A:6126,offset:2018,leaf block:245
B:81911,A:21386,offset:302,leaf block:31
B:82717,A:10268,offset:1622,leaf block:228
B:83172,A:46978,offset:3206,leaf block:364
B:83602,A:89835,offset:830,leaf block:192
B:84063,A:98697,offset:2964,leaf block:353
B:84571,A:9764,offset:1578,leaf block:226
B:85033,A:59776,offset:3074,leaf block:358
B:85471,A:83888,offset:478,leaf block:53
B:86224,A:39031,offset:2172,leaf block:252
B:86980,A:61579,offset:1006,leaf block:244
B:87142,A:42336,offset:1974,leaf block:200
B:87905,A:39935,offset:2656,leaf block:339
B:88630,A:74063,offset:258,leaf block:21
B:89335,A:42509,offset:2810,leaf block:346
B:90013,A:10693,offset:1490,leaf block:222
B:90671,A:21582,offset:2920,leaf block:351
B:91364,A:38857,offset:720,leaf block:128
B:92062,A:98705,offset:2414,leaf block:328
B:92852,A:82534,offset:1270,leaf block:212
B:93665,A:57106,offset:1820,leaf block:237
B:94450,A:36182,offset:456,leaf block:41
B:95109,A:82179,offset:2854,leaf block:348
B:95803,A:39017,offset:1468,leaf block:221
B:96562,A:53131,offset:2832,leaf block:347
B:97236,A:69746,offset:808,leaf block:132
B:97963,A:33843,offset:2766,leaf block:344
B:98709,A:78567,offset:1512,leaf block:223
B:99367,A:91536,offset:2942,leaf block:352


这里B代表是ID1的值，A代表是ID2的值，那么这里我们
证明了在B+数的分支节点存储了组合索引的全部键值，
这里存储的是在叶子结点中物理位置的开始值
随便查询一下：
mysql> select * from testzh where id2=23;
+-------+------+-------+---------+
| id    | id2  | id3   | name    |
+-------+------+-------+---------+
| 99428 |   23 |  9079 | gaopeng |
|   378 |   23 | 59613 | gaopeng |
| 90301 |   23 | 93864 | gaopeng |
+-------+------+-------+---------+
3 rows in set (0.00 sec)


mysql> select * from testzh where id2=736;
+-------+------+-------+---------+
| id    | id2  | id3   | name    |
+-------+------+-------+---------+
|  2716 |  736 | 31951 | gaopeng |
| 95328 |  736 | 53286 | gaopeng |
| 75440 |  736 | 85983 | gaopeng |
+-------+------+-------+---------+
3 rows in set (0.00 sec)


mysql> select * from testzh where id2=1591;
+-------+------+-------+---------+
| id    | id2  | id3   | name    |
+-------+------+-------+---------+
| 10056 | 1591 | 58218 | gaopeng |
+-------+------+-------+---------+


可以看到这些值都是有的。


然后我们更进一步，为了算出每一行占用的空间，我需要取出根结点实际物理空间排序使用：
./dataview testzh.ibd 4 |awk -F ',' '{print $3}'|awk -F ':' '{print $2}'|sort -n
得出
126
148
170
192
214
236
258
........
3074
3096
3118
3140
3162
3184
3206


因为字段字节固定那么算出根结点中，一行数据占用的空间为22字节，6字节的行头开销，8字节的数据开销，剩下的8字节
为一个指针(我只是猜测最后2个字节为块号其他6个字节未知)，那么我们可以验证查看指针指向块的数据是否包含了我们根
结点的指向数据，我们知道B+树中，所有叶子结点包含了分支节点的全部数据，一般来讲只要我们验证分支节点的指向数据
在叶子结点存在，则证明指向的块正确。查询的话，一旦定位到了页块那么剩下的工作就是二分法进行查找数据了。
因为我并没搞懂指针所有位数的含义，但是最后几个字节看起来像块号，所以这样验证。
(当然懂得源码的人来看我的方法太LOW了,我自己现在确实没有能力看懂源码)。


我们来验证一下指向的块中是否包含了指向的数据，这里leaf block没有意义因为是叶子结点
写一个脚本：
test.sh 
./dataview testzh.ibd 8     |grep B:23,A:59613
./dataview testzh.ibd 249   |grep B:736,A:31951
./dataview testzh.ibd 203   |grep B:1591,A:58218
./dataview testzh.ibd 324   |grep B:2390,A:34725
./dataview testzh.ibd 54    |grep B:3231,A:16448
./dataview testzh.ibd 360   |grep B:3647,A:95182
./dataview testzh.ibd 235   |grep B:4050,A:42271
./dataview testzh.ibd 201   |grep B:4751,A:62810
./dataview testzh.ibd 240   |grep B:5614,A:53731
./dataview testzh.ibd 34    |grep B:6482,A:29372
./dataview testzh.ibd 333   |grep B:7216,A:26606
......
一共验证141行


[root@testmy test]# sh test.sh 
B:23,A:59613,offset:126,leaf block:24
B:736,A:31951,offset:126,leaf block:24
B:1591,A:58218,offset:126,leaf block:24
B:2390,A:34725,offset:126,leaf block:24
B:3231,A:16448,offset:126,leaf block:24
B:3647,A:95182,offset:126,leaf block:24
B:4050,A:42271,offset:126,leaf block:24
B:4751,A:62810,offset:126,leaf block:24
B:5614,A:53731,offset:126,leaf block:24
B:6482,A:29372,offset:126,leaf block:24
B:7216,A:26606,offset:126,leaf block:24
B:8028,A:78263,offset:126,leaf block:24
B:8777,A:61401,offset:126,leaf block:24
B:9562,A:34379,offset:126,leaf block:24
B:10353,A:93823,offset:126,leaf block:24
B:11114,A:50825,offset:126,leaf block:24
........
输出141
如下：
[root@testmy test]# sh test.sh |wc -l
141
[root@testmy test]# cat test.sh|wc -l
141


那我们可以发现所有的根结点的记录在叶子结点都找到了，而且我们可以看到偏移量都是物理偏移量的
126，及物理上的第一个记录（因为我这里没有delete过数据）则我们也验证了分支节点存储是物理位置
上块中的第一个值，及这里的偏移量126的值。

最后我们取5个值画一个图，也许大家就明白了 上面是根结点，下面是叶子结点：
（注意叶子节点都是取得开始值，千万不要以为叶子节点就一个值，并且实际上同一层B+树是一个双向链表
关于双向链表参考：http://blog.itpub.net/7728585/viewspace-2125114/
）


水平有限 如果有误请告知


 ./b_level 工具源码：

点击(此处)折叠或打开

1. #include<stdio.h>
   
2. #include<stdlib.h>
   
3. #include<string.h>
   
4. 
   
5. 
   
6. void* reverse(void* p,int length) //Little_endian reverse
   
7. {
   
8.         int i;
   
9.         char* s= (char*)(p);
   
10.         char* temp = (char*)calloc(1,length);
    
11.         memcpy(temp,s,length);
    
12. 
    
13. 
    
14.         for(i=0;i<length;i++)
    
15.         {
    
16.         s[i] = temp[length-1-i];
    
17.         }
    
18.         free(temp);
    
19.         temp=NULL;
    
20.         return p;
    
21. }
    
22. 
    
23. 
    
24. int main(int argc, char* argv[])
    
25. {
    
26.         FILE* fd;
    
27.         short n_lev;
    
28.         long n_idx_id;
    
29.         unsigned short b_type;
    
30.         int i=0;
    
31.         long ed;
    
32. 
    
33.         if(argc != 2 )
    
34.         {
    
35.                         printf("USEAGE ERROR useage:./tool dbf \n");
    
36.                         exit(2);
    
37.         }
    
38. 
    
39.         if(!(fd = fopen(argv[1],"r")))
    
40.     {
    
41.             perror("error:");
    
42.             exit(1);
    
43.     }
    
44.                 fseek(fd,0,SEEK_END);
    
45.                 ed=ftell(fd);
    
46.                 fseek(fd,0,0);
    
47.         printf("file size is %ld\n",ed);
    
48.         while(ftell(fd)<ed)
    
49.                 {
    
50.                         fseek(fd,24,1);
    
51.                         //printf("%d\n",ftell(fd));
    
52.                         fread(&b_type,2,1,fd);
    
53.                         fseek(fd,38,1);
    
54.                         //printf("%d\n",ftell(fd));
    
55.                         fread(&n_lev,2,1,fd);
    
56.                         fread(&n_idx_id,8,1,fd);
    
57.                         i++;
    
58.                         fseek(fd,16384*i,0);
    
59.                         reverse(&n_idx_id,8);
    
60.                         reverse(&n_lev,2);
    
61.                         if(b_type == (unsigned short)0Xbf45)
    
62.                                 {
    
63.                                         printf("Block id is %d:Index page no is %ld : B+ Tree Level is %d\n",i-1,n_idx_id,n_lev);
    
64.                                 }
    
65.                 }
    
66.          
    
67. }

 ./dataview 源码，此工具只对2列init数据类型 的组合索引读取分支节点有效，因为写死了