mycat初探

1:安装客户端

yum install mysql 

2:安装服务端

yum install mysql-server 

3:mycat要求不区分大小写

my.cnf(/etc/my.cnf)的[mysqld]区段下增加: lower_case_table_names=1

4:启动mysql

service mysqld start 

5:创建用户

mysqladmin -u root password 110110 

6:登陆mysql

mysql -u root; 

7:赋予远程登陆权限

GRANT ALL PRIVILEGES ON *.* TO 'root'@'%' IDENTIFIED BY '110110' WITH GRANT OPTION;
flush privileges; 

8:上传mycat到服务器(java要求1.7以上)

9:启动mycat

chmod 777 ./* 在bin目录下

./startup_nowrap.sh

10:修改配置文件

	<dataHost name="localhost1" maxCon="1000" minCon="10" balance="0"
		writeType="0" dbType="mysql" dbDriver="native" switchType="1"  slaveThreshold="100">
		<heartbeat>select user()</heartbeat>
		<!-- can have multi write hosts -->
		<writeHost host="hostM1" url="10.97.190.27:3306" user="qidian"
			password="qidian">
			<!-- can have multi read hosts -->

		</writeHost>
		<!-- 
		<writeHost host="hostS1" url="localhost:3316" user="root"
			password="123456" />
		<writeHost host="hostM2" url="localhost:3316" user="root" password="123456"/> -->
	</dataHost>

链接自己的mysql服务器

11:登陆mysql建立数据库。数据库名字db1,db2,db3

12:登陆mycat管理

 mysql -utest -ptest -h127.0.0.1 -P9066

show @@help

其中reload配置文件需要reload @@config_all,类似于nginx的reload

conf/server.xml 存储mycat的账户,和mysql账户没有关系

conf/schema.xml 逻辑表 

13:登陆mycat

mysql -utest -ptest -h127.0.0.1 -P8066 -DTESTDB

14:分片

1:全局表

<table name="company" primaryKey="ID" type="global" dataNode="dn1,dn2,dn3" />

每行记录在每个分片上同时存在

 

2:枚举

schema.xml

<table name="employee" primaryKey="ID" dataNode="dn1,dn2" rule="sharding-by-intfile" />

rule.xml

<tableRule name="sharding-by-intfile">
<rule>
<columns>sharding_id</columns>
<algorithm>hash-int</algorithm>
</rule>
</tableRule>

<function name="hash-int"
class="org.opencloudb.route.function.PartitionByFileMap">
<property name="mapFile">partition-hash-int.txt</property>
</function>

partition-hash-int.txt

10000=0
10010=1

DEFAULT_NODE=1

如果你输入

insert into employee(id,name,sharding_id) values(4, 'mydog',10011);则出错因为分片策略没有枚举10011的分片位置

上面columns 标识将要分片的表字段,algorithm 分片函数,
其中分片函数配置中,mapFile标识配置文件名称,type默认值为0,0表示Integer,非零表示String,
所有的节点配置都是从0开始,及0代表节点1
/**
*  defaultNode 默认节点:小于0表示不设置默认节点,大于等于0表示设置默认节点
* 
默认节点的作用:枚举分片时,如果碰到不识别的枚举值,就让它路由到默认节点
*                如果不配置默认节点(defaultNode值小于0表示不配置默认节点),碰到
*                不识别的枚举值就会报错,
*                like this:can't find datanode for sharding column:column_name val:ffffffff    
*/

 

3:父子表

		<table name="customer" primaryKey="ID" dataNode="dn1,dn2"
			rule="sharding-by-intfile">
			<childTable name="orders" primaryKey="ID" joinKey="customer_id"
				parentKey="id">
				<childTable name="order_items" joinKey="order_id"
					parentKey="id" />
			</childTable>
			<childTable name="customer_addr" primaryKey="ID" joinKey="customer_id"
				parentKey="id" />
		</table>

 

explain create table customer(id int not null primary key,name varchar(100),company_id int not null,sharding_id int not null);
explain insert into customer (id,name,company_id,sharding_id )values(1,'wang',1,10000);  
explain insert into customer (id,name,company_id,sharding_id )values(2,'xue',2,10010);  
explain insert into customer (id,name,company_id,sharding_id )values(3,'feng',3,10000); 
explain Select * from  customer; 


create table orders (id int not null primary key ,customer_id int not null,sataus int ,note varchar(100) );
        insert into orders(id,customer_id) values(1,1); //stored in db1 because customer table with id=1 stored in db1   
        insert into orders(id,customer_id) values(2,2); //stored in db2 because customer table with id=1 stored in db2    
        explain insert into orders(id,customer_id) values(2,2); 
        select customer.name ,orders.* from customer ,orders where customer.id=orders.customer_id; 

4:范围约定

<table name="travelrecord" dataNode="dn1,dn2,dn3" rule="auto-sharding-long" />

	<tableRule name="auto-sharding-long">
		<rule>
			<columns>id</columns>
			<algorithm>rang-long</algorithm>
		</rule>
	</tableRule>

	<function name="rang-long"
		class="org.opencloudb.route.function.AutoPartitionByLong">
		<property name="mapFile">autopartition-long.txt</property>
	</function>

# range start-end ,data node index
# K=1000,M=10000.
0-500M=0
500M-1000M=1
1000M-1500M=2

5:固定分片hash算法

<tableRule name="rule1">
    <rule>
      <columns>user_id</columns>
      <algorithm>func1</algorithm>
    </rule>
</tableRule>

  <function name="func1" class="org.opencloudb.route.function.PartitionByLong">
    <property name="partitionCount">2,1</property>
    <property name="partitionLength">256,512</property>
  </function>

配置说明:
上面columns 标识将要分片的表字段,algorithm 分片函数,
partitionCount 分片个数列表,partitionLength 分片范围列表
分区长度:默认为最大2^n=1024 ,即最大支持1024分区
约束 :
count,length两个数组的长度必须是一致的。
1024 = sum((count[i]*length[i])). count和length两个向量的点积恒等于1024
用法例子:
        本例的分区策略:希望将数据水平分成3份,前两份各占25%,第三份占50%。(故本例非均匀分区)
        // |<---------------------1024------------------------>|
        // |<----256--->|<----256--->|<----------512---------->|
        // | partition0 | partition1 | partition2 |
        // | 共2份,故count[0]=2 | 共1份,故count[1]=1 |
        int[] count = new int[] { 2, 1 };
        int[] length = new int[] { 256, 512 };
        PartitionUtil pu = new PartitionUtil(count, length);

        // 下面代码演示分别以offerId字段或memberId字段根据上述分区策略拆分的分配结果
        int DEFAULT_STR_HEAD_LEN = 8; // cobar默认会配置为此值
        long offerId = 12345;
        String memberId = "qiushuo";

        // 若根据offerId分配,partNo1将等于0,即按照上述分区策略,offerId为12345时将会被分配到partition0中
        int partNo1 = pu.partition(offerId);

        // 若根据memberId分配,partNo2将等于2,即按照上述分区策略,memberId为qiushuo时将会被分到partition2中
        int partNo2 = pu.partition(memberId, 0, DEFAULT_STR_HEAD_LEN);

如果需要平均分配设置:平均分为4分片,partitionCount*partitionLength=1024
<function name="func1" class="org.opencloudb.route.function.PartitionByLong">
    <property name="partitionCount">4</property>
    <property name="partitionLength">256</property>
  </function>

6:求模法

<tableRule name="mod-long">
    <rule>
      <columns>user_id</columns>
      <algorithm>mod-long</algorithm>
    </rule>
  </tableRule>
  <function name="mod-long" class="org.opencloudb.route.function.PartitionByMod">
   <!-- how many data nodes  -->
    <property name="count">3</property>
  </function> 
配置说明:
上面columns 标识将要分片的表字段,algorithm 分片函数,
此种配置非常明确即根据id进行十进制求模预算,相比方式1,此种在批量插入时需要切换数据源,id不连续

7:日期列分区

<tableRule name="sharding-by-date">
      <rule>
        <columns>create_time</columns>
        <algorithm>sharding-by-date</algorithm>
      </rule>
   </tableRule>  
<function name="sharding-by-date" class="org.opencloudb.route.function.PartitionByDate">
    <property name="dateFormat">yyyy-MM-dd</property>
    <property name="sBeginDate">2014-01-01</property>
    <property name="sPartionDay">10</property>
  </function>
配置说明:
上面columns 标识将要分片的表字段,algorithm 分片函数,
配置中配置了开始日期,分区天数,即默认从开始日期算起,分隔10天一个分区


Assert.assertEquals(true, 0 == partition.calculate("2014-01-01"));
Assert.assertEquals(true, 0 == partition.calculate("2014-01-10"));
Assert.assertEquals(true, 1 == partition.calculate("2014-01-11"));
Assert.assertEquals(true, 12 == partition.calculate("2014-05-01"));

8:通配取模

<tableRule name="sharding-by-pattern">
      <rule>
        <columns>user_id</columns>
        <algorithm>sharding-by-pattern</algorithm>
      </rule>
   </tableRule>
<function name="sharding-by-pattern" class="org.opencloudb.route.function.PartitionByPattern">
    <property name="patternValue">256</property>
    <property name="defaultNode">2</property>
    <property name="mapFile">partition-pattern.txt</property>

  </function>
partition-pattern.txt 
# id partition range start-end ,data node index
###### first host configuration
1-32=0
33-64=1
65-96=2
97-128=3
######## second host configuration
129-160=4
161-192=5
193-224=6
225-256=7
0-0=7

配置说明:
上面columns 标识将要分片的表字段,algorithm 分片函数,patternValue 即求模基数,defaoultNode 默认节点,如果配置了默认,则不会按照求模运算
mapFile 配置文件路径
配置文件中,1-32 即代表id%256后分布的范围,如果在1-32则在分区1,其他类推,如果id非数据,则会分配在defaoultNode 默认节点

String idVal = "0";
Assert.assertEquals(true, 7 == autoPartition.calculate(idVal));
idVal = "45a";
Assert.assertEquals(true, 2 == autoPartition.calculate(idVal));

9:ASCII码求模通配

<tableRule name="sharding-by-prefixpattern">
      <rule>
        <columns>user_id</columns>
        <algorithm>sharding-by-prefixpattern</algorithm>
      </rule>
   </tableRule>
<function name="sharding-by-pattern" class="org.opencloudb.route.function.PartitionByPattern">
    <property name="patternValue">256</property>
    <property name="prefixLength">5</property>
    <property name="mapFile">partition-pattern.txt</property>

  </function>

partition-pattern.txt

# range start-end ,data node index
# ASCII
# 48-57=0-9
# 64、65-90=@、A-Z
# 97-122=a-z
###### first host configuration
1-4=0
5-8=1
9-12=2
13-16=3
###### second host configuration
17-20=4
21-24=5
25-28=6
29-32=7
0-0=7
配置说明:
上面columns 标识将要分片的表字段,algorithm 分片函数,patternValue 即求模基数,prefixLength ASCII 截取的位数
mapFile 配置文件路径
配置文件中,1-32 即代表id%256后分布的范围,如果在1-32则在分区1,其他类推 

此种方式类似方式6只不过采取的是将列种获取前prefixLength位列所有ASCII码的和进行求模sum%patternValue ,获取的值,在通配范围内的
即 分片数,
/**
* ASCII编码:
* 48-57=0-9阿拉伯数字
* 64、65-90=@、A-Z
* 97-122=a-z
*
*/
如 

String idVal="gf89f9a";
Assert.assertEquals(true, 0==autoPartition.calculate(idVal));

idVal="8df99a";
Assert.assertEquals(true, 4==autoPartition.calculate(idVal));

idVal="8dhdf99a";
Assert.assertEquals(true, 3==autoPartition.calculate(idVal));

10:编程指定

<tableRule name="sharding-by-substring">
      <rule>
        <columns>user_id</columns>
        <algorithm>sharding-by-substring</algorithm>
      </rule>
   </tableRule>
<function name="sharding-by-substring" class="org.opencloudb.route.function.PartitionDirectBySubString">
    <property name="startIndex">0</property> <!-- zero-based -->
    <property name="size">2</property>
    <property name="partitionCount">8</property>
    <property name="defaultPartition">0</property>
  </function>
配置说明:
上面columns 标识将要分片的表字段,algorithm 分片函数 
此方法为直接根据字符子串(必须是数字)计算分区号(由应用传递参数,显式指定分区号)。
例如id=05-100000002
在此配置中代表根据id中从startIndex=0,开始,截取siz=2位数字即05,05就是获取的分区,如果没传默认分配到defaultPartition

11:字符串拆分hash解析

<tableRule name="sharding-by-stringhash">
      <rule>
        <columns>user_id</columns>
        <algorithm>sharding-by-stringhash</algorithm>
      </rule>
   </tableRule>
<function name="sharding-by-substring" class="org.opencloudb.route.function.PartitionDirectBySubString">
    <property name=length>512</property> <!-- zero-based -->
    <property name="count">2</property>
    <property name="hashSlice">0:2</property>
  </function>
配置说明:
上面columns 标识将要分片的表字段,algorithm 分片函数 
函数中length代表字符串hash求模基数,count分区数,hashSlice hash预算位

即根据子字符串 hash运算

	

hashSlice : 0 means str.length(), -1 means str.length()-1

/**
     * "2" -> (0,2)<br/>
     * "1:2" -> (1,2)<br/>
     * "1:" -> (1,0)<br/>
     * "-1:" -> (-1,0)<br/>
     * ":-1" -> (0,-1)<br/>
     * ":" -> (0,0)<br/>
     */
例子:
String idVal=null;
 rule.setPartitionLength("512");
 rule.setPartitionCount("2");
 rule.init();
 rule.setHashSlice("0:2");
//		idVal = "0";
//		Assert.assertEquals(true, 0 == rule.calculate(idVal));
//		idVal = "45a";
//		Assert.assertEquals(true, 1 == rule.calculate(idVal));

 
 
 //last 4
 rule = new PartitionByString();
 rule.setPartitionLength("512");
 rule.setPartitionCount("2");
 rule.init();
 //last 4 characters
 rule.setHashSlice("-4:0");
 idVal = "aaaabbb0000";
 Assert.assertEquals(true, 0 == rule.calculate(idVal));
 idVal = "aaaabbb2359";
 Assert.assertEquals(true, 0 == rule.calculate(idVal));

12:一致性hash

<tableRule name="sharding-by-murmur">
      <rule>
        <columns>user_id</columns>
        <algorithm>murmur</algorithm>
      </rule>
   </tableRule>
<function name="murmur" class="org.opencloudb.route.function.PartitionByMurmurHash">
      <property name="seed">0</property><!-- 默认是0-->
      <property name="count">2</property><!-- 要分片的数据库节点数量,必须指定,否则没法分片-->
      <property name="virtualBucketTimes">160</property><!-- 一个实际的数据库节点被映射为这么多虚拟节点,默认是160倍,也就是虚拟节点数是物理节点数的160倍-->
      <!--
      <property name="weightMapFile">weightMapFile</property>
                     节点的权重,没有指定权重的节点默认是1。以properties文件的格式填写,以从0开始到count-1的整数值也就是节点索引为key,以节点权重值为值。所有权重值必须是正整数,否则以1代替 -->
      <!--
      <property name="bucketMapPath">/etc/mycat/bucketMapPath</property>
                      用于测试时观察各物理节点与虚拟节点的分布情况,如果指定了这个属性,会把虚拟节点的murmur hash值与物理节点的映射按行输出到这个文件,没有默认值,如果不指定,就不会输出任何东西 -->
  </function>
一致性hash预算有效解决了分布式数据的扩容问题,前1-9中id规则都多少存在数据扩容难题,而10规则解决了数据扩容难点

 

posted @ 2016-01-06 19:51  李占卫  阅读(1665)  评论(0编辑  收藏  举报