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18. sharding-jdbc源码之复杂路由实现

18. sharding-jdbc源码之复杂路由实现

作者: 阿飞的博客 | 来源:发表于2018-05-08 11:32 被阅读248次

    阿飞Javaer,转载请注明原创出处,谢谢!

    路由条件

    ParsingSQLRouter.java中决定是简单路由还是复杂路由的条件如下;

    private RoutingResult route(final List<Object> parameters, final SQLStatement sqlStatement) {
        Collection<String> tableNames = sqlStatement.getTables().getTableNames();
        RoutingEngine routingEngine;
        if (1 == tableNames.size()
                || shardingRule.isAllBindingTables(tableNames)
                || shardingRule.isAllInDefaultDataSource(tableNames)) {
            routingEngine = new SimpleRoutingEngine(shardingRule, parameters, tableNames.iterator().next(), sqlStatement);
        } else {
            // TODO config for cartesian set
            routingEngine = new ComplexRoutingEngine(shardingRule, parameters, tableNames, sqlStatement);
        }
        return routingEngine.route();
    }
    
    • 是否只有一张表--tableNames.size()

    说明:这个"一张表"并不是指SQL中只有一张表,而是有分库分表规则的表数量,例如下面这段构造ShardingRule的源码,tableRules()有两个表,所以tableNames.size()的值为2;如果(Arrays.asList(orderTableRule))即只有1个表,那么tableNames.size()的值为1;

    ShardingRule.builder()
    .dataSourceRule(dataSourceRule)
    .tableRules(Arrays.asList(orderTableRule, userTableRule))
    .databaseShardingStrategy(*** ***).tableShardingStrategy(*** ***) .build();
    
    • 是否都是绑定表--shardingRule.isAllBindingTables(tableNames)

    说明:isAllBindingTables(tableNames)判断tableNames是否都属于绑定表,例如下面这段构造ShardingRule的源码,.bindingTableRules()里的参数就是绑定表集合,这里是t_order和t_order_item都是绑定表,那么:SELECT od.user_id, od.order_id, oi.item_id, od.status FROM t_order od join t_order_item oi on od.order_id=oi.order_id这个SQL只有t_order和t_order_item两个表且都是绑定表,那么shardingRule.isAllBindingTables(tableNames)为true;

    ShardingRule.builder()
    .dataSourceRule(dataSourceRule)
    .tableRules(Arrays.asList(orderTableRule, orderItemTableRule, userTableRule))
    .bindingTableRules(Collections.singletonList(new BindingTableRule(Arrays.asList(orderTableRule, orderItemTableRule))))
    . *** ***;       
    
    • 是否都在默认数据源中--shardingRule.isAllInDefaultDataSource(tableNames)

    说明:sharding-jdbc判断逻辑源码如下,即只要在表规则集合中能够匹配到逻辑表,就认为不属于默认数据源中(默认数据源不分库分表),例如ShardingRule.builder().dataSourceRule(dataSourceRule).tableRules(Arrays.asList(orderTableRule, orderItemTableRule, userTableRule)),根据tableRules参数可知,主要SQL中有t_usert_ordert_order_item三个表的任意一个表,那么shardingRule.isAllInDefaultDataSource(tableNames)都为false;

    public boolean isAllInDefaultDataSource(final Collection<String> logicTables) {
        for (String each : logicTables) {
            if (tryFindTableRule(each).isPresent()) {
                return false;
            }
        }
        return !logicTables.isEmpty();
    }
    
    public Optional<TableRule> tryFindTableRule(final String logicTableName) {
        for (TableRule each : tableRules) {
            if (each.getLogicTable().equalsIgnoreCase(logicTableName)) {
                return Optional.of(each);
            }
        }
        return Optional.absent();
    }   
    

    构造复杂路由

    综上分析,如果三个条件都不满足就走复杂路由ComplexRoutingEngine,构造这种场景:
    t_order和t_order_item分库分表且绑定表关系,加入一个新的分库分表t_user;ShardingRule如下:

    ShardingRule shardingRule = ShardingRule.builder()
            .dataSourceRule(dataSourceRule)
            .tableRules(Arrays.asList(orderTableRule, orderItemTableRule, userTableRule))
            .bindingTableRules(Collections.singletonList(new BindingTableRule(Arrays.asList(orderTableRule, orderItemTableRule))))
            .databaseShardingStrategy(new DatabaseShardingStrategy("user_id", new ModuloDatabaseShardingAlgorithm()))
            .tableShardingStrategy(new TableShardingStrategy("order_id", new ModuloTableShardingAlgorithm()))
            .build();
    

    执行的SQL为:

    SELECT od.user_id, od.order_id, oi.item_id, od.status 
    FROM `t_user` tu 
    join t_order od on tu.user_id=od.user_id 
    join t_order_item oi on od.order_id=oi.order_id 
    where tu.`status`='VALID' and tu.user_id=?
    

    构造的这个场景:tableNames.size()=3(三张表t_user,t_order,t_order_item都有分库分表规则,所以值为3),shardingRule.isAllBindingTables(tableNames)为false(t_user表不属于绑定表范围);shardingRule.isAllInDefaultDataSource(tableNames)为false(三张表都不属于默认数据源中的表);所以这个SQL会走复杂路由的逻辑;

    ComplexRoutingEngine

    复杂路由引擎的核心逻辑就是拆分成多个简单路由,然后求笛卡尔积,复杂路由核心源码如下:

    @RequiredArgsConstructor
    @Slf4j
    public final class ComplexRoutingEngine implements RoutingEngine {
        
        // 分库分表规则
        private final ShardingRule shardingRule;
        
        // SQL请求参数,猪油一个user_id的值为10
        private final List<Object> parameters;
        
        // 逻辑表集合:t_order,t_order_item,t_user,三个逻辑表
        private final Collection<String> logicTables;
        
        // SQL解析结果
        private final SQLStatement sqlStatement;
        
        // 复杂路由的核心逻辑
        @Override
        public RoutingResult route() {
            Collection<RoutingResult> result = new ArrayList<>(logicTables.size());
            Collection<String> bindingTableNames = new TreeSet<>(String.CASE_INSENSITIVE_ORDER);
            // 遍历逻辑表集合
            for (String each : logicTables) {
                Optional<TableRule> tableRule = shardingRule.tryFindTableRule(each);
                // 如果遍历的表配置了分库分表规则
                if (tableRule.isPresent()) {
                    // 如果绑定关系表已经处理过,那么不需要再处理,例如t_order处理过,由于t_order_item与其是绑定关系,那么不需要再处理
                    if (!bindingTableNames.contains(each)) {
                        // 根据当前遍历的逻辑表构造一个简单路由规则
                        result.add(new SimpleRoutingEngine(shardingRule, parameters, tableRule.get().getLogicTable(), sqlStatement).route());
                    }
    
                    // 根据当前逻辑表,查找其对应的所有绑定表,例如根据t_order就能够查询出t_order和t_order_item;假如配置了.bindingTableRules(***t_point, t_point_detail***),那么,根据t_point能查询出t_point和t_point_detail,其目的是N个绑定表只需要路由一个绑定表即可,因为绑定表之间的路由关系完全一致。
                    Optional<BindingTableRule> bindingTableRule = shardingRule.findBindingTableRule(each);
                    if (bindingTableRule.isPresent()) {
                        bindingTableNames.addAll(Lists.transform(bindingTableRule.get().getTableRules(), new Function<TableRule, String>() {
                            
                            @Override
                            public String apply(final TableRule input) {
                                return input.getLogicTable();
                            }
                        }));
                    }
                }
            }
            log.trace("mixed tables sharding result: {}", result);
            // 如果是复杂路由,但是路由结果为空,那么抛出异常
            if (result.isEmpty()) {
                throw new ShardingJdbcException("Cannot find table rule and default data source with logic tables: '%s'", logicTables);
            }
            // 如果结果的size为1,那么直接返回即可
            if (1 == result.size()) {
                return result.iterator().next();
            }
            // 对刚刚的路由结果集合计算笛卡尔积,就是最终复杂的路由结果
            return new CartesianRoutingEngine(result).route();
        }
    }
    

    由上面源码分析可知,会分别对t_user和t_order构造简单路由(t_order_item和t_order是绑定关系,二者取其一即可);

    • t_user只分库不分表(因为构造TableRule时逻辑表和实际表一致),且请求参数为user_id=10,所以t_user这个逻辑表的简单路由结果为:数据源ds_jdbc_0,实际表t_user;
    • t_order分库分表,且请求参数user_id被解析为t_user的条件(笛卡尔积路由引擎会处理),所以t_order的简单路由结果为:数据源ds_jdbc_0和ds_jdbc_1,实际表t_order_0和t_order_1;

    debug的result如下:


    result detail

    CartesianRoutingEngine

    如上分析,求得简单路由结果集后,求笛卡尔积就是复杂路由的最终路由结果,笛卡尔积路由引擎CartesianRoutingEngine的核心源码如下:

    @RequiredArgsConstructor
    @Slf4j
    public final class CartesianRoutingEngine implements RoutingEngine {
        
        private final Collection<RoutingResult> routingResults;
        
        @Override
        public CartesianRoutingResult route() {
            CartesianRoutingResult result = new CartesianRoutingResult();
            // getDataSourceLogicTablesMap()的分析参考下面的分析
            for (Entry<String, Set<String>> entry : getDataSourceLogicTablesMap().entrySet()) {
                // 根据数据源&逻辑表,得到实际表集合,即[["t_user"],["t_order_0","t_order_1"]]
                List<Set<String>> actualTableGroups = getActualTableGroups(entry.getKey(), entry.getValue());
                // 把逻辑表名封装,TableUnit的属性有:数据源名称,逻辑表名,实际表名(这三个属性才能确定最终访问的表)
                List<Set<TableUnit>> tableUnitGroups = toTableUnitGroups(entry.getKey(), actualTableGroups);
                // 计算所有实际表的笛卡尔积
                result.merge(entry.getKey(), getCartesianTableReferences(Sets.cartesianProduct(tableUnitGroups)));
            }
            log.trace("cartesian tables sharding result: {}", result);
            return result;
        }
        
        // 得到数据源-逻辑表集合组成的Map
        private Map<String, Set<String>> getDataSourceLogicTablesMap() {
            // 这里很关键,是得到数据源的交集(上面分析时t_user逻辑表路由到数据源ds_jdbc_0,而t_order表路由到数据源ds_jdbc_0和ds_jdbc_1,数据源交集就是ds_jdbc_0)
            Collection<String> intersectionDataSources = getIntersectionDataSources();
            Map<String, Set<String>> result = new HashMap<>(routingResults.size());
            for (RoutingResult each : routingResults) {
                for (Entry<String, Set<String>> entry : each.getTableUnits().getDataSourceLogicTablesMap(intersectionDataSources).entrySet()) {
                    if (result.containsKey(entry.getKey())) {
                        result.get(entry.getKey()).addAll(entry.getValue());
                    } else {
                        result.put(entry.getKey(), entry.getValue());
                    }
                }
            }
            // 得到的最终结果为数据源-逻辑表集合组成的Map,这里就是{"ds_jdbc_0":["t_order", "t_user"]}
            return result;
        }
        ... ...
    }
    

    计算得到的笛卡尔积结果如下:


    image.png

    sql.show结果如下,可以看到重写后的2条实际SQL:t_user&t_order_0,以及t_user&t_order_1(t_order_item与t_order是绑定表,保持一致即可):

    [INFO ] 2018-05-08 11:13:02,044 --main-- [Sharding-JDBC-SQL] Logic SQL: SELECT od.user_id, od.order_id, oi.item_id, od.status FROM `t_user` tu join t_order od on tu.user_id=od.user_id join t_order_item oi on od.order_id=oi.order_id where tu.`status`='VALID' and tu.user_id=? 
    ... ...
    [INFO ] 2018-05-08 11:13:02,059 --main-- [Sharding-JDBC-SQL] Actual SQL: ds_jdbc_0 ::: SELECT od.user_id, od.order_id, oi.item_id, od.status FROM t_user tu join t_order_0 od on tu.user_id=od.user_id join t_order_item_0 oi on od.order_id=oi.order_id where tu.`status`='VALID' and tu.user_id=? ::: [10] 
    [INFO ] 2018-05-08 11:13:02,059 --main-- [Sharding-JDBC-SQL] Actual SQL: ds_jdbc_0 ::: SELECT od.user_id, od.order_id, oi.item_id, od.status FROM t_user tu join t_order_1 od on tu.user_id=od.user_id join t_order_item_1 oi on od.order_id=oi.order_id where tu.`status`='VALID' and tu.user_id=? ::: [10] 
    

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