有两张表t1和t2，t1和t2两张表大部分字段都相同但是t2比t1多了一个status状态字段，现在需要将两个表合起来查询，且要求设置t1表所有信息的status状态都为00。如果两张表有id重复的数据则展示t2数据库中的信息，该如何书写sql

你可以使用SQL的UNION ALL操作符来合并两张表，并通过CASE语句设置t1表的状态。如果你想要优先显示t2表的信息，当id在两表中存在重复时，则可以这样编写：

SELECT id,
       CASE WHEN t1.id IS NULL THEN t2.*
            ELSE COALESCE(t1.*, (SELECT * FROM t2 WHERE t2.id = t1.id AND t1.status = 0)) -- 如果t1的status为00，则保留；否则取t2的信息
       END AS combined_data
FROM t1
LEFT JOIN t2 ON t1.id = t2.id
UNION ALL
SELECT id,
       t2.*
FROM t2
WHERE NOT EXISTS (SELECT 1 FROM t1 WHERE t1.id = t2.id)

这个查询首先做了一个左连接(t1 LEFT JOIN t2)，然后在结果集中检查t1是否存在。如果id在t1中并且status为00，就保留t1的所有列；否则，从t2中选择对应的数据。同时，还包含了一个UNION ALL后的部分，用于处理只有在t2中存在的id。

相关问题

用MySQL8.0写代码，表stock_k_data AS T1 中字段有 ID,open, high, low, close,price,status用游标形式循环逐行计算， T1.字段名表示当前行，T2.字段名表示上一行， t1.ID= 2，循环到最后一行 ， IF t2.status = "买入" AND t1.low<T2.price*0.95,THEN SET t1.status="卖出", t1.price=T2.price*0.95 IF t2.status = "买入" ,THEN t1.status="持有",t1.price=GREATEST(T2.price,T1.high) ELSEIF t2.status= "持有" AND t1.low<GREATEST(T2.price,T1.high)*0.95 ,THEN t1.status="卖出",t1.price=GREATEST(T2.price,T1.high)*0.95 ELSEIF t2.status= "持有" ,THEN t1.status="持有", t1.price=GREATEST(T2.price,T1.high) ELSEIF t2.status = "卖出" AND t1.high>T2.price*1.05,THEN t1.status="买入", t1.price=T2.price*1.05 ELSEIF t2.status = "卖出" ,THEN t1.status="空仓", t1.price=LEAST(T2.price,low) ELSEIF t2.status = "空仓" AND t1.high>LEAST(T2.price,low)*1.05,THEN t1.status="买入", t1.price=LEAST(T2.price,low)*1.05 ELSEIF t2.status = "卖出" ,THEN t1.status="空仓", t1.price=LEAST(T2.price,low) ENDIF

好的，下面是用MySQL8.0实现上述功能的代码：

DECLARE done INT DEFAULT FALSE;
DECLARE cur_id INT;
DECLARE cur_open FLOAT;
DECLARE cur_high FLOAT;
DECLARE cur_low FLOAT;
DECLARE cur_close FLOAT;
DECLARE cur_price FLOAT;
DECLARE cur_status VARCHAR(10);
DECLARE prev_id INT;
DECLARE prev_open FLOAT;
DECLARE prev_high FLOAT;
DECLARE prev_low FLOAT;
DECLARE prev_close FLOAT;
DECLARE prev_price FLOAT;
DECLARE prev_status VARCHAR(10);

DECLARE cur CURSOR FOR SELECT ID, open, high, low, close, price, status FROM stock_k_data WHERE ID >= 2 ORDER BY ID ASC;
DECLARE CONTINUE HANDLER FOR NOT FOUND SET done = TRUE;

OPEN cur;

FETCH cur INTO cur_id, cur_open, cur_high, cur_low, cur_close, cur_price, cur_status;

WHILE NOT done DO
    IF prev_status = "买入" AND cur_low < prev_price * 0.95 THEN
        SET cur_status = "卖出", cur_price = prev_price * 0.95;
    ELSEIF prev_status = "买入" THEN
        SET cur_status = "持有", cur_price = GREATEST(prev_price, cur_high);
    ELSEIF prev_status = "持有" AND cur_low < GREATEST(prev_price, cur_high) * 0.95 THEN
        SET cur_status = "卖出", cur_price = GREATEST(prev_price, cur_high) * 0.95;
    ELSEIF prev_status = "持有" THEN
        SET cur_status = "持有", cur_price = GREATEST(prev_price, cur_high);
    ELSEIF prev_status = "卖出" AND cur_high > prev_price * 1.05 THEN
        SET cur_status = "买入", cur_price = prev_price * 1.05;
    ELSEIF prev_status = "卖出" THEN
        SET cur_status = "空仓", cur_price = LEAST(prev_price, cur_low);
    ELSEIF prev_status = "空仓" AND cur_high > LEAST(prev_price, cur_low) * 1.05 THEN
        SET cur_status = "买入", cur_price = LEAST(prev_price, cur_low) * 1.05;
    ELSEIF prev_status = "空仓" THEN
        SET cur_status = "空仓", cur_price = LEAST(prev_price, cur_low);
    END IF;

    UPDATE stock_k_data SET status = cur_status, price = cur_price WHERE ID = cur_id;

    SET prev_id = cur_id, prev_open = cur_open, prev_high = cur_high, prev_low = cur_low, prev_close = cur_close, prev_price = cur_price, prev_status = cur_status;

    FETCH cur INTO cur_id, cur_open, cur_high, cur_low, cur_close, cur_price, cur_status;
END WHILE;

CLOSE cur;

该代码通过游标形式循环逐行计算表stock_k_data的数据，并按照给定的条件更新状态和价格。其中，使用了IF-ELSEIF-ELSE语句对不同的状态进行判断和处理。在每次更新完当前行的数据后，将当前行的数据赋值给prev_XXX变量，作为下一次循环时的上一行数据。最后，关闭游标。

SELECT t1.organization_id, t1.organization_name, t1.road_name, SUM( t2.sj_amount ) xfSjAmount, SUM( t2.jl_amount ) xfJlAmount, SUM( t3.xfGeneralSectionAmount ) xfGeneralSectionAmount, SUM( t3.xfGeneralSectionLength ) xfGeneralSectionLength FROM `main_project` t1 LEFT JOIN mat_approval_project t2 ON t2.main_id = t1.id LEFT JOIN ( SELECT project_id, data_type, SUM( money ) xfGeneralSectionAmount, SUM( mileage ) xfGeneralSectionLength FROM mat_project_pavement_approval GROUP BY project_id ) t3 ON t3.project_id = t2.id WHERE t1.project_general_name = '修复性养护' AND t1.project_category_name = '路面工程' AND t1.process_status = '100' AND t3.data_type = '1' AND t1.road_name = '常祁高速' AND t1.project_year = '2023' 这条sql中t1.road_name多条数据只会展示一个，导致根据t1.road_name查询不到数据

这个问题的原因可能是使用了`LEFT JOIN`，导致t1表中的多个记录与t2和t3表中的记录进行了匹配。如果t1表中有多个记录的`t1.road_name`值相同，那么在`SELECT`语句中使用`SUM`函数对t2和t3表中的记录进行了合并，使得最终查询结果中只有一个记录。这可能导致你根据`t1.road_name`无法查到完整的数据。

为了解决这个问题，你可以考虑在`SELECT`语句中增加`t1.road_name`字段，并在`GROUP BY`子句中包含该字段，以确保t1表中的每个记录都能够被正确匹配和展示。例如：

SELECT
  t1.organization_id,
  t1.organization_name,
  t1.road_name,
  SUM(t2.sj_amount) AS xfSjAmount,
  SUM(t2.jl_amount) AS xfJlAmount,
  SUM(t3.xfGeneralSectionAmount) AS xfGeneralSectionAmount,
  SUM(t3.xfGeneralSectionLength) AS xfGeneralSectionLength
FROM
  main_project t1
  LEFT JOIN mat_approval_project t2 ON t2.main_id = t1.id
  LEFT JOIN (
    SELECT
      project_id,
      data_type,
      SUM(money) AS xfGeneralSectionAmount,
      SUM(mileage) AS xfGeneralSectionLength
    FROM
      mat_project_pavement_approval
    GROUP BY
      project_id
  ) t3 ON t3.project_id = t2.id
WHERE
  t1.project_general_name = '修复性养护'
  AND t1.project_category_name = '路面工程'
  AND t1.process_status = '100'
  AND t3.data_type = '1'
  AND t1.road_name = '常祁高速'
  AND t1.project_year = '2023'
GROUP BY
  t1.organization_id,
  t1.organization_name,
  t1.road_name;

这样，就可以确保每个`t1.road_name`值都能够正确地展示在查询结果中。