我要两张表计算

Question

我想使用 2 个表中的数据进行相加和计算。

我想要的结果如下。

但是，如果我发送如下查询：

SELECT COUNT(IF(treatment_fees_check_division = 'test', treatment_fees_check_division, NULL)) AS COUNT, SUM(CASE WHEN treatment_fees_check_division = '测试' THEN treatment_fees_difference END) + SUM(advenced_amount) AS if_treatment_fees_check_division, SUM(advenced_amount) 作为 sum_init_amount FROM ( SELECT treatment_fees_check_division，treatment_fees_difference， advenced_amount FROM hospital_payment_data，advenced_pa​​yment WHERE hospital_payment_data.id = advenced_pa​​yment.chart_num ) 作为一个

会出现这些结果。

创建医院支付数据和数据表

  CREATE TABLE `hospital_payment_data` (
        `id` INT(11) NOT NULL AUTO_INCREMENT,
        `date` TIMESTAMP NOT NULL DEFAULT current_timestamp(),
        `chart_num` INT(11) NOT NULL ,
        `chart_name` VARCHAR(50) NOT NULL,
        `visit` DATE NOT NULL,
        `non_payment_sales` VARCHAR(50) NOT NULL,
        `total_medical_bills` VARCHAR(50) NOT NULL,
        `total_amount` VARCHAR(50) NOT NULL,
        `amount_asked` VARCHAR(50) NOT NULL,
        `medical_bills_payment` VARCHAR(50) NOT NULL,
        `personal_liability_amount` VARCHAR(50) NOT NULL,
        `non_payment` VARCHAR(50) NOT NULL,
        `insurance_division` VARCHAR(50) NOT NULL,
        `division` VARCHAR(50) NOT NULL,
        `cash_amount_received` VARCHAR(50) NOT NULL,
        `cash_receipt` VARCHAR(50) NOT NULL,
        `cash_receipt_non_payment` VARCHAR(50) NOT NULL,
        `cash_receipt_payment` VARCHAR(50) NOT NULL,
        `card_amount_received` VARCHAR(50) NOT NULL,
        `card_non_payment` VARCHAR(50) NOT NULL,
        `card_payment` VARCHAR(50) NOT NULL,
        `medical_bills_check` VARCHAR(50) NOT NULL,
        `medical_bills_check_modify` VARCHAR(50) NOT NULL DEFAULT 'N',
        `treatment_fees_difference` VARCHAR(50) NOT NULL,
        `init_amount` VARCHAR(50) NOT NULL DEFAULT '0',
        `treatment_fees_check_division` VARCHAR(50) NOT NULL DEFAULT '-',
        `treatment_fees_check` VARCHAR(50) NOT NULL,
        `treatment_fees_check_modify` VARCHAR(50) NOT NULL DEFAULT 'N',
        `treatment_fees_check_memo` VARCHAR(50) NOT NULL,
        PRIMARY KEY (`id`)
    )
    COLLATE='euckr_korean_ci'
    ENGINE=InnoDB
    AUTO_INCREMENT=18
    ;

创建advenced_pa​​yment和数据表

CREATE TABLE `advenced_payment` (`id` INT(11) NOT NULL AUTO_INCREMENT,
`date` TIMESTAMP NULL DEFAULT NULL,
`chart_num` VARCHAR(50) NULL DEFAULT NULL,
`chart_name` VARCHAR(50) NULL DEFAULT NULL,
`advenced_amount` VARCHAR(50) NULL DEFAULT NULL,
PRIMARY KEY (`id`),
UNIQUE INDEX `chart_num` (`chart_num`)) COLLATE='euckr_korean_ci' ENGINE=InnoDB AUTO_INCREMENT=2 ;

如何修复我的查询？？

Answer 1

您的查询有很多问题。这不会解决您的问题，但我建议将其写为：

SELECT SUM(hpd.treatment_fees_check_division = 'test') AS COUNT,
       (SUM(CASE WHEN hpd.treatment_fees_check_division = 'test' THEN hpd.treatment_fees_difference END) +
        SUM(advenced_amount)
       ) AS if_treatment_fees_check_division,
       SUM(advenced_amount) AS sum_init_amount
FROM hospital_payment_data hpd JOIN
     advanced_payment ap
     ON hpd.id = ap.chart_num;

这简化并修复了一些问题：

• JOIN 使用正确、明确、标准、可读的语法。 切勿在FROM 子句中使用逗号。
• 子查询完全没有必要，可能会对性能产生不利影响。
• 所有表都有表别名。
• 所有列引用都是合格的。
• 使用sum(<boolean>) 可以大大简化条件计数。

我注意到treatment_fees_difference 是一个字符串，但您使用的是SUM()。这真的非常糟糕。

无论如何，您的问题是其中一张表有多行。很难知道是哪一个，但我猜是advanced_payment。解决方法是在JOIN之前预聚合：

SELECT SUM(hpd.treatment_fees_check_division = 'test') AS COUNT,
       (SUM(CASE WHEN hpd.treatment_fees_check_division = 'test' THEN hpd.treatment_fees_difference END) +
        SUM(ap.sum_init_amount)
       ) AS if_treatment_fees_check_division,
       SUM(ap.sum_init_amount) AS sum_init_amount
FROM hospital_payment_data hpd JOIN
     (SELECT ap.chart_num, COUNT(*) as cnt, SUM(advenced_amount) as sum_init_amount
      FROM advanced_payment ap
      GROUP BY ap.chart_num
     ) ap
     ON hpd.id = ap.chart_num;