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在select stamenet-MySQL中填充字段_Mysql_Sql_Phpmyadmin - Fatal编程技术网

在select stamenet-MySQL中填充字段

在select stamenet-MySQL中填充字段,mysql,sql,phpmyadmin,Mysql,Sql,Phpmyadmin,我试图进行计算并在select语句中填充一个字段。看起来是这样的: CREATE VIEW SALES_REPORT AS( SELECT INVOICENO, INVOICEDATE, CLIENTID, CONTACT, INVOICEJOBNO, ADDCHARGES, CHARGESINFO, EMPLOYEEID, USUALPAY, VAT, SUBTOTAL (SELECT(USUALPAY * COUNT(*) AS SUBTOTAL FROM SALES_REPORT

我试图进行计算并在select语句中填充一个字段。看起来是这样的:

    CREATE VIEW SALES_REPORT AS(
SELECT 
INVOICENO,
INVOICEDATE,
CLIENTID,
CONTACT,
INVOICEJOBNO,
ADDCHARGES,
CHARGESINFO,
EMPLOYEEID,
USUALPAY,
VAT,
SUBTOTAL (SELECT(USUALPAY * COUNT(*) AS SUBTOTAL FROM SALES_REPORT)),
TOTAL = (SUBTOTAL * VAT)
FROM SALES_REPORT_JOINS_CONFIG
GROUP BY INVOICENO ORDER BY INVOICEDATE DESC);
任何帮助都会很好,谢谢

TOTAL = (SUBTOTAL * VAT)
应该是

(SUBTOTAL * VAT) AS TOTAL
现在,它将返回相等比较的布尔真/假结果。您没有将乘法结果赋给“total”字段,而是将总计值与乘法结果进行比较

这是一个明显的语法错误:

SUBTOTAL (SELECT(USUALPAY * COUNT(*) AS SUBTOTAL FROM SALES_REPORT)),

那么,当您这样做的时候,您是否只是得到了错误,或者这是一个内容是动态的,在运行时确定的东西?我没有得到任何错误,但是小计应该被填充并设置为null,phpmyadmin认为我的select语句是字段名的一部分。非常感谢,这解决了我的问题!