Php 错误：您的SQL语法有错误；查看与您的MySQL服务器版本对应的手册，了解要使用的正确语法

为什么我在尝试将tbl_面试计划加入tbl_预订信息id时总是出现这样的错误

SELECT CI.*, HM.*, TP.*, RI.*, IS.*
FROM tbl_client_info AS CI

INNER JOIN tbl_reservation_info AS RI
ON RI.client_info_id = CI.client_info_id

INNER JOIN tbl_house_model AS HM
ON HM.house_model_id = RI.house_model_id

INNER JOIN tbl_type_of_payment AS TP
ON TP.type_of_payment_id = RI.type_of_payment_id

INNER JOIN tbl_interview_schedule AS IS
ON IS.reservation_info_id = RI.reservation_info_id

WHERE RI.reservationstatus = 'For Interview' AND TP.paymenttype = 'Cash'

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IS

是MySQL中的保留字。

在帖子正文中包含错误消息…IS

IS

是您没有转义的关键字吗？

原样

-IS是保留字。尝试使用其他别名。谢谢你的否决权。。。如果不能发布…明白了，伙计们！谢谢！：）