Question

我已经查看过几个连接问题，似乎无法让这些解决方案适合我。

case "tblContestant":
            $sql = "SELECT e.id as Id,@n := @n + 1 n, replace(c.name, ',', ' ') as name, c.address, NULL, c.city, c.state, c.phonenumber, c.email,  c.sanctionbody,     concat(c.id,c.sanctionnumber) as sanctionnumber, mcrc.amount,   c.zip, c.jrboater
                    FROM contestants c 
                    LEFT JOIN main_contestant_register_class mcrc ON mcrc.contestant_id = c.id 
                    LEFT JOIN events e ON e.id = mcrc.event_id, (SELECT @n := 0) m  
                    WHERE e.id = $id";

上面给出了将c.id与c.sanctionnumber（142155）结合起来的结果，但是我想在c.id和c.sanctionnumber之间有一个空格。（14 - 2155）或（14-2155）如果不能选择空格。

提前致谢！

Answer 1

case "tblContestant":
            $sql = "SELECT e.id as Id,@n := @n + 1 n, replace(c.name, ',', ' ') as name, c.address, NULL, c.city, c.state, c.phonenumber, c.email,  c.sanctionbody,     concat(c.id,' - '+ c.sanctionnumber) as sanctionnumber, mcrc.amount,   c.zip, c.jrboater
                    FROM contestants c 
                    LEFT JOIN main_contestant_register_class mcrc ON mcrc.contestant_id = c.id 
                    LEFT JOIN events e ON e.id = mcrc.event_id, (SELECT @n := 0) m  
                    WHERE e.id = $id";

Answer 2

Avi感谢您的帮助，问题得到了解决！我只是将+改为a，现在它给了我想要的结果。

concat(c.id,' - '+ c.sanctionnumber) as sanctionnumber, gives me 142155
concat(c.id,' - ', c.sanctionnumber) as sanctionnumber, gives me 14 - 2155

你能和空间结合吗？

2 个答案: