我正在使用C ++而不想使用SELECT*
FROM (SELECT @rowsnum := @rowsnum + 1 AS ae1num,
Concat(ae1.a, ae1.b) AS id,
ae1.*
FROM A as ae1,
(SELECT @rowsnum := 0) r
ORDER BY ae1.a,
ae1.b) ae1
LEFT JOIN (SELECT aenum,
Concat(ae111.a, ae111.b) AS id,
aecm.*
FROM (SELECT @rownum := @rownum + 1 AS aenum,
Concat(ae11.a, ae11.b) AS id,
ae11.*
FROM A as ae11,
(SELECT @rownum := 0) a
ORDER BY ae11.a,
ae11.b) ae111
LEFT JOIN B as aecm
ON ae111.a = aecm.a
AND ae111.b = aecm.b
ORDER BY ae111.a,
ae111.b) aec
ON ae1.a = aec.a
AND ae1.b = aec.b
AND aec.aenum = ae1.ae1num
。我试图根据条件的值设置C样式字符串的值。这可能吗？
以下是最低工作示例：
std::string此程序没有错误，但将非#include <iostream>
void print_thing(int a)
{
char* message;
switch (a)
{
case 1:
message = "case 1";
break;
default:
message = "case 2";
break;
}
std::cout << message << std::endl;
}
int main()
{
int a{ 5 };
print_thing(a);
}
const设置为字符串文字似乎不好。有更好的方法吗？
答案 0 :(得分：4)
代码很好，除了C ++中的字符串文字类型为const char[]（与C char[]不同）。因此，您的变量必须声明为const char* message;。