关于获取上一步insert的SQL语句的id的问题
用mysql_insert_id函数的话,当错作流程是:A页面插入了一条,id=1,B页面插入了一条,id=2,A页面执行mysql_insert_id获取到的id是1还是2?
回复讨论(解决方案)
A页面当然是 1
A页面当然是 1
我可以理解为是返回`当前页面`的上一步执行的inser语句的id吗,w3school上说得不太清楚
如果不是1 ，就麻烦了，那么写所有的东西，都要操心并发的问题了。
可以对mysql服务器发送查询
mysql_query("select LAST_INSERT_ID()"); A页面的mysql_insert_id返回是1
MySQL的mysql_insert_id是针对MySQL进程来进行的，对于服务器来说mysql_query都是一个原子操作，不是MAX(id),就算在大并发的情况下也不会出现不准确的情况。
可以看看MySQL的
mysql_insert_id是这么定义的
my_ulonglong STDCALL mysql_insert_id(MYSQL *mysql)
{
return mysql->;last_used_con->;insert_id;
}
MYSQL参数是个结构体，里面包括数据库链接和一些当前数据库链接的状态值，
其中在MYSQL结构体里面有insert_id，mysql_insert_id函数返回的就是结构体里面的找个值。
肯定是1,自己可以实际操作一下,就明白了
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