为什么mysql_insert_id()返回0? [英] Why does mysql_insert_id() return 0?

问题描述

当我运行这段代码时

$sql_select = "INSERT INTO `database`.`table`(Columns) VALUES (Values)... ";
$mysqlid = mysql_insert_id($sql_select->db);
echo ($mysqlid);

我收到错误消息

mysql_insert_id(): supplied argument is not a valid MySQL-Link 

我已经尝试过这种变化；

I have tried this variation;

$mysqlid = mysql_insert_id();
echo ($mysqlid);

，但返回0，根据文档，这意味着未找到auto_increment字段.我唯一能想到的是，我没有在$ sql_select中调用auto_increment列，但是在其中有一个auto_increment列.会影响mysql_insert_id的行为吗?

but that returns a 0 which, according to the documentation, means an auto_increment field was not found. The only thing I can think of is that I am not calling the auto_increment column in the $sql_select, but there is an auto_increment column in there; will that affect the behavior of mysql_insert_id?

推荐答案

您需要首先实际运行查询:

you need to actually run the query first:

$sql= "INSERT INTO `database`.`table`(Columns) VALUES (Values)...";
$result = mysql_query($sql);

然后在那之后:

$id = mysql_insert_id();
echo $id

如果您仍有问题，请告诉我.

Let me know if you have still problems.

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