-3

我从“last_insert_id()”获取值时遇到问题。我想将相同的 id 号插入到表“memo”的 id 中。但是,问题是在 foreach 之后,last_insert_id() 获得了一个新值,所以,我尝试创建一个变量并从 'last_insert_id()' 中获取一个值,但它没有用..

$diary_id = 'last_insert_id()';

foreach ($_POST['memo'] as $memo) {
     echo "You selected: $diary_id <br>";
     echo "You selected: $memo <br>";

 $query_test = "insert into memo(memo_no,id,memo)
    values(NULL,$diary_id,'$memo')";
 mysql_query($query_test, $connect);

}

4

2 回答 2

0

我假设 $connect 是一个数据库连接链接,所以使用 mysql_insert_id($connect) 然后尝试:

<?php 

$diary_id = mysql_insert_id($connect);

foreach ($_POST['memo'] as $memo) {
     echo "You selected: $diary_id <br>";
     echo "You selected: $memo <br>";

 $query_test = "insert into memo(memo_no,id,memo) values(NULL,$diary_id,'$memo')";
 mysql_query($query_test, $connect);
}

?>
于 2014-08-20T09:20:09.483 回答
-1

使用mysql_insert_idphp 中的函数来获取上一个 AUTO INCREMENTED 生成的 id。使用下面的代码

$diary_id = mysql_insert_id();

foreach ($_POST['memo'] as $memo) {
     echo "You selected: $diary_id <br>";
     echo "You selected: $memo <br>";

 $query_test = "insert into memo(memo_no,id,memo)
    values(NULL,$diary_id,'$memo')";
 mysql_query($query_test, $connect);

}

希望这可以帮助你

于 2014-08-20T09:20:16.303 回答