选择 LAST_INSERT_ID() *更新

发布于 2024-10-11 21:41:04 字数 888 浏览 5 评论 0原文

我正在寻找使用 SELECT LAST_INSERT_ID() 我使用表单来让用户输入值。对于第一次插入，我需要获取下一次插入的最后插入的 id...我还没有弄清楚如何获取最后选择的 id，然后将其传递到我的第二个插入语句中

我已经更新了我的代码，但我仍然不能获取要发布到表中的 id

include("config.inc.php");
$link = mysql_connect($db_host,$db_user,$db_pass);
if(!$link) die ('Could not connect to database: '.mysql_error());
mysql_select_db($db_name,$link);
$query = "INSERT into `".$db_table."` (producer_id,series_id,lang_id,title_name,title_public_access) VALUES ('" . $_POST['producer_id'] . "','" . $_POST['series_id'] . "','" . $_POST['lang_id'] . "','" . $_POST['title_name'] . "','" . $_POST['title_public_access'] . "')";

$last_id = mysql_insert_id();

$query = "INSERT into `".$db_table2."` (seg_id, file_video_UNC,file_video_URL) VALUES ('" . '$last_id' . "','" . $_POST['file_video_UNC'] . "','" . $_POST['file_video_URL'] . "')";

mysql_query($query);
mysql_close($link);

原文

I'm looking to use SELECT LAST_INSERT_ID()
Am using a form to have a user input values. With the first insert I need to get the last inserted id for the next insert... I have not figured out how to get the last selected id and then pass it into my 2nd insert statement

I have updated my code though I still can not get the id to post into the table

include("config.inc.php");
$link = mysql_connect($db_host,$db_user,$db_pass);
if(!$link) die ('Could not connect to database: '.mysql_error());
mysql_select_db($db_name,$link);
$query = "INSERT into `".$db_table."` (producer_id,series_id,lang_id,title_name,title_public_access) VALUES ('" . $_POST['producer_id'] . "','" . $_POST['series_id'] . "','" . $_POST['lang_id'] . "','" . $_POST['title_name'] . "','" . $_POST['title_public_access'] . "')";

$last_id = mysql_insert_id();

$query = "INSERT into `".$db_table2."` (seg_id, file_video_UNC,file_video_URL) VALUES ('" . '$last_id' . "','" . $_POST['file_video_UNC'] . "','" . $_POST['file_video_URL'] . "')";

mysql_query($query);
mysql_close($link);

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眼泪淡了忧伤 2024-10-18 21:41:04

有一个函数可以实现此目的，名为 mysql_insert_id()。

... first query here ...
$last_id = mysql_insert_id();
$sql = "INSERT INTO $db_table SET 
    file_video = " . $_POST['file_video_UNC'].",
    file_video_URL = " . $_POST['file_video_URL'] . ",
    insert_id_of_first_query = $last_id";
...

您更新的代码不会将查询发送到数据库 - 因此没有 INSERT，因此也没有 LAST_INSERT_ID

$query = "INSERT into ".$db_table." 
    (producer_id,series_id,lang_id,title_name,title_public_access) VALUES
    ('" . $_POST['producer_id'] . "','" 
        . $_POST['series_id'] . "','" 
        . $_POST['lang_id'] . "','" . $_POST['title_name'] . "','" 
        . $_POST['title_public_access'] . "')";

mysql_query($query); /* YOU FORGOT THIS PART */
$last_id = mysql_insert_id();

There's a function for that, called mysql_insert_id().

... first query here ...
$last_id = mysql_insert_id();
$sql = "INSERT INTO $db_table SET 
    file_video = " . $_POST['file_video_UNC'].",
    file_video_URL = " . $_POST['file_video_URL'] . ",
    insert_id_of_first_query = $last_id";
...

Your updated code doesn't send the query to database - as a result no INSERT, so no LAST_INSERT_ID

$query = "INSERT into ".$db_table." 
    (producer_id,series_id,lang_id,title_name,title_public_access) VALUES
    ('" . $_POST['producer_id'] . "','" 
        . $_POST['series_id'] . "','" 
        . $_POST['lang_id'] . "','" . $_POST['title_name'] . "','" 
        . $_POST['title_public_access'] . "')";

mysql_query($query); /* YOU FORGOT THIS PART */
$last_id = mysql_insert_id();

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佞臣 2024-10-18 21:41:04

您不能仅将查询转储到 PHP 行中的字符串中。您应该在第二个查询中使用 LAST_INSERT_ID() ，或者更好的是，使用 PHP 的 mysql_insert_id() 函数来包装这在 API 中为您提供。

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紅太極 2024-10-18 21:41:04

在该行中：

$query = "INSERT into `".$db_table2."` (seg_id, file_video_UNC,file_video_URL) VALUES ('" . '$last_id' . "','" . $_POST['file_video_UNC'] . "','" . $_POST['file_video_URL'] . "')";

我认为 VALUES ('" . '$last_id' . "', 应该只是 VALUES ('" . $last_id . "', 没有单个变量周围用引号引起来。

In the line:

$query = "INSERT into `".$db_table2."` (seg_id, file_video_UNC,file_video_URL) VALUES ('" . '$last_id' . "','" . $_POST['file_video_UNC'] . "','" . $_POST['file_video_URL'] . "')";

I think VALUES ('" . '$last_id' . "', should just be VALUES ('" . $last_id . "', without the single quotes around the variable.

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