我正在从另一个页面收集数据,并将其发送到处理数据并将其发送到我的数据库表。这是代码。
require "config.php";
if ( $_POST['insertPost'] == true ){
$postTitle = $_POST['title'];
$postContent = $_POST['content'];
$postImage = $_POST['image'];
$postDate = $_POST['datetime'];
$categories = $_POST['categories'];
$c->query( "INSERT INTO blog ( title, content, image, datetime, categories ) VALUES ( '$postTitle', '$postContent', '$postImage','$postDate', '$categories')" );
//echo "Inserted";
}
如何获取插入数据的ID信息并将其存储到变量中?
最佳答案
使用 mysqli 进行数据库操作,因为 mysql 现已弃用。所以,试试这个:
$con=mysqli_connect("localhost","my_user","my_password","my_db");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
mysqli_query($con, "INSERT INTO blog (title, content, image, datetime, categories) VALUES ('$postTitle', '$postContent', '$postImage','$postDate', '$categories')");
// store auto-generated id
$lastInsertedId = mysqli_insert_id($con);
mysqli_close($con);
关于php - 获取php和mysqli中插入数据的Id,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/33998131/