mysql-insert-id

How reliable is mysql_insert_id() in a competitive condition? I mean when multiple users are inserting data at the same time, will this function return the true ID or it will return other user's inserted data ID? The table engine is MyISAM. mysql_insert_id is completely multi-user safe.It is entirely relies upon the database connection, and you can only have one user per connection....So, as per your question, it returns true id.... From MySql documentation , The ID that was generated is maintained in the server on a per-connection basis. This means that the value returned by the function to a

I've found a few answers for this using mySQL alone, but I was hoping someone could show me a way to get the ID of the last inserted or updated row of a mysql DB when using PHP to handle the inserts/updates. Currently I have something like this, where column3 is a unique key, and there's also an id column that's an autoincremented primary key: $query ="INSERT INTO TABLE (column1, column2, column3) VALUES (value1, value2, value3) ON DUPLICATE KEY UPDATE SET column1=value1, column2=value2, column3=value3"; mysql_query($query); $my_id = mysql_insert_id(); $my_id is correct on INSERT, but

I know there are a lot of topics with the same title. But mostly it's the query that's been inserted in the wrong place. But I think I placed it right. So the problem is, that I still get 0 even when the data is inserted in the db. Does someone knows an answer where I could be wrong? here's my code: mysql_query('SET NAMES utf8'); $this->arr_kolommen = $arr_kolommen; $this->arr_waardes = $arr_waardes; $this->tabel = $tabel; $aantal = count($this->arr_kolommen); //$sql="INSERT INTO `tbl_photo_lijst_zoekcriteria` ( `PLZ_FOTO` , `PLZ_ZOEKCRITERIA`,`PLZ_CATEGORIE`)VALUES ('$foto', '$zoekje','

When I run the following code, I get the error saying Fatal error: Uncaught exception 'mysqli_sql_exception' with message 'No index used in query/prepared statement' $mysql = new mysqli(DB_SERVER, DB_USER, DB_PASSWORD, DB_NAME) or die('There was a problem connecting to the database'); if (mysqli_connect_errno()) { printf("DB error: %s", mysqli_connect_error()); exit(); } $get_emp_list = $mysql->prepare("SELECT id, name FROM calc"); if(!$get_emp_list){ echo "prepare failed\n"; echo "error: ", $mysql->error, "\n"; return; } $get_emp_list->execute(); $get_emp_list->bind_result($id, $emp_list);