LAST_INSERT_ID() MySQL

Question

You could store the last insert id in a variable :

INSERT INTO table1 (title,userid) VALUES ('test', 1); 
SET @last_id_in_table1 = LAST_INSERT_ID();
INSERT INTO table2 (parentid,otherid,userid) VALUES (@last_id_in_table1, 4, 1);    

Or get the max id from table1 (EDIT: Warning. See note in comments from Rob Starling about possible errors from race conditions when using the max id)

INSERT INTO table1 (title,userid) VALUES ('test', 1); 
INSERT INTO table2 (parentid,otherid,userid) VALUES (LAST_INSERT_ID(), 4, 1); 
SELECT MAX(id) FROM table1;  

(Warning: as Rob Starling points out in the

Since you actually stored the previous LAST_INSERT_ID() into the second table, you can get it from there:

INSERT INTO table1 (title,userid) VALUES ('test',1); 
INSERT INTO table2 (parentid,otherid,userid) VALUES (LAST_INSERT_ID(),4,1);
SELECT parentid FROM table2 WHERE id = LAST_INSERT_ID();

This enables you to insert a row into 2 different tables and creates a reference to both tables too.

START TRANSACTION;
INSERT INTO accounttable(account_username) 
    VALUES('AnAccountName');
INSERT INTO profiletable(profile_account_id) 
    VALUES ((SELECT account_id FROM accounttable WHERE account_username='AnAccountName'));
    SET @profile_id = LAST_INSERT_ID(); 
UPDATE accounttable SET `account_profile_id` = @profile_id;
COMMIT;

I had the same problem in bash and i'm doing something like this:

mysql -D "dbname" -e "insert into table1 (myvalue) values ('${foo}');"

which works fine:-) But

mysql -D "dbname" -e "insert into table1 (myvalue) values ('${foo}');set @last_insert_id = LAST_INSERT_ID();"
mysql -D "dbname" -e "insert into table2 (id_tab1) values (@last_insert_id);"

don't work. Because after the first command, the shell will be logged out from mysql and logged in again for the second command, and then the variable @last_insert_id isn't set anymore. My solution is:

lastinsertid=$(mysql -B -N -D "dbname" -e "insert into table1 (myvalue) values ('${foo}');select LAST_INSERT_ID();")
mysql -D "dbname" -e "insert into table2 (id_tab1) values (${lastinsertid});"

Maybe someone is searching for a solution an bash :-)

We only have one person entering records, so I execute the following query immediately following the insert:

$result = $conn->query("SELECT * FROM corex ORDER BY id DESC LIMIT 1");

while ($row = $result->fetch_assoc()) {

            $id = $row['id'];

}

This retrieves the last id from the database.

It would be possible to save the last_id_in_table1 variable into a php variable to use it later?

With this last_id I need to attach some records in another table with this last_id, so I need:

1) Do an INSERT and get the last_id_in_table1

INSERT into Table1(name) values ("AAA"); 
SET @last_id_in_table1 = LAST_INSERT_ID();

2) For any indeterminated rows in another table, UPDATING these rows with the last_id_insert generated in the insert.

$element = array(some ids)    
foreach ($element as $e){ 
         UPDATE Table2 SET column1 = @last_id_in_table1 WHERE id = $e 
    }

Instead of this LAST_INSERT_ID() try to use this one

mysqli_insert_id(connection)