MySql and inserting data in a link table, how to? - php

What is the trick in MySql (Version 4) to write from PHP thtree Inserts and to be sure that we can link tables properly.
[Table1]-[LinkTable]-[Table2]
In the PHP code I create an Insert that add 1 row in table 1 and an other Insert that add an other row in table 2. I get both rows PK (primary key) with two SELECTs that return me the last row of each tables. This way I get the PK (auto-increment) from Table 1 and Table 2 and I insert it on the link table.
The problem is that is not atomic and now that I get a lot of transaction something it doesn't link properly.
How can I make a link between 2 tables in MySQL 4 that preserve data? I cannot use any stored procedure.


Well, if you have the ability to use InnoDB tables (instead of MyISAM) then you can use transactions. Here's some rough code
<?php
// Start a transaction
mysql_query( 'begin' );
// Execute the queries
if ( mysql_query( "insert into table_one (col1, col2) values('hello','world')" ) )
{
$pk1 = mysql_insert_id();
if ( mysql_query( "insert into table_two (col1, col2) values('foo', 'bar')" ) )
{
$pk2 = mysql_insert_id();
$success = mysql_query( "insert into link_table (fk1, fk2) values($pk1, $pk2)" );
}
}
// Complete the transaction
if ( $success )
{
mysql_query( 'commit' );
} else {
mysql_query( 'rollback' );
}
If you can't use InnoDB tables then I suggest looking into the Unit Of Work pattern.


your problem is here:
SELECTs that return me the last row of each tables.
if you did something like this: SELECT MAX(id) FROM table you can get a wrong id for your transaction.
This is a common many-to-many relationship
as BaileyP say you should use the function mysql_insert_id().
at http://ar2.php.net/mysql_insert_id
you can read
Retrieves the ID generated for an AUTO_INCREMENT column by the previous INSERT query.
so dont care if other process insert a row in the same table. you always will get the last id for your current connection.

Related

INSERT MULTIPLE ROWS in Gerund Table using Insert Into Select

I used INSERT INTO SELECT to copy values (multiple rows) from one table to another. Now, my problem is how do I insert rows with its corresponding IDs from different tables (since it's normalized) into a gerund table because it only outputs one row in my gerund table. What should I do to insert multiple rows and their corresponding IDs in the gerund table.
My code for the gerund table goes like this.
$insert = "INSERT INTO table1 SELECT * FROM sourcetable"; // where id1 is pk of table1.
$result =mysqli_query($conn,$insert)
$id1=mysqli_insert_id($conn);
Now table 1 has inserted multiple rows same as the other 2 tables.
Assuming id.. are the foreign keys
INSERT INTO gerundtable (pk, id1,id2,id3) VALUES ($id1,$id2,$id3);
My problem is it doesn't yield multiple rows.

According to MySql documentation:
For a multiple-row insert, LAST_INSERT_ID() and mysql_insert_id() actually return the AUTO_INCREMENT key from the first of the inserted rows. This enables multiple-row inserts to be reproduced correctly on other servers in a replication setup.
So, grab the number of records being copied, and the LAST_INSERT_ID() and you should be able to map exact IDs with each copied row.
In the lines of:
$mysqli->query("Insert Into dest_table Select * from source_table");
$n = $mysqli->affected_rows; // number of copied rows
$id1 = $mysqli->insert_id; // new ID of the first copied row
$id2 = $mysqli->insert_id + 1; // new ID of the second copied row
$id3 = $mysqli->insert_id + 2; // new ID of the third copied row
...
$mysqli->query("INSERT INTO gerundtable (pk, id1,id2,id3) VALUES ($id1,$id2,$id3)");

Thank you for trying to understand and also answering my question. I resolved my own code. I used while loop to get the ids of every row and didn't use INSERT INTO SELECT.
Here is the run down. SInce I'm just using my phone bare with my way posting.
$sqlselect = SELECT * FROM table1;
While($row=mysqli_fetch_array(table1){
$insertquery...
$id1=mysqli_insert_id($conn)
$insertgerundtable = INSERT INTO gerundtable VALUES ( $id1, $id2);
}

pdo update multiple rows in one query [duplicate]

I know that you can insert multiple rows at once, is there a way to update multiple rows at once (as in, in one query) in MySQL?
Edit:
For example I have the following
Name id Col1 Col2
Row1 1 6 1
Row2 2 2 3
Row3 3 9 5
Row4 4 16 8
I want to combine all the following Updates into one query
UPDATE table SET Col1 = 1 WHERE id = 1;
UPDATE table SET Col1 = 2 WHERE id = 2;
UPDATE table SET Col2 = 3 WHERE id = 3;
UPDATE table SET Col1 = 10 WHERE id = 4;
UPDATE table SET Col2 = 12 WHERE id = 4;

Yes, that's possible - you can use INSERT ... ON DUPLICATE KEY UPDATE.
Using your example:
INSERT INTO table (id,Col1,Col2) VALUES (1,1,1),(2,2,3),(3,9,3),(4,10,12)
ON DUPLICATE KEY UPDATE Col1=VALUES(Col1),Col2=VALUES(Col2);

Since you have dynamic values, you need to use an IF or CASE for the columns to be updated. It gets kinda ugly, but it should work.
Using your example, you could do it like:
UPDATE table SET Col1 = CASE id
WHEN 1 THEN 1
WHEN 2 THEN 2
WHEN 4 THEN 10
ELSE Col1
END,
Col2 = CASE id
WHEN 3 THEN 3
WHEN 4 THEN 12
ELSE Col2
END
WHERE id IN (1, 2, 3, 4);

The question is old, yet I'd like to extend the topic with another answer.
My point is, the easiest way to achieve it is just to wrap multiple queries with a transaction. The accepted answer INSERT ... ON DUPLICATE KEY UPDATE is a nice hack, but one should be aware of its drawbacks and limitations:
As being said, if you happen to launch the query with rows whose primary keys don't exist in the table, the query inserts new "half-baked" records. Probably it's not what you want
If you have a table with a not null field without default value and don't want to touch this field in the query, you'll get "Field 'fieldname' doesn't have a default value" MySQL warning even if you don't insert a single row at all. It will get you into trouble, if you decide to be strict and turn mysql warnings into runtime exceptions in your app.
I made some performance tests for three of suggested variants, including the INSERT ... ON DUPLICATE KEY UPDATE variant, a variant with "case / when / then" clause and a naive approach with transaction. You may get the python code and results here. The overall conclusion is that the variant with case statement turns out to be twice as fast as two other variants, but it's quite hard to write correct and injection-safe code for it, so I personally stick to the simplest approach: using transactions.
Edit: Findings of Dakusan prove that my performance estimations are not quite valid. Please see this answer for another, more elaborate research.

Not sure why another useful option is not yet mentioned:
UPDATE my_table m
JOIN (
SELECT 1 as id, 10 as _col1, 20 as _col2
UNION ALL
SELECT 2, 5, 10
UNION ALL
SELECT 3, 15, 30
) vals ON m.id = vals.id
SET col1 = _col1, col2 = _col2;

All of the following applies to InnoDB.
I feel knowing the speeds of the 3 different methods is important.
There are 3 methods:
INSERT: INSERT with ON DUPLICATE KEY UPDATE
TRANSACTION: Where you do an update for each record within a transaction
CASE: In which you a case/when for each different record within an UPDATE
I just tested this, and the INSERT method was 6.7x faster for me than the TRANSACTION method. I tried on a set of both 3,000 and 30,000 rows.
The TRANSACTION method still has to run each individually query, which takes time, though it batches the results in memory, or something, while executing. The TRANSACTION method is also pretty expensive in both replication and query logs.
Even worse, the CASE method was 41.1x slower than the INSERT method w/ 30,000 records (6.1x slower than TRANSACTION). And 75x slower in MyISAM. INSERT and CASE methods broke even at ~1,000 records. Even at 100 records, the CASE method is BARELY faster.
So in general, I feel the INSERT method is both best and easiest to use. The queries are smaller and easier to read and only take up 1 query of action. This applies to both InnoDB and MyISAM.
Bonus stuff:
The solution for the INSERT non-default-field problem is to temporarily turn off the relevant SQL modes: SET SESSION sql_mode=REPLACE(REPLACE(##SESSION.sql_mode,"STRICT_TRANS_TABLES",""),"STRICT_ALL_TABLES",""). Make sure to save the sql_mode first if you plan on reverting it.
As for other comments I've seen that say the auto_increment goes up using the INSERT method, this does seem to be the case in InnoDB, but not MyISAM.
Code to run the tests is as follows. It also outputs .SQL files to remove php interpreter overhead
<?php
//Variables
$NumRows=30000;
//These 2 functions need to be filled in
function InitSQL()
{
}
function RunSQLQuery($Q)
{
}
//Run the 3 tests
InitSQL();
for($i=0;$i<3;$i++)
RunTest($i, $NumRows);
function RunTest($TestNum, $NumRows)
{
$TheQueries=Array();
$DoQuery=function($Query) use (&$TheQueries)
{
RunSQLQuery($Query);
$TheQueries[]=$Query;
};
$TableName='Test';
$DoQuery('DROP TABLE IF EXISTS '.$TableName);
$DoQuery('CREATE TABLE '.$TableName.' (i1 int NOT NULL AUTO_INCREMENT, i2 int NOT NULL, primary key (i1)) ENGINE=InnoDB');
$DoQuery('INSERT INTO '.$TableName.' (i2) VALUES ('.implode('), (', range(2, $NumRows+1)).')');
if($TestNum==0)
{
$TestName='Transaction';
$Start=microtime(true);
$DoQuery('START TRANSACTION');
for($i=1;$i<=$NumRows;$i++)
$DoQuery('UPDATE '.$TableName.' SET i2='.(($i+5)*1000).' WHERE i1='.$i);
$DoQuery('COMMIT');
}
if($TestNum==1)
{
$TestName='Insert';
$Query=Array();
for($i=1;$i<=$NumRows;$i++)
$Query[]=sprintf("(%d,%d)", $i, (($i+5)*1000));
$Start=microtime(true);
$DoQuery('INSERT INTO '.$TableName.' VALUES '.implode(', ', $Query).' ON DUPLICATE KEY UPDATE i2=VALUES(i2)');
}
if($TestNum==2)
{
$TestName='Case';
$Query=Array();
for($i=1;$i<=$NumRows;$i++)
$Query[]=sprintf('WHEN %d THEN %d', $i, (($i+5)*1000));
$Start=microtime(true);
$DoQuery("UPDATE $TableName SET i2=CASE i1\n".implode("\n", $Query)."\nEND\nWHERE i1 IN (".implode(',', range(1, $NumRows)).')');
}
print "$TestName: ".(microtime(true)-$Start)."<br>\n";
file_put_contents("./$TestName.sql", implode(";\n", $TheQueries).';');
}

UPDATE table1, table2 SET table1.col1='value', table2.col1='value' WHERE table1.col3='567' AND table2.col6='567'
This should work for ya.
There is a reference in the MySQL manual for multiple tables.

Use a temporary table
// Reorder items
function update_items_tempdb(&$items)
{
shuffle($items);
$table_name = uniqid('tmp_test_');
$sql = "CREATE TEMPORARY TABLE `$table_name` ("
." `id` int(10) unsigned NOT NULL AUTO_INCREMENT"
.", `position` int(10) unsigned NOT NULL"
.", PRIMARY KEY (`id`)"
.") ENGINE = MEMORY";
query($sql);
$i = 0;
$sql = '';
foreach ($items as &$item)
{
$item->position = $i++;
$sql .= ($sql ? ', ' : '')."({$item->id}, {$item->position})";
}
if ($sql)
{
query("INSERT INTO `$table_name` (id, position) VALUES $sql");
$sql = "UPDATE `test`, `$table_name` SET `test`.position = `$table_name`.position"
." WHERE `$table_name`.id = `test`.id";
query($sql);
}
query("DROP TABLE `$table_name`");
}

Why does no one mention multiple statements in one query?
In php, you use multi_query method of mysqli instance.
From the php manual
MySQL optionally allows having multiple statements in one statement string. Sending multiple statements at once reduces client-server round trips but requires special handling.
Here is the result comparing to other 3 methods in update 30,000 raw. Code can be found here which is based on answer from #Dakusan
Transaction: 5.5194580554962
Insert: 0.20669293403625
Case: 16.474853992462
Multi: 0.0412278175354
As you can see, multiple statements query is more efficient than the highest answer.
If you get error message like this:
PHP Warning: Error while sending SET_OPTION packet
You may need to increase the max_allowed_packet in mysql config file which in my machine is /etc/mysql/my.cnf and then restart mysqld.

There is a setting you can alter called 'multi statement' that disables MySQL's 'safety mechanism' implemented to prevent (more than one) injection command. Typical to MySQL's 'brilliant' implementation, it also prevents user from doing efficient queries.
Here (http://dev.mysql.com/doc/refman/5.1/en/mysql-set-server-option.html) is some info on the C implementation of the setting.
If you're using PHP, you can use mysqli to do multi statements (I think php has shipped with mysqli for a while now)
$con = new mysqli('localhost','user1','password','my_database');
$query = "Update MyTable SET col1='some value' WHERE id=1 LIMIT 1;";
$query .= "UPDATE MyTable SET col1='other value' WHERE id=2 LIMIT 1;";
//etc
$con->multi_query($query);
$con->close();
Hope that helps.

You can alias the same table to give you the id's you want to insert by (if you are doing a row-by-row update:
UPDATE table1 tab1, table1 tab2 -- alias references the same table
SET
col1 = 1
,col2 = 2
. . .
WHERE
tab1.id = tab2.id;
Additionally, It should seem obvious that you can also update from other tables as well. In this case, the update doubles as a "SELECT" statement, giving you the data from the table you are specifying. You are explicitly stating in your query the update values so, the second table is unaffected.

You may also be interested in using joins on updates, which is possible as well.
Update someTable Set someValue = 4 From someTable s Inner Join anotherTable a on s.id = a.id Where a.id = 4
-- Only updates someValue in someTable who has a foreign key on anotherTable with a value of 4.
Edit: If the values you are updating aren't coming from somewhere else in the database, you'll need to issue multiple update queries.

No-one has yet mentioned what for me would be a much easier way to do this - Use a SQL editor that allows you to execute multiple individual queries. This screenshot is from Sequel Ace, I'd assume that Sequel Pro and probably other editors have similar functionality. (This of course assumes you only need to run this as a one-off thing rather than as an integrated part of your app/site).

And now the easy way
update my_table m, -- let create a temp table with populated values
(select 1 as id, 20 as value union -- this part will be generated
select 2 as id, 30 as value union -- using a backend code
-- for loop
select N as id, X as value
) t
set m.value = t.value where t.id=m.id -- now update by join - quick

Yes ..it is possible using INSERT ON DUPLICATE KEY UPDATE sql statement..
syntax:
INSERT INTO table_name (a,b,c) VALUES (1,2,3),(4,5,6)
ON DUPLICATE KEY UPDATE a=VALUES(a),b=VALUES(b),c=VALUES(c)

use
REPLACE INTO`table` VALUES (`id`,`col1`,`col2`) VALUES
(1,6,1),(2,2,3),(3,9,5),(4,16,8);
Please note:
id has to be a primary unique key
if you use foreign keys to
reference the table, REPLACE deletes then inserts, so this might
cause an error

I took the answer from #newtover and extended it using the new json_table function in MySql 8. This allows you to create a stored procedure to handle the workload rather than building your own SQL text in code:
drop table if exists `test`;
create table `test` (
`Id` int,
`Number` int,
PRIMARY KEY (`Id`)
);
insert into test (Id, Number) values (1, 1), (2, 2);
DROP procedure IF EXISTS `Test`;
DELIMITER $$
CREATE PROCEDURE `Test`(
p_json json
)
BEGIN
update test s
join json_table(p_json, '$[*]' columns(`id` int path '$.id', `number` int path '$.number')) v
on s.Id=v.id set s.Number=v.number;
END$$
DELIMITER ;
call `Test`('[{"id": 1, "number": 10}, {"id": 2, "number": 20}]');
select * from test;
drop table if exists `test`;
It's a few ms slower than pure SQL but I'm happy to take the hit rather than generate the sql text in code. Not sure how performant it is with huge recordsets (the JSON object has a max size of 1Gb) but I use it all the time when updating 10k rows at a time.

The following will update all rows in one table
Update Table Set
Column1 = 'New Value'
The next one will update all rows where the value of Column2 is more than 5
Update Table Set
Column1 = 'New Value'
Where
Column2 > 5
There is all Unkwntech's example of updating more than one table
UPDATE table1, table2 SET
table1.col1 = 'value',
table2.col1 = 'value'
WHERE
table1.col3 = '567'
AND table2.col6='567'

UPDATE tableName SET col1='000' WHERE id='3' OR id='5'
This should achieve what you'r looking for. Just add more id's. I have tested it.

UPDATE `your_table` SET
`something` = IF(`id`="1","new_value1",`something`), `smth2` = IF(`id`="1", "nv1",`smth2`),
`something` = IF(`id`="2","new_value2",`something`), `smth2` = IF(`id`="2", "nv2",`smth2`),
`something` = IF(`id`="4","new_value3",`something`), `smth2` = IF(`id`="4", "nv3",`smth2`),
`something` = IF(`id`="6","new_value4",`something`), `smth2` = IF(`id`="6", "nv4",`smth2`),
`something` = IF(`id`="3","new_value5",`something`), `smth2` = IF(`id`="3", "nv5",`smth2`),
`something` = IF(`id`="5","new_value6",`something`), `smth2` = IF(`id`="5", "nv6",`smth2`)
// You just building it in php like
$q = 'UPDATE `your_table` SET ';
foreach($data as $dat){
$q .= '
`something` = IF(`id`="'.$dat->id.'","'.$dat->value.'",`something`),
`smth2` = IF(`id`="'.$dat->id.'", "'.$dat->value2.'",`smth2`),';
}
$q = substr($q,0,-1);
So you can update hole table with one query

Inserting data into multiple tables not functioning correctly

I have the following two tables
Table player:
player_id (int)(primary)
player_name (varchar)
player_report_count (int)
Table report:
report_id (int)(primary)
player_id
report_description
report_location
Firstly I ask the user for the player_name and insert it into the player database. From here the player is given an id.
Then I tried to grab the value of the players report count and increment the current value by one (which isn't working).
This is followed by grabbing the playerId from the player table and then inserting into the corresponding column from the report table (also does not work).
When I insert some values into the database, the names, description and report are added to the database however the playerID remains at 0 for all entries and the player_report_count remains at a consistent 0.
What is the correct way to make these two features function? And also is there a more efficient way of doing this?
<?php
$records = array();
if(!empty($_POST)){
if(isset($_POST['player_name'],
$_POST['report_description'],
$_POST['report_location'])){
$player_name = trim($_POST['player_name']);
$report_description = trim($_POST['report_description']);
$report_location = trim($_POST['report_location']);
if(!empty($player_name) && !empty($report_description) && !empty($report_location)){
$insertPlayer = $db->prepare("
INSERT INTO player (player_name)
VALUES (?)
");
$insertPlayer->bind_param('s', $player_name);
$reportCount = $db->query("
UPDATE player
SET player_report_count = player_report_count + 1
WHERE
player_name = $player_name
");
$getPlayerId = $db->query("
SELECT player_id
FROM player
WHERE player_name = $player_name
");
$insertReport = $db->prepare("
INSERT INTO report (player_id, report_description, report_location)
VALUES (?, ?, ?)
");
$insertReport->bind_param('iss', $getPlayerId, $report_description, $report_location);
if($insertPlayer->execute()
&& $insertReport->execute()
){
header('Location: insert.php');
die();
}
}
}

Main issue here is you are getting player details before inserting it. $getPlayerId will return empty result always.
Please follow the order as follows.
Insert player details in to player table and get payerid with mysql_insert_id. After binding you need to execute to insert details to the table.
Then bind and execute insert report .
Then update the player table by incrementing report count with playerid which you got in step 1.
Note : use transactions when inserting multiple table. This will help you to rollback if any insert fails.
MySQL Query will return result object. Refer it from here https://stackoverflow.com/a/13791544/3045153
I hope it will help you

If you need to catch the ID of the last insterted player, This is the function you need if you're using PDO or if it's a custom Mysql Class, you need the return value of mysql_insert_id() (or mysqli_insert_id()) and then directly use it in the next INSERT INTO statement

Insert id from from one table and insert it into another one. Mysql + PHP [duplicate]

This question already has answers here:
How to get the last field in a Mysql database with PHP?
(5 answers)
Closed 9 years ago.
I am working on a register user form and I have two tables in mysql. What I want to do is when a new user has registered, take the id (which primary key) of that user and insert it into another table. What is the best way to do that?
Thanks in advance.

You need to use mysql_insert_id for this purpose. Here is an example:
<?php
$link = mysql_connect('localhost', 'mysql_user', 'mysql_password');
if (!$link) {
die('Could not connect: ' . mysql_error());
}
mysql_select_db('mydb');
mysql_query("INSERT INTO mytable (product) values ('kossu')");
printf("Last inserted record has id %d\n", mysql_insert_id());
?>

First insert the user details into users table and get inserted user id using mysql_insert_id. and use that user id to insert into another table.

AS ON GETTING IT ON PHP
GET LAST INSERT ID HERE
BUT IF YOU INTEND TO GET IT USING MYSQL QUERY
use stored procedure to store last insert id to a variable then generete your second query
INSERT INTO T1 (col1,col2) VALUES (val1,val2);
SET #last_id_in_T1 = LAST_INSERT_ID();
INSERT INTO T2 (col1,col2) VALUES (#last_id_in_T1,val2);
or direct insert after your first insert
INSERT INTO T1 (col1,col2) VALUES (val1,val2);
INSERT INTO T2 (col1,col2) VALUES (LAST_INSERT_ID(),val2);

Use any of transaction query for writing:
Following is in CI pattern:
$this->db->trans_start();
$this->db->query('AN SQL QUERY...');
$this->db->query('AN SQL QUERY...');
if(!$this->db->trans_complete()){
$this->db->trans_rollback();
}

$query1= "INSERT INTO employee ( username, email,...)
VALUES ('".$_POST["username"]."', ...)";
if($result1 = mysql_query($query1))
{
$emp_id = mysql_insert_id(); // last created id by above query
$query2= "INSERT INTO dept ( emp_id, dept_name, ...)
VALUES ('".$emp_id."', '".$_POST["dept_name"]."',...)";
if($result2 = mysql_query($query2))
{
//success msg
}
}

Another neat way to do it at do it at Database level itself is to used Stored Procedure
Look at this solution to see an example of how to do it. You will have to check how Stored procedures work in your specific database to get the specific syntax. This makes it error free even if someone refactors or moves around the code and more efficient.

Using trigger the Mysql on database-level. For example, I have two tables:
user(id int primary key, nombre varchar(50));
replication(id_r int primary key, nombre_r varchar(50));
Using the trigger:
create trigger user_r after insert on user
for each row
insert into replication(id_r, nombre_r)
select u.id, u.nombre
from user u
where u.id=NEW.id and u.nombre=NEW.nombre;

How do I get all the ids of the row created by one multiple row insert statement

I'm new to php. So, please forgive me if this seems like a dumb question.
Say i have a MySQL insert statement insert into table (a,b) values (1,2),(3,4),(5,6). table 'table' has a auto increment field called 'id'.
how can I retrieve all the ids created by the insert statement above?
It will be great if i get an example that uses mysqli.

You can't. I would suggest that you maintain your own ids (using guid or your own auto-increment table) and use it when you insert into the table.
But it's possible to get the auto-increment value for the last inserted using LAST_INSERT_ID():
http://dev.mysql.com/doc/refman/5.0/en/getting-unique-id.html

AngeDeLaMort's answer is almost right. Certainly, the most appropriate way to deal with the problem is to insert one row at a time and poll the insert_id or generate the sequence elsewhere (which has additional benefits in terms of scalability).
I'd advise strongly against trying to determine the last insert_id and comparing this the most recent insert_id after the insert - there's just too may ways this will fail.
But...an alternative approach would be:
....
"INSERT INTO destn (id, data, other, trans_ref)
SELECT id, data, other, connection_id() FROM source";
....
"SELECT id FROM destn WHERE trans_ref=connection_id()";
....
"UPDATE destn SET trans_ref=NULL where trans_ref=connection_id()";
The second query will return the ids generated (note that this assumes that you use the same connection for all 3 queries). The third query is necessary because connection ids to go back into the pool when you disconnect (i.e. are reused).
C.

In some cases, if you have another identifier of sort such as a UserID, you could filter your query by UniqueID's greater than or equal to mysql_insert_id(), limit by the number of affected rows and only display those by the user. This would really only work inside of a transaction.
$SQL = "INSERT INTO Table
(UserID, Data)
VALUES
(1,'Foo'),
(1,'Bar'),
(1,'FooBar')";
$Result = mysql_query($SQL);
$LastID = mysql_insert_id();
$RowsAffected = mysql_affected_rows();
$IDSQL = "SELECT RecordID
FROM Table
WHERE UserID = 1
AND RecordID >= '$LastID'
LIMIT '$RowsAffected'";
$IDResult = mysql_query($IDSQL);

as a follow up to AngeDeLaMort:
You could seperate your inserts and do it something like this:
$data = array (
array(1,2),
array(3,4),
array(5,6)
);
$ids = array();
foreach ($data as $item) {
$sql = 'insert into table (a,b) values ('.$item[0].','.$item[1].')';
mysql_query ($sql);
$id[] = mysql_insert_id();
}
Now all your new id's are in the $id array.

Maybe I can do this
$insert = "insert into table (a,b) values (1,2),(3,4),(5,6)";
$mysqli->query($insert);
$rows_to_be_inserted=3;
$inserted_id = $mysqli->insert_id // gives me the id of the first row in my list
$last_row_id = ($inserted_id+$rows_to_be_inserted)-1;
$mysql->query("select * from table where id between $inserted_id and $last_row_id");
what to you guys say?

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