I need to port some data from tables on a development database into identical tables on the production database, but the production already has records with primary keys that match the dev database so I can't dump the data in with primary keysbundleRenderer.renderToStream
In this case, item_id is the primary key in the parent record, which is used to relate child records to it. Doing the insert of parent records will create a new primary key, so I need child inserts to also have the newly created primary key so that the relationship is maintained on the production databasebundleRenderer.renderToStream
my script so far:
<?php
$DB2connPROD = odbc_connect("schema","user", "pass");
$DB2connDEV = odbc_connect("schema","user", "pass");
//Get itemt records from dev
$getDevitems = "
select item_id,item_typet_id,item_identifier,expiration_timestamp
from development.itemt where item_typet_id in (2,3)
";
//$getDevitems will get records that have a primary key item_id which is used to get the records in the following select queries
foreach($getDevitems as $items){
//Get all comments
$getComments = "
select tc.item_id, tc.comment, tc.comment_type_id from development.item_commentt tc
inner join development.itemt t on tc.item_id = t.item_id
where t.item_id = {item_id_from_getDevitems}
";
$insertitem = "INSERT into production (item_identifier,expiration_timestamp)
values (item_identifier,expiration_timestamp)";
$insertComment = "INSERT into productionComment (item_id, comment, comment_type_id)
values (item_id, comment, comment_type_id)";
}
?>
So if $getDevitems returns
item_id | item_typet_id | item_identifier | expiration_timestamp
----------------------------------------------------------------------------
123 1 544 '2020-03-01 12:00:00'
I would want it to now run the comment select with 123 as the ID in the where clause:
select tc.item_id, tc.comment, tc.comment_type_id from development.item_commentt tc
inner join development.itemt t on tc.item_id = t.item_id
where t.item_id = 123
Now for my legacy parent record I have all of the parent data and all of the relational child data. so I want to insert the new parent record into the database, creating the new ID, and inserting the child record with the newly created primary key/ID. So for the new parent record I would do:
$insertitem = "INSERT into production (item_identifier,expiration_timestamp)
values (544,'2020-03-01 12:00:00')";
Let's say that creates the new record with item_id = 43409. I want my comment insert to be:
$insertComment = "INSERT into productionComment (item_id, comment, comment_type_id)
values (43409, comment, comment_type_id)";
Bottom LIne: I need to take relational data (all based on item_id) from a development database, and insert these into a new database which creates a new primary key but I need to keep the relationship.
How can I properly finish this to do what I need and make sure I maintain the full relationship for each originally selected item?
Given that your inserts are:
$insertitem = "INSERT into production (item_identifier,expiration_timestamp)
values (item_identifier,expiration_timestamp)";
$insertComment = "INSERT into productionComment (item_id, comment, comment_type_id)
values (item_id, comment, comment_type_id)";
It looks like you are using an identity column for item_id. You can retrieve the most recent generated identity value using the IDENTITY_VAL_LOCAL() function so the second insert should be:
$insertComment = "INSERT into productionComment (item_id, comment, comment_type_id)
values (IDENTITY_VAL_LOCAL(), comment, comment_type_id)";
I cannot help with PHP but with DB2 for IBMi you have different solutions :
If i understand it correctly item_id is a GENERATED ALWAYS as IDENTITY column
You can get newly created item_id using
select item_id from final table (
INSERT into production (item_identifier,expiration_timestamp)
values (544,'2020-03-01 12:00:00')
)
Or you can force value of item_id with dev value or your own increment
INSERT into production (idtem_id, item_identifier,expiration_timestamp)
values (<your value>, 544,'2020-03-01 12:00:00')
OVERRIDING SYSTEM VALUE
In this case you will have to set next value for item_id by issueing
alter table production alter column item_id restart with <restart value>
Related
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, $id2);
}
I want to use one form to insert into two different Microsoft sql tables. I tryed to use 2 inserts, but didnt work.
if (isset($_GET['submit'])) {
$sth = $connection->prepare("INSERT INTO DB.dbo.Fehler (QualiID, TestaufstellungID, ModulinfoID, failAfter, Datum, Verbleib, DUTNr) VALUES ($QualiID, $TestaufstellungID,$ModulinfoID,'$failAfter','$Datum','$Verbleib','$DUTNr')");
echo "INSERT INTO DB.dbo.Fehler (QualiID, TestaufstellungID, ModulinfoID, failAfter, Datum, Verbleib, DUTNr) VALUES ($QualiID, $TestaufstellungID,$ModulinfoID,'$failAfter',$Datum,'$Verbleib','$DUTNr')";
$sth->execute();
if($sth)
{
echo "";
}
else
{
echo sqlsrv_errors();
}
$MID = $connection->prepare("MAX(MID) as MID FROM DB.dbo.Fehler WHERE DB.dbo.Fehler.TestaufstellungID = '". $TestaufstellungID . "'");
$MID->execute();
$sth2 = $connection->prepare("INSERT INTO DB.dbo.Fehlerinfo (MID, Tester, Test, Ausfallbedingungen, Fehlerbeschreibung, Ersteller) VALUES ($MID, '$Tester','$Test','$Ausfallbedingungen','$Fehlerbeschreibung','$Ersteller')");
$sth2->execute();
To understand MID is the Primary key of table Fehler and ist the foreign key in the second table Fehlerinfo
Thats why i have the select work around to get the last MID and want to save it in a variable $MID to insert it into the second table.
Is there a smarter solution possible?
As I mentioned in the comments, generally the better way is to do the insert in one batch. This is very over simplified, however, should put you in the right direction. Normally you would likely be passing the values for the Foreign Table in a Table Value Parameter (due to the Many to One relationship) and would encapsulate the entire thing in a TRY...CATCH and possibly a stored procedure.
I can't write this in PHP, as my knowledge of it is rudimentary, but this should get you on the right path to understanding:
USE Sandbox;
--Couple of sample tables
CREATE TABLE dbo.PrimaryTable (SomeID int IDENTITY(1,1),
SomeString varchar(10),
CONSTRAINT PK_PTID PRIMARY KEY NONCLUSTERED (SomeID));
CREATE TABLE dbo.ForeignTable (AnotherID int IDENTITY(1,1),
ForeignID int,
AnotherString varchar(10),
CONSTRAINT PK_FTID PRIMARY KEY NONCLUSTERED(AnotherID),
CONSTRAINT FK_FTPT FOREIGN KEY (ForeignID)
REFERENCES dbo.PrimaryTable(SomeID));
GO
--single batch example
--Declare input parameters and give some values
--These would be the values coming from your application
DECLARE #SomeString varchar(10) = 'abc',
#AnotherString varchar(10) = 'def';
--Create a temp table or variable for the output of the ID
DECLARE #ID table (ID int);
--Insert the data and get the ID at the same time:
INSERT INTO dbo.PrimaryTable (SomeString)
OUTPUT inserted.SomeID
INTO #ID
SELECT #SomeString;
--#ID now has the inserted ID(s)
--Use it to insert into the other table
INSERT INTO dbo.ForeignTable (ForeignID,AnotherString)
SELECT ID,
#AnotherString
FROM #ID;
GO
--Check the data:
SELECT *
FROM dbo.PrimaryTable PT
JOIN dbo.ForeignTable FT ON PT.SomeID = FT.ForeignID;
GO
--Clean up
DROP TABLE dbo.ForeignTable;
DROP TABLE dbo.PrimaryTable;
As i mentioned the answer how it works for me fine atm.
if (isset($_GET['submit'])) {
$failInsert = ("INSERT INTO DB.dbo.Fehler (QualiID, TestaufstellungID, ModulinfoID, failAfter, Datum, Verbleib, DUTNr) VALUES ($QualiID, $TestaufstellungID,$ModulinfoID,'$failAfter','$Datum','$Verbleib','$DUTNr')");
$failInsert .= ("INSERT INTO DB.dbo.Fehlerinfo (MID, Tester, Test, Ausfallbedingungen, Fehlerbeschreibung, Ersteller) VALUES (NULL, '$Tester','$Test','$Ausfallbedingungen','$Fehlerbeschreibung','$Ersteller')");
$failInsert .= ("UPDATE DB.dbo.Fehlerinfo SET DB.dbo.Fehlerinfo.MID = i.MID FROM (SELECT MAX(MID)as MID FROM DB.dbo.Fehler) i WHERE DB.dbo.Fehlerinfo.TestID = ( SELECT MAX(TestID) as TestID FROM DB.dbo.Fehlerinfo)");
$sth = $connection->prepare($failInsert);
$sth->execute();
}
I currently have a script that runs every 5 minutes and selects the data from a table on server 1 and an identical table on server2. This is a workaround for replication, essentially, since we don't have that option currently.
The script is successful but I've realized that it misses records sometimes, for whatever reason. The current script selects all records from the destination table, stores the max primary key, selects all data from the source table and then inserts anything with a greater Primary key into the dest. table.
I'd like to modify the script slightly and instead of using max id, just say "if a row has an primary key that doesn't exist in the destination table, insert that row there."
Again these are cloned tables so the structure is the same and they both use AI Primary Keys.
Here's the current working script:
$latest_result = $conn2->query("SELECT MAX(`SESSIONID`) FROM
`ambition`.`session`");
$latest_row = $latest_result->fetch_row();
$latest_session_id = $latest_row[0];
//Select All rows from the source phone database
$source_data = mysqli_query($conn, "SELECT * FROM
`cdrdb`.`session` WHERE `SESSIONID` > $latest_session_id");
// Loop on the results
while($source = $source_data->fetch_assoc()) {
// Check if row exists in destination phone database
$row_exists = $conn2->query("SELECT SESSIONID FROM
ambition.session WHERE SESSIONID = '".$source['SESSIONID']."' ") or
die(mysqli_error($conn2));
//if query returns false, rows don't exist with that new ID.
if ($row_exists->num_rows == 0){
//Insert new rows into ambition.session
$stmt = $conn2->prepare("INSERT INTO ambition.session (SESSIONID,
SESSIONTYPE,CALLINGPARTYNO,FINALLYCALLEDPARTYNO,
DIALPLANNAME,TERMINATIONREASONCODE //etc. There are a lot of columns so I
ommitted the others
Is there a way I can slightly modify this to just insert what doesn't exist rather than relying on the MAX ID?
Or is there something here that would be a culprit as to why it's missing records?
You could use INSERT INTO SELECT and check if value is already in target:
INSERT INTO trg_table (cols)
SELECT cols
FROM src_table s
WHERE NOT EXISTS (SELECT 1 FROM trg_table t WHERE t.id = s.id);
After looking around on stackoverflow, I'm still having a little trouble understanding the one-to-many relationship in mysql. I have a request coming in from the user (form submission) which will be stored in one table. This is a dynamic form that lets the user add extra fields therefore those will be stored in a separate table. So in short, in my db design, there will be one table for the users with PRIMARY KEY AUTO INCREMENT and there will be another table for the hostnames PER user (multiple fields -array) and using a foreign key that references to the primary key in the user table. Sorry if this is long but trying to make this a good question.
Example:
User Table: (ONE)
1. John Doe, blah, 11-12-15
2. Sally Po, blah, 11-14-15
3. John Doe, blah, 11-15-15
(these are three separate requests)
(numbers are primary key auto incr.)
Host Name Table: (MANY)
1. www.johndoe.com
1. www.johndoe2.com
1. www.johndoe3.com
2. www.sallypo.com
2. www.sallypo2.com
(these numbers (foreign key) should match the primary key for each request)
Code (Leaving out the actual queries + pretty sure I shouln't be using last_id):
$sql = "CREATE TABLE IF NOT EXISTS userTable (
id int AUTO_INCREMENT,
firstName VARCHAR(30) NOT NULL,
date DATE NOT NULL,
PRIMARY KEY (id)
)";
//query
$sql = "CREATE TABLE IF NOT EXISTS hostNamesTable (
id int NOT NULL,
hostName VARCHAR(90) NOT NULL,
FOREIGN KEY (id) REFERENCES userTable(id)
)";
//query
$sql = "INSERT INTO userTable (firstName, date)
VALUES ('$firstName', '$date')";
//query
$last_id = mysqli_insert_id();
for($i = 0; $i < sizeof($hostName); $i++){
$sql = "INSERT INTO hostNamesTable (id, hostName)
VALUES ('$last_id', '$hostName[$i]')";
//query
}
What am I doing wrong? (is this the right way to go about it?)
note: I was trying to get the last_id of the user Table so that I can use it in the hostName table as the foreign key
EDIT: I'm using MySQLi with php
EDIT 2:
After the changes, this is the error I am getting now: Cannot add or update a child row: a foreign key constraint fails (d9832482827984hb28397429.hostNamesTable, CONSTRAINT hostNamesTable_ibfk_1 FOREIGN KEY (id) REFERENCES userTable (id))Error: INSERT INTO hostNamesTable (id, hostName, ) VALUES ('', 'secondhost.net')
--Looks like the $last_id isn't even being recorded?
EDIT 3: Started working. Not sure what it was but I think it was because of some type.
why dont you just add an extra column in the hostNames table which is called "ref_user" and contains the ID of the user you are reffering to? So you can use unique IDs in both tables.
Make a query like:
SELECT * FROM hostNames WHERE ref_user = (SELECT id FROM userTable WHERE <uniqueColumn> = <uniqueIdentifierOfUser>);
But the included request must return only one line from users.
try adding mysqli $link as a parameter in your mysqli_insert_id
$last_id = mysqli_insert_id($link);
i presume you have this somewhere in your code
$link = mysqli_connect("localhost", "mysql_user", "mysql_password", "mysql_db");
if this doesn't work, try using mysql LAST_INSERT_ID() function
$last_id = $mysqli->query("SELECT LAST_INSERT_ID() AS last_id")->fetch_object()->last_id;
I recently asked a question about writing to multiple tables: PHP/MySQL insert into multiple data tables on submit
I have now tried out this code and there are no errors produced in the actual code but the results I am getting are strange. When a user clicks register this 'insert.php' page is called and the code can be found below.
<?php
$username = $_POST["username"];
$password = $_POST["password"];
$institution = $_POST["institution"];
$conn = pg_connect("database connection information"); //in reality this has been filled
$result = pg_query($conn, "INSERT INTO institutions (i_id, name) VALUES (null, '$institution') RETURNING i_id");
$insert_row = pg_fetch_row($result);
$insti_id = $insert_row[0];
// INSTITUTION SAVED AND HAS ITS OWN ID BUT NO MEMBER OF STAFF ID
$resultTwo = pg_query($conn, "INSERT INTO staff VALUES (NULL, '$username', '$password', '$insti_id'");
$insert_rowTwo = pg_fetch_row($resultTwo);
$user_id = $insert_rowTwo[0];
// USER SAVED WITH OWN ID AND COMPANY ID
// ASSIGN AN INSTITUTION TO A STAFF MEMBER IF THE STAFF'S $company_id MATCHES THAT OF THE
// INSTITUION IN QUESTION
$update = pg_query($conn, "UPDATE institutions SET u_id = '$user_id' WHERE i_id = '$insti_id'");
pg_close($conn);
?>
What the result of this is just the browser waiting for a server response but there it just constantly waits. Almost like an infinite loop I'm assuming. There are no current errors produced so I think it may be down to a logic error. Any ideas?
The errors:
RETURNING clause is missing in the second INSERT statement.
Provide an explicit list of columns for your second INSERT statement, too.
Don't supply NULL in the INSERT statements if you want the column default (serial columns?) to kick in. Use the keyword DEFAULT or just don't mention the column at all.
The better solution:
Use data-moidifying CTE, available since PostgreSQL 9.1 to do it all in one statement and save a overhead and round trips to the server. (MySQL knows nothing of the sort, not even plain CTEs).
Also, skip the UPDATE by re-modelling the logic. Retrieve one id with nextval(), and make do with just two INSERT statements.
Assuming this data model (you should have supplied that in your question):
CREATE TABLE institutions(i_id serial, name text, u_id int);
CREATE TABLE staff(user_id serial, username text, password text, i_id int);
This one query does it all:
WITH x AS (
INSERT INTO staff(username, password, i_id) -- provide column list
VALUES ('$username', '$password', nextval('institutions_i_id_seq'))
RETURNING user_id, i_id
)
INSERT INTO institutions (i_id, u_id, name)
SELECT x.i_id, x.user_id, '$institution'
FROM x
RETURNING u_id, i_id; -- if you need the values back, else you are done
Data model
You might think about changing your data model to a classical n:m relationship.
Would include these tables and primary keys:
staff (u_id serial PRIMARY KEY, ...)
institution (i_id serial PRIMARY KEY, ...)
institution_staff (i_id, u_id, ..., PRIMARY KEY(i_id, u_id)) -- implements n:m
You can always define institution_staff.i_id UNIQUE, if a user can only belong to one institution.