I have this PHP/MySQL script which adds a comment into my DB:
$SQL = "INSERT INTO blog_comments (article_id, author_name, comment, path, posted, status) ";
$SQL .= "VALUES (:article_id, :name, :comment, :next_path, Now(), 'Live');";
$STH = $DBH->prepare($SQL);
$STH->bindParam(':article_id', $article_id);
$STH->bindParam(':name', $name);
$STH->bindParam(':comment', $comment);
$STH->bindParam(':next_path', $next_path);
$STH->execute();
Is there any way to modify this so that it doesn't insert the same [author_name], [article_id] and [comment] into this table? I know it's possible for one column by adding UNIQUE to my table, but not sure about multiple columns.
you can add a UNIQUE constraint for multiple columns
CREATE TABLE
(
.....
CONSTRAINT tab_uq UNIQUE (author_name, comment)
)
or by altering the existing table,
ALTER TABLE myTable ADD UNIQUE (author_name, comment);
call those columns as composite primary key..
that way, they will have a unique entry.
You can make a unique key of multiple fields
I guess you could check the db for a result before inserting data
you can also check INSERT ... ON DUPLICATE KEY UPDATE dev.mysql.com
Related
I want to use one form to insert into two different Microsoft sql tables. I tryed to use 2 inserts, but didnt work.
if (isset($_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 am using a three table structure to deal with a many-to-many relationship. I have one table that has a list of people and another that has a list of items. Sometimes multiple people have the same item and sometimes multiples items are linked to the same person so I set up the following table structure:
CREATE TABLE people (
id int(11) NOT NULL PRIMARY KEY AUTO_INCREMENT,
fname varchar(128) NOT NULL,
lname varchar(128) NOT NULL,
);
CREATE TABLE items (
id int(11) NOT NULL PRIMARY KEY AUTO_INCREMENT,
name varchar(128) NOT NULL UNIQUE,
);
The UNIQUE prevents the item name from repeating.
CREATE TABLE people_items (
pid int(11) NOT NULL,
iid int(11) NOT NULL,
FOREIGN KEY (pid) REFERENCES people(id)
ON UPDATE CASCADE
ON DELETE CASCADE
FOREIGN KEY (iid) REFERENCES items(id)
ON UPDATE CASCADE
ON DELETE CASCADE
);
This allows me to link multiple items to multiple people and vice versa. It also allows me to delete unnecessary records from the intermediate table.
All works fine so long as a new item is entered, but if an existing item is entered, the intermediate table is not updated, even though the people table is. I also do not get any errors.
Items are a comma delimited text entry which are exploded and lower cased into $items.
First I insert any new items and retrieve the id:
for ($i=0;$i<count($items);$i++){
$sql="INSERT IGNORE INTO items (name) VALUES (?);";
$stmt=$conn->prepare($sql);
$stmt->bind_param('s',$items[$i]);
$stmt->execute();
$itemid=$stmt->insert_id;
If a new id is returned the following is executed:
if ($itemid){
$sql="INSERT INTO people_items (pid,iid) VALUES (?,?);";
$stmt=$conn->prepare($sql);
$stmt->bind_param('ii',$peopleid,$itemid);//the $peopleid is acquired the same way that the $itemid is acquired above
$stmt->execute();
}
Up to here, everything works just fine. At this point if an existing item is already in the items table, is where my intermediate table does not update. The people table however updates just fine, and the items table does not need to update as it already has the item in it.
Here is where I tried two different approaches to update the intermediate table.
First I kept the select and insert queries separate.
elseif(!$itemid){
$sql="SELECT id,name FROM items WHERE name=?;";
$stmt=$conn->prepare($sql);
$stmt->bind_param('s',$items[$i]);
$stmt->execute();
$stmt->store_result();
$stmt->bind_result($iid,$name);
$stmt->fetch();
$sql="INSERT INTO people_items (pid,iid) VALUES (?,?);";
$stmt=$conn->prepare($sql);
$stmt->bind_param('ii',$pid,$iid);
$stmt->execute();
}
Here is my alternative approach which also does not update the intermediate table:
elseif(!$itemid){
$sql="INSERT INTO people_items (pid, iid) SELECT id,name FROM items WHERE name IN (?);";
$stmt=$conn->prepare($sql);
$stmt->bind_param('s',$items[$i]);
$stmt->execute();
}
What am I doing wrong?
It's because you are using INSERT IGNORE, so if the item already exists, it won't insert anything, and you will get either 0 or the ID of the last successful insert as the ID (per the docs). So what you need to check is the affected number of rows. If it's 0, then it was ignored and you can SELECT it from the DB. If it's not 0, then you can safely insert the last ID.
for ($i=0;$i<count($items);$i++){
$sql="INSERT IGNORE INTO items (name) VALUES (?);";
$stmt=$conn->prepare($sql);
$stmt->bind_param('s',$items[$i]);
$stmt->execute();
$itemid=$stmt->insert_id;
$affected_rows = $stmt->affected_rows;
if ($affected_rows !== 0) {
// insert
}
else {
// select and insert
}
}
Sidenote:
The way you are inserting the items in a loop is terrible for performance, especially if the DB is not in the same host as the server (think of going to the supermarket: do you go N times and get each item by itself, or do you just go once and get your N items?). What you should be doing, is inserting all of the elements in one query. Clearly it's not that clear cut in your case since you the IDs and what not, but you can read about this here to get started.
Sorry I didn't get to this sooner, but I have figured it out and was busy with work in the meantime. I figure I'd provide the answer in case anyone has the same issue in the future. Anyway, the issue was with my SQL query.
Where I had this:
elseif(!$itemid){
$sql="INSERT INTO people_items (pid, iid) SELECT id,name FROM items WHERE name IN (?);";
$stmt=$conn->prepare($sql);
$stmt->bind_param('s',$items[$i]);
$stmt->execute();
}
I should have had this:
elseif(!$itemid){
$sql="INSERT INTO people_items (pid,iid) VALUES (?,(SELECT id FROM items WHERE name IN (?)));";
$stmt=$conn->prepare($sql);
$stmt->bind_param('ss',$peopleid,$items[$i]);
$stmt->execute();
}
I have a database of Users and another table for Teachers. Teachers have all the properties as a user but also an e-mail address. When inserting into the DB how can I insert the info, ensuring that the ID is the same for both?
the ID currently is on automatic incrament.
this is what I have at the moment:
$sqlQuery="INSERT INTO user(firstName,lastName,DOB,title,password,classRoomID)
VALUES('$myFirstName','$myLastName','$myDOB','$myTitle','$newPassword','$myClassRoom')";
$result=mysql_query($sqlQuery);
$sqlQuery = "INSERT INTO teacher(email) VALUES ('$myEmail')";
$result=mysql_query($sqlQuery);
thank you!
use MYSQL function LAST_INSERT_ID()
OR php mysql http://ro1.php.net/manual/en/function.mysql-insert-id.php
why to use separate table for teachers. instead, you can have email field with in user table and additional field with flag (T ( for teacher) and U (for user). Default can be a U. This have following Pros.
Will Not increase table size as email would be varchar
Remove extra overhead of maintaining two tables.
Same Id can be used
If you want to have that as separate table then answer you selected is good one but make sure last insert id is called in same connection call.
After the first insert, fetch the last inserted id:
$last_id = mysqli_insert_id(); // or mysql_insert_id() if you're using old code
Or you could expand your second query and use mysql's integrated LAST_INSERT_ID() function:
$sqlQuery = "INSERT INTO teacher(id, email) VALUES ((SELECT LAST_INSERT_ID()), '$myEmail')";
Try this:
$sqlQuery="INSERT INTO user(firstName,lastName,DOB,title,password,classRoomID)
VALUES('$myFirstName','$myLastName','$myDOB','$myTitle','$newPassword','$myClassRoom')";
$result=mysql_query($sqlQuery);
$id = mysql_insert_id();
$sqlQuery = "INSERT INTO teacher(id, email) VALUES (' $id ','$myEmail')";
$result=mysql_query($sqlQuery);
Insert data into two tables & using the same ID
First method
$sqlQuery1 = "INSERT INTO user(firstName,lastName,DOB,title,password,classRoomID) VALUES('$myFirstName','$myLastName','$myDOB','$myTitle','$newPassword','$myClassRoom')";
$result1 = mysqli_query($conn, $sqlQuery1);
$lastID = mysqli_insert_id($conn);
$sqlQuery2 = "INSERT INTO teacher(email, lastID) VALUES ('$myEmail', 'lastID')";
$result2 = mysqli_query($conn, $sqlQuery2);
If the first method not work then this is the second method for you
$sqlQuery1 = "INSERT INTO user(firstName,lastName,DOB,title,password,classRoomID) VALUES('$myFirstName','$myLastName','$myDOB','$myTitle','$newPassword','$myClassRoom')";
$result1 = mysqli_query($sqlQuery1);
$sqlQuery2 = "INSERT INTO teacher(email) VALUES ('$myEmail')";
$result2 = mysqli_query($sqlQuery2);
You can set the Foreign Key in your database table (phpMyAdmin/ MySQL Workbench) to let the Foreign Key follow the Primary Key (ID). Then the data after insert will auto-follow the Primary Key ID.
Example here,
Teachers table set ID - Primary Key
Users table set UserID - Foreign Key (will follow the Teachers table ID)
if you're using MySQL WorkBench, you can refer to this link to set a foreign key.
https://dev.mysql.com/doc/workbench/en/wb-table-editor-foreign-keys-tab.html
Hope I can help any of you.
Though you can use the LAST_INSERT_ID() function in order to get the last insert id, the best approach in this case is to create a column reference to user id table.
teacher
id | user_id | email
So the teacher.id could be anyting, but the user_id column is the real reference to user table.
If you use InnoDB table, you can make the database consistent using Foreign keys
You Should Use A Transaction In MySQL. First insert In One Table And GET LAST_INSERT_ID().
Insert LAST_INSERT_ID() In Second Table.
$sqlQuery="INSERT INTO user(firstName,lastName,DOB,title,password,classRoomID)
VALUES('$myFirstName','$myLastName','$myDOB','$myTitle','$newPassword','$myClassRoom')";
$sqlQuery = "INSERT INTO teacher(LAST_INSERT_ID(), email) VALUES (' $id ','$myEmail')";
$result=mysql_query($sqlQuery);
I have two tables cw_users and ref_users, both have a column named id.
I'm using ISAM so can't use a foreign key.
So now I wanted to insert id from cw_users into ref_users if it didn't exist.
This is what I did, but didn't help:
$id = $_SESSION['id'];
$ref_code=md5($id);
mysql_query("INSERT INTO ref_users (id) VALUES ('$id') WHERE NOT EXISTS (SELECT FROM cw_users where id='$id')");
The correct syntax is INSERT IGNORE INTO
INSERT IGNORE INTO ref_users (id)
VALUES ('$id')
It will insert if the value does not exist, and ignore the statement if it does.
Note that this will only work if id is the Primary Key
EDIT: It seems from your comments that you would be much better off using ON DUPLICATE KEY UPDATE.
Try this query
INSERT INTO ref_users(id, ref_code)
VALUES ('$id', '$ref_code')
ON DUPLICATE KEY UPDATE
ref_code = '$ref_code'
...WHERE NOT EXISTS (SELECT id FROM ...
why not run normal insert, it will fail if row exists?
Your query is
"INSERT INTO ref_users (id) VALUES ('$id') WHERE NOT EXISTS (SELECT FROM cw_users where id='$id')"
You can't use were case in insert query . you can use normal insert .
Instead of finding that a row exists where 'bookName' equals 'bookName' and just updating, it creates a brand new row. What is wrong with my command? thanks!
$query = mysql_query(
"INSERT INTO custombookinfo (userId, sendToAddress, work, caseStudies, url, entryPoint, date, bookName)
VALUES ('$userId', '$emailAddress', '$work', '$caseStudies', '$url', '$entryPoint', '$date', '$bookName')
ON DUPLICATE KEY UPDATE bookName = 'bookName'"
);
the INSERT query is correct but I think you have failed to define UNIQUE constraint on column bookName. Execute the following statement:
ALTER TABLE custombookinfo ADD CONSTRAINT tb_uq UNIQUE (bookName);
Depending on what you want to be UNIQUE in the table, you should create an index (PRIMARY or UNIQUE) for whether user_id, or bookName.
More details about INSERT ... ON DUPLICATE KEY UPDATE http://dev.mysql.com/doc/refman/5.0/en/insert-on-duplicate.html