I'm using a for loop to add a specific number of rows to the table Beds as placeholders. The code doesn't throw any error, but only the last numbered bed appears in the database. I.e.: if you try to add 10 beds, only bed 10 gets added.
Here is the for loop I am using:
$roomName = mysqli_real_escape_string($con, $_POST['roomName']);
$numberBeds = mysqli_real_escape_string($con, $_POST['numberBeds']);
for($x=1; $x <= $numberBeds; $x++) {
$sql = "INSERT INTO Beds (roomName, bedNumber, patientID) VALUES('$roomName','$x', NULL)";
}
And here is the accompanying table:
DROP TABLE IF EXISTS `ChildrensHospital`.`Beds` ;
CREATE TABLE IF NOT EXISTS `ChildrensHospital`.`Beds` (
`roomID` INT(20) NOT NULL AUTO_INCREMENT,
`bedNumber` INT(10) NOT NULL,
`patientID` INT(20) NULL,
PRIMARY KEY (`roomID`, `bedNumber`),
INDEX `fk_Beds_PatientPersonalInformation1_idx` (`patientID` ASC),
CONSTRAINT `fk_Beds_PatientPersonalInformation1`
FOREIGN KEY (`patientID`)
REFERENCES `ChildrensHospital`.`PatientPersonalInformation` (`patientID`)
ON DELETE NO ACTION
ON UPDATE NO ACTION)
ENGINE = InnoDB;
I'm not sure if it's a syntax error in my PHP code, or something about the table itself.
Any help would be appreciated.
The problem with your code is that you are not executing your query. Try it like this:
for($x=1; $x <= $numberBeds; $x++){
$sql = "INSERT INTO Beds (roomName, bedNumber, patientID) VALUES('$roomName','$x', NULL)";
mysqli_query($con,$sql);
}
I guess that you are only executing your query after the loop and thus, of course only the last query gets executed instead of each.
Related
I want to use one form to insert into two different Microsoft sql tables. I tryed to use 2 inserts, but didnt work.
if (isset($_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 was trying to find out what was wrong with my code.
This is the error I'm recieving
"Cannot add or update a child row: a foreign key constraint fails "
This is my code
<?php
$sql = "INSERT INTO stasjon (navn) VALUES ('skogen', 'voksenlia') ";
$resultat = $kobling->query ($sql);
$sql ="SELECT * FROM stasjon WHERE navn = ('skogen')";
$resultat = $kobling->query ($sql);
while ($rad = $resultat->fetch_assoc()) {
$stasjon_id = $rad['stasjon_id'];
}
$sql = "INSERT INTO linjestasjon (linje_nr, stasjon_id) VALUES ('1','$stasjon_id')";
$resultat = $kobling->query ($sql);
if($kobling->query($sql)) {
echo "Spoerringen $sql ble gjennomfoert.";
} else {
echo "Noe gikk galt med spoerringen $sql ($kobling->error).";
?>
Some of it is in Norwegian because That's the language of the database I'm making. I was trying to add values to two different tables (that had stasjon_id as a foreign key) Thanks in advance
Foreign Key constraints checks in the value that you are inserting or updating to a particular filed exists in some other filed of another table. Suppose I have 2 tables as follows
CREATE TABLE TableA
(
SeqNo INT PRIMARY KEY,
Name VARCHAR(500
)
CREATE TABLE TableB
(
SeqNo INT NULL FOREIGN KEY REFERENCES TableA(SeqNo),
Name VARCHAR(50)
)
So when you insert a new record to the Tableb.SeqNo filed, The value should either be NULL or some value that exists in the TableA.SeqNo.
So Before inserting the values make sure that you are inserting the value that satisfy your foreign key constraint
I have a question about the update function in mysqli.
For school, I'm trying to create a click counter for my website which counts how many times a user has visited a certain page.
So far I've come up with this:
<?php
/*
* ToDo: Check why number of clicks goes back to two when completely
* refreshing page.
*
*/
include("init.php");
session_start();
//Count variable
$clicks = 0;
//Query for checking if there are any entry's in the database
$query = "SELECT * FROM `beoordelingen`.`clickcounter` WHERE `game_id`={$id}";
$result = $conn->query($query);
//If query returns false
if (!mysqli_num_rows($result)) {
//Create entry in database
$insert = "INSERT INTO `beoordelingen`.`clickcounter` (`ID`, `game_id`, `clicks`) VALUES (NULL, '1', '1');";
$createEntry = $conn->query($insert);
}
//If query returns true
else {
//Setting the number of clicks equal to $clicks
while ($data = $result->fetch_assoc()) {
$clicks = $data['clicks'];
}
//Insert new number into database
$sql="insert into `clickcounter` set `clicks`='{$clicks}', `game_id`='{$id}'
on duplicate key update
`clicks`=`clicks`+1;";
$insertInto = $conn->query($sql);
//Echo current number of clicks
echo $clicks;
}
?>
The actual problem is that my update statement doesn't seem to work properly. If anyone would be able to spot why it doesn't work I'd be very happy.
The database is as following;
Beoordelingen <- Database
clickcounter <- Table which has the following three columns:
1. ID
2. game_id
3. clicks
The scripts does add an entry into the databse with click count 2. So when I reload the page it says 2. And when refreshing it counts up, but doesn't update the table.
Thanks! If anything is unclear please ask me!
Theoretically you should be able to do all of it in one query if game_id is unique.
Given the following table structure the sql query below will insert if the relevant record does not exists and then update if it does.
create table `clickcounter` (
`id` int(10) unsigned not null auto_increment,
`game_id` int(10) unsigned not null default '0',
`clicks` int(10) unsigned not null default '0',
primary key (`id`),
unique index `game_id` (`game_id`)
)
engine=innodb;
The trick is setting the indices on your table correctly ~ initially you don't know the value of the ID and I would guess that is an auto increment primary key? So, set a unique key on game_id...I hope it helps!
/* Could even change `clicks`='{$clicks}' to `clicks`=1 in initial insert */
$sql="insert into `clickcounter` set `clicks`='{$clicks}', `game_id`='{$id}'
on duplicate key update
`clicks`=`clicks`+1;";
<?php
include("init.php");
session_start();
/* Where / how is "$id" defined? */
/* insert new record / update existing */
$sql="insert into `clickcounter` set `clicks`=1, `game_id`='{$id}'
on duplicate key update
`clicks`=`clicks`+1;";
$result = $conn->query( $sql );
/* retrieve the number of clicks */
$sql="select `clicks` from `clickcounter` where `game_id`='{$id}';";
$result = $conn->query( $sql );
while( $rs=$result->fetch_object() ) $clicks=intval( $rs->clicks );
echo 'Total clicks: '.$clicks;
?>
I have a form to edit a record (specimen). On the form is a multiple select list which contains records from a table (topic). This select list shows topics as selected that exist for the specimen (as identified in the specimen_topic lookup table) as well as those that can be added to the specimen (from the topic table).
I want to be able to add topics not selected in the list to the lookup table where the topic_fk does not already exist for the specimen_fk:
CREATE TABLE IF NOT EXISTS `specimen_topic_lookup` (
`specimen_topic_lookup_pk` int(6) NOT NULL AUTO_INCREMENT,
`specimen_fk` int(6) NOT NULL,
`topic_fk` int(3) NOT NULL,
PRIMARY KEY (`specimen_topic_lookup_pk`)
) ENGINE=MyISAM DEFAULT CHARSET=utf8 COLLATE=utf8_swedish_ci AUTO_INCREMENT=8 ;
Any ideas how I can do this?
UPDATE
I have made the fields specimen_fk and topic_fk UNIQUE. Using the code below, only one record is created in specimen_table lookup, when two records should have been created (before making the fields UNIQUE, two records were created OK...). I assume this is because $specimen_pk is the same value for each insert.
foreach($topics as $topic){
$query_topics = "INSERT IGNORE INTO specimen_topic_lookup(specimen_fk, topic_fk)
VALUES ('$specimen_pk', '$topic')";
$result_topics = mysql_query($query_topics, $connection) or die(mysql_error());
}
Looks like having UNIQUE is stopping having a record made with the same value (which is at least what I expected...)
THIS WORKS
Without having to make specimen_fk OR topic_fk UNIQUE...
foreach($topics as $topic){
$query_topics = "INSERT INTO specimen_topic_lookup(specimen_fk, topic_fk)
SELECT '$specimen_pk', '$topic'
FROM DUAL
WHERE NOT EXISTS (SELECT 1
FROM specimen_topic_lookup
WHERE specimen_fk = '$specimen_pk' AND topic_fk = '$topic')";
$result_topics = mysql_query($query_topics, $connection) or die(mysql_error());
Create a unique index on the table and use insert ignore or on duplicate key update:
create unique index specimen_topic_lookup(specimen_fk, topic_fk);
insert ignore into specimen_topic_lookup(specimen_fk, topic_fk)
select $speciment_fk, $topic_fk;
Or, alternatively, you can just do the following without the unique index:
insert into specimen_topic_lookup(specifmen_fk, topic_fk)
select $speciment_fk, $topic_fk
from dual
where not exists (select 1
from specimen_topic_lookup
where specimen_fk = $specimen_fk and topic_fk = $topic_fk
);
Use an INSERT IGNORE statement. This will insert any rows that do not violate the unique key, and ignore the ones that do.
create table cmu_patient
( patient_id character varying(13) NOT NULL,
patient_hn character varying(7),
patient_fname character varying(50),
patient_lname character varying(50),
home_id integer,
CONSTRAINT cmu_patient_pkey PRIMARY KEY (patient_id),
CONSTRAINT Fk_home FOREIGN KEY(home_id)
REFERENCES cmu_home(home_id)
);
create table cmu_treatment
( treatment_id serial NOT NULL,
treatment_date date,
treatment_time time without time zone,
treatment_typecome character varying(100),
treatment_detail text,
patient_id character varying(13),
appointment_id character varying(5),
transfer_id character varying(5),
res_users_id integer,
CONSTRAINT cmu_treatment_pkey PRIMARY KEY (treatment_id),
CONSTRAINT Fk_patient FOREIGN KEY(patient_id)
REFERENCES cmu_patient(patient_id),
CONSTRAINT Fk_user_id FOREIGN KEY(res_users_id)
REFERENCES res_users(id)
);
$treatment_date = $GET_[...];
$treatment_time = $GET_[...];
$treatment_typecome = $GET_[...];
$treatment_note = $GET_[...];
$CID = $GET_[...];
this code -------- it's incorrect
INSERT INTO cmu_treatment(treatment_id, treatment_date, treatment_time,
treatment_typecome, treatment_detail, patient_id, appointment_id,transfer_id, res_users_id)
VALUES(NULL,'".$tratment_date."','".$treatment_time."','".
$treatment_typecome."','".$treatment_note."','".$CID."',NULL,NULL,NULL)
WHERE cmu_patient.patient_id = cmu_treatment.patient_id ;
i think that's wrong
i don't know if i want to write insert data into table with where cause i should write sql ?
thank :)
I suspect what you really want is an update, to change existing values in an existing record:
update cmu_treatment
set treatment_date = $treatment_date,
treatment_time = $treatment_time,
treatment_detail = $treatment_typecome,
treatment_note = $treatment_note
where patient_id = $CID;
(I'm leaving out the NULL values on the assumption that those shouldn't really change.)
If you do indeed want a new record, you can do:
INSERT INTO cmu_treatment(treatment_id, treatment_date, treatment_time,
treatment_typecome, treatment_detail, patient_id, appointment_id,
transfer_id, res_users_id
)
select NULL,'".$tratment_date."', '".$treatment_time."','".
$treatment_typecome."','".$treatment_note."','".$CID."', NULL, NULL, NULL;
You can write an INSERT statement populating target table with a SELECT statement. In the SELECT statement you can use WHERE condition.
So instead this query:
INSERT INTO table VALUES (....)
You must write:
INSERT INTO table
SELECT fields
FROM anothertable
WHERE condition
In your case, I think you must use an INSERT without WHERE condition if you want to insert only a row in your treatment table.
Tell me if you want to know further info
EDIT After comment
IMHO your statement must be:
INSERT INTO cmu_treatment
(treatment_id, treatment_date, treatment_time,
treatment_typecome, treatment_detail, patient_id, appointment_id,
transfer_id, res_users_id)
VALUES
(NULL,'".$tratment_date."','".$treatment_time."',
'".$treatment_typecome."','".$treatment_note."','".$CID."',NULL,NULL,NULL)
INSERT INTO `cmu_treatment`(`treatment_id`, `treatment_date`, `treatment_time`,
`treatment_typecome`, `treatment_detail`, `patient_id`, `appointment_id`,`transfer_id`, `res_users_id`)
VALUES(NULL,'".$tratment_date."','".$treatment_time."','".
$treatment_typecome."','".$treatment_note."','".$CID."',NULL,NULL,NULL)
WHERE `cmu_patient.patient_id` = `cmu_treatment.patient_id` ;
And you don't need (table name).(column).
Is this Inside "" ? If yes then you don't need '".$tratment_date."' you can use only '' so your code will look like this.
INSERT INTO cmu_treatment(treatment_id, treatment_date, treatment_time,
treatment_typecome, treatment_detail, patient_id, appointment_id,transfer_id, res_users_id)
VALUES(NULL,'$tratment_date','$treatment_time','
$treatment_typecome','$treatment_note','$CID',NULL,NULL,NULL)
WHERE `patient_id` = patient_id ;
And finally what is patient_id? Is it variable? If not IT MUST BE. Don't give same names to different things.