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','$CID',NULL,NULL,NULL)
WHERE `patient_id` = patient_id ;
And finally what is patient_id? Is it variable? If not IT MUST BE. Don't give same names to different things.
Related
I want to use one form to insert into two different Microsoft sql tables. I tryed to use 2 inserts, but didnt work.
if (isset($_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();
}
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 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 = $topic_fk
);
Use an INSERT IGNORE statement. This will insert any rows that do not violate the unique key, and ignore the ones that do.
Let me explain what I need and canot get :(
I have to DB one i main the other is just getting part of data from the firs one.
This is my code:
foreach($id_product_array AS $id_product) {
$resultf = mysql_query("SELECT * FROM db1_available_product WHERE id_product='".$id_product."'");
while($rowi = mysql_fetch_array($resultf)) {
$aa1=$rowi['id_product'];
$aa2=$rowi['date'];
$aa3=$rowi['available'];
$aa4=$rowi['published'];
mysql_query("INSERT INTO aa_bb.db2_available_product (`id_product`, `date`, `available`, `published`) VALUES ('".$aa1."','".$aa2."', '".$aa3."', '".$aa4."') ON DUPLICATE KEY UPDATE `id_product` = '".$aa1."', `date` = '".$aa2."', `available` = '".$aa3."', `published` = '".$aa4."'");
}
The problem is that this multiples the record in DB2 so I am now in millions!!!
Its set up as cron job on 1h basis.
What I need is ether it checks what is existing and don't touch it or if need on update or insert.
The other solution would be to delete the whole table in DB2 then to insert a fresh one from DB1
You can simplify your query like so:
INSERT INTO tbl2 (column1, column2)
SELECT column1, column2 FROM tbl1
ON DUPLICATE ...
See the documentation
You are looking for MySQL's proprietary REPLACE command. It has the same syntax as a regular INSERT, but it checks for duplicate primary key before inserting, and if it is found it will do an UPDATE instead:
REPLACE works exactly like INSERT, except that if an old row in the
table has the same value as a new row for a PRIMARY KEY or a UNIQUE
index, the old row is deleted before the new row is inserted.
Of course you will have to define a unique PK/index on your table that allows this functionality to work.
here is an update!
I solved the problem :)
Thanks s to Niels because he made me rethink my strategy so the solution was simple.
In the DB1 and DB2 there is and ID filed
A added
$aa5=$rowi['id'];
so that made ON DUPLICATE KEY UPDATE work correctly!
foreach($id_product_array AS $id_product) {
$resultf = mysql_query("SELECT * FROM db1_available_product WHERE id_product='".$id_product."'");
while($rowi = mysql_fetch_array($resultf)) {
$aa5=$rowi['id'];
$aa1=$rowi['id_product'];
$aa2=$rowi['date'];
$aa3=$rowi['available'];
$aa4=$rowi['published'];
mysql_query("INSERT INTO aa_bb.db2_available_product (`id`,`id_product`, `date`, `available`, `published`) VALUES ('".$aa5."','".$aa1."','".$aa2."', '".$aa3."', '".$aa4."') ON DUPLICATE KEY UPDATE `id` = '".$aa5."',`id_product` = '".$aa1."', `date` = '".$aa2."', `available` = '".$aa3."', `published` = '".$aa4."'");
}
and it seams that it is working OK!
:)
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 .