In my Document table I have:
id (auto int index),
user_id (P.key and links to other table),
Doc_Name,
abstract
When I use the code below, it just inserts another row so I have two user_id's the same when it should have updated. Obviously the id just carries on in number as it is auto int and I'm not sure if this has something do with why it won't work.
$the_query = sprintf("INSERT INTO `document` (`user_id`,`Doc_Name`,`abstract`)
VALUES ('%d','%s','%s')",'$user_id', '$Doc_Name', '$abstract')
ON DUPLICATE KEY UPDATE
user_id=user_id+'$user_id',
Doc_Name=Doc_Name+'$Doc_Name',
abstract=abstract+'$abstract' "
);
Cargo cult programming? Using sprintf() without any % placeholders is just.... WRONG. As well, why the addition on the updated fields?
MySQL uses concat() for concatenation. + is purely a mathematical operation. Doing 'a' + 'a' does NOT give you aa, you'll get 0.
if user_id is unique field in your table, your query should look like this:
$query = sprintf("
INSERT INTO
`document`(`user_id`, `Doc_Name`, `abstract`)
VALUES
('%s', '%s', '%s')
ON DUPLICATE KEY UPDATE
`Doc_Name` = VALUES(`Doc_Name`),
`abstract` = VALUES(`abstract`)
", $user_id, $Doc_Name, $abstract);
Related
I want to use one form to insert into two different Microsoft sql tables. I tryed to use 2 inserts, but didnt work.
if (isset($_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;
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.
I am inserting multiple rows using one query and, obviously, the ID column auto increments each row. I want to create another ID column and have the ID remain the same for all rows inserted during the query. So if I insert 10 rows during one query, I want all 10 rows to have the id "1". How can this be done? Thanks for any help
If I understood your question correctly, you want to supply an ID for the specific group of INSERT statements.
Assumming you have this schema
CREATE TABLE TableName
(
RecordID INT AUTO_INCREMENT PRIMARY KEY,
OtherColumn VARCHAR(25) NOT NULL,
GroupID INT NOT NULL
)
You can have two statements for this:
1.) Getting the last GroupID and increment it by 1.
SELECT COALESCE(MAX(GroupID), 0) + 1 AS newGroupID FROM TableName
2.) once you have executed it, store the value in a variable. Use this variable for all the insert statement,
$groupID = row['newGroupID'];
$insert1 = "INSERT INTO TableName(OtherColumn, GroupID) VALUES ('a', $groupID)";
$insert2 = "INSERT INTO TableName(OtherColumn, GroupID) VALUES ('b', $groupID)";
$insert3 = "INSERT INTO TableName(OtherColumn, GroupID) VALUES ('c', $groupID)";
UPDATE 1
SQLFiddle Demo
I have a simple INSERT statement which looks like this...
mysql_query("INSERT INTO comments (`user_id`, `profile_id`, `comment`) VALUES ('{$_SESSION['user_id']}', ('$problemID'), ('$comment'))") or die(mysql_error());
Everything is being inserted fine apart from the $problemID variable. In the MySql table it is just returning a 0. The table is set up to receive integers up to 11 characters.
The variable itself is set on a different page but is retrieved using this...
$problemID = intval( $_GET["problem"]);
If I echo the $problemID I get the correct number so I'm unsure as to why it won't just insert this number into my table. Any pointers would be great.
Make sure that your comment is more clearly sanitized; Try something like this:
mysql_query( sprintf(
"INSERT INTO
comments (`user_id`, `profile_id`, `comment`)
VALUES
(%s, %s, '%s')",
intval( $_SESSION['user_id'] ),
intval( $problemID ),
mysql_real_escape_string( $comment )
)) or die( mysql_error() );
Just to be thorough, make sure that your table has a separate primary index (aka entry ID) with auto-increment tacked on. It could be that your MySQL insertion is working fine, however, the receiving table doesn't know that it should keep appending entries.
My hunch is that your INSERT query is referring to the wrong column in your comments table, as you have the following:
INSERT INTO comments (`user_id`, `profile_id`, `comment`)
but you're referring to a variable named $problemID, so my guess is that you meant something like this:
INSERT INTO comments (`user_id`, `problem_id`, `comment`)
Perhaps you copied and pasted the query code but forgot to change the column name in the projection?
Remove the brackets and try to add the variables rather than including them into the string.
mysql_query("INSERT INTO comments (`user_id`, `profile_id`, `comment`)
VALUES ('".$_SESSION['user_id']."', ".$problemID.", '".mysql_real_escape_string($comment)."')") or die(mysql_error());