Check if exists, if so, update by 1++, if not insert - php

Hey guys quick question, I currently have an insert statement
$query= "INSERT into new_mail VALUES ('$to1', '0')"; where fields are username, and message_number
Currently what I would do to check if the entry exists, is do a select query then check the number of rows with mysql_num_rows (php). If rows==1 then I get the current message_number and set it equal to
$row['message_number']+1.
Then I update that entry with another query.
Is there an easier way to do all this in just mysql with just one query (check if exists, if not insert, if so update message_number, increase by 1)?

Depending on how your table is structured, you may be able to use the ON DUPLICATE KEY UPDATE (link to the MySQL manual) feature of INSERT:
INSERT into new_mail VALUES ('$to1', '0') ON DUPLICATE KEY UPDATE message_number=message_number+1

Use INSERT...ON DUPLICATE KEY UPDATE. The MySQL manual has an example which does almost exactly what you need:
INSERT INTO table (a,b,c) VALUES (1,2,3)
ON DUPLICATE KEY UPDATE c=c+1;
To make this work you need to add a UNIQUE index on the column that you use to check for duplicates. There is one important warning though:
In general, you should try to avoid using an ON DUPLICATE KEY clause on tables with multiple unique indexes.

Got a little confused by your question and your table structures but I think you want something like this.
INSERT INTO new_mail (username, message_number)
VALUES ($username, $message_number)
ON DUPLICATE KEY UPDATE message_number=message_number + 1;
This is presuming username is your primary key (more likely something like userid). Hope this helps.

EDIT: The ON DUPLICATE KEY UPDATE answers are better, but you could do this (eludes the select query):
Assuming you're using the mysqli extenson:
$db = //Some construction of mysqli object;
$sql = 'UPDATE tablename SET RowValue = RowValue + 1 WHERE message_number = ?';
$updateStatement = $db->prepare($sql);
$updateStatement->bind_param('i', $message_number);
$message_number = //Set message number;
$updateStatement->execute();
if ($updateStatement->affectedRows == 0) {
$sql = 'INSERT INTO tablename (RowValue, message_number) VALUES (?, ?)';
$insertStatement = $db->prepare($sql);
$insertStatement->bind_param('ii', $rowValue, $messageNumber);
$rowValue = something;
$messageNumber = something;
$insertStatement->execute();
}

Related

INSERT ...ON DUPLICATE KEY UPDATE when key is not always known?

I'm trying to update a record if the key is known else I want to insert it and get the inserted id, currently I have:
if(isset($data['applicationId']))
{
//update
$sql="
UPDATE myTable SET data='jsonstring' WHERE id = {$data['applicationId']}
";
}
else
{
//insert and get id
$sql="
INSERT INTO myTable SET data='jsonstring'
";
}
Is it possible to simplify the above to one query using INSERT ...ON DUPLICATE KEY UPDATE even when the key is not always known ?
I've tried this:
INSERT INTO myTable
(
id,
data
)
VALUES
(
?, # <- I may not know this!!
'jsonstring'
)
ON DUPLICATE KEY UPDATE
data = 'jsonstring'
Thanks for any suggestions.
Yes, you can do that, assumed id is your primary key and auto_increment. You will have two different queries, one if you know the applicationId and one when you not knowing it.
The first, when you know it:
INSERT INTO myTable
(
id,
data
)
VALUES
(
1337, # <- insert id
'jsonstring'
)
ON DUPLICATE KEY UPDATE
data = 'jsonstring';
And the one if the applicationId is unknown:
INSERT INTO myTable
(
id,
data
)
VALUES
(
NULL, # <- This will cause mysql to use a auto_increment value
'jsonstring'
)
ON DUPLICATE KEY UPDATE
data = 'jsonstring';
So you can conclude this to:
$sql="INSERT INTO myTable
(
id,
data
)
VALUES
(" .
isset($data['applicationId']) ? $data['applicationId'] : 'NULL'
.",
'jsonstring'
)
ON DUPLICATE KEY UPDATE
data = 'jsonstring';
";
But be aware of How can I prevent SQL-injection in PHP?
Happy coding
Please forgive because your question is not 100% clear. However, the concept I can tell is that you want to be able to ask more than 1 query on 1 sql statement. That can be done with a multi-query command. However, if you want some of your data from a query placed in your next query I do not think it will work. Link provided for multi_query
http://php.net/manual/en/mysqli.quickstart.multiple-statement.php
First, Simple update query will run. If it runs successfully, it will not go to if condition and your ID will be the one which was used in updating.
And, if that ID is not available (means update query fails, $Query will be false), so pointer jumps to if condition and insert the query. Now, new Inserted ID we can get.
$ID=$data['applicationId'];
$Query=mysql_query("UPDATE myTable SET data='jsonstring' WHERE id='$ID' ");
if(!$Query)
{
$InsertQuery=mysql_query("INSERT INTO myTable SET data='jsonstring'");
$ID=mysql_insert_id();
}
So, $ID will be your ID.(either updated or currently inserted)

Inserting record if the ID is not in the database and update it elsewise

I'm trying to insert data into my database. I'm able to insert it using INSERT INTO. However I want to check, if the database already has the ID and if it has I want it to just update the records.
$check = $db->query("SELECT Fuid FROM users WHERE Fuid='$fuid_db'");
//IF the user is new
if(empty($check))
{
$insert = $db->query("INSERT INTO users (Fuid, Ffname, Femail) VALUES ('$fuid_db', '$ffname_db', '$femail_db'");
}
else
{
//ELSE update the information
$update = $db->query("UPDATE users SET Ffname='$ffname_db', Femail='$femail_db' WHERE Fuid='$fuid_db'");
}
How come this is not working? I'm using ERROR_REPORTING(E_ALL),
but I'm not getting any errors. It seems like the issue is the "checking" before inserting. Without the checking part it works fine but now it's not inserting anything.
The Problem
$db->query() returns an (maybe empty) mysqli_result object. So empty($check) evaluates to false (since empty($var) returns true only if $var is zero, false, an empty string, an empty array or null) and thus an UPDATE is performed, no matter if there is a corresponding id or not.
To check if there is a record you have to use $check->num_rows to get the number of records containing the id.
Better
The ON DUPLICATE KEY UPDATE command should work for you.
INSERT INTO users (Fuid, Ffname, Femail)
VALUES ('$fuid_db', '$ffname_db', '$femail_db')
ON DUPLICATE KEY UPDATE users
SET Ffname = '$ffname_db', Femail = '$femail_db'
This way you don't have to make a SELECT first, so you reduce the number of database calls.
Notice: This only works, if Fuid is set as a unique key (or primary key).
Security
The way you use the SQL commands is vulnerable to SQL injection. You should use prepared statements.
Using prepared statements, your code would look like this:
$stmt = $db->prepare("INSERT INTO users (Fuid, Ffname, Femail)
VALUES (?, ?, ?)
ON DUPLICATE KEY UPDATE users
SET Ffname = ?, Femail = ?");
$stmt->bind_param('issss',$fuid_db,$ffname_db, $femail_db,$ffname_db, $femail_db)
$stmt->execute();
First and foremost, for security reasons you should not insert value that way.
But if we ignore it for now, you can instead use ON DUPLICATE KEY UPDATE as below provided that Fuid is a primary key
INSERT INTO users (Fuid, Ffname, Femail) VALUES ('$fuid_db', '$ffname_db', '$femail_db')
ON DUPLICATE KEY UPDATE users SET Ffname = '$ffname_db', Femail = '$femail_db'

MySQL stops running queries after if statement

I've been stuck on this for a few hours now ...
Here's my code:
$SQLQuery1 = $db_info->prepare("SELECT COUNT(ID) FROM menusize WHERE typesize=:typesize");
$SQLQuery1->bindValue(':typesize',$_POST['typesize'],PDO::PARAM_STR);
$SQLQuery1->execute();
if($SQLQuery1->fetchColumn() > 0) {
$SQLQuery2 = $db_info->prepare("INSERT INTO menucatagorysize (menucatagory_ID,menusize_ID) VALUES (:catagoryid,(SELECT ID FROM menusize WHERE typesize=:typesize))");
$SQLQuery2->bindValue(':typesize',$_POST['typesize'],PDO::PARAM_STR);
$SQLQuery2->bindValue(':catagoryid',$_POST['catagoryid'],PDO::PARAM_STR);
$SQLQuery2->execute();
} else {
$SQLQuery2 = $db_info->prepare("INSERT INTO menusize (typesize) VALUES (:typesize);
SET #menusizeid=LAST_INSERT_ID();
INSERT INTO menucatagorysize (menusize_ID,menucatagory_ID) VALUES (#menusizeid,:catagoryid)");
$SQLQuery2->bindValue(':typesize',$_POST['typesize'],PDO::PARAM_STR);
$SQLQuery2->bindValue(':catagoryid',$_POST['catagoryid'],PDO::PARAM_STR);
$SQLQuery2->execute();
}
$SQLQuery3 = $db_info->prepare("SELECT DISTINCT(menuitem_ID) FROM menuprice WHERE menucatagory_ID=:catagoryid");
$SQLQuery3->bindValue(':catagoryid',$_POST['catagoryid'],PDO::PARAM_STR);
$SQLQuery3->execute();
$rows = $SQLQuery3->fetchAll(PDO::FETCH_ASSOC);
So, it will run through the if statement fine, running $SQLQuery1 and $SQLQuery2 (Which ever one is required) without any problems, errors or warnings. But, if it runs the else { part of the code, it will not run $SQLQuery3. Any thoughts?
Thanks :D
EDIT: Got it to work by doing $SQLQuery2=NULL in the else statement ... Sucks that I still cant figure out why it wouldnt work the original way.
It appears that you're trying to enforce a uniqueness constraint over the typesize column of your menusize table from within your application code. However, the database can do this for you—which will make your subsequent operations much simpler:
ALTER TABLE menusize ADD UNIQUE (typesize)
Now, one can simply attempt to insert the posted value into the table and the database will prevent duplicates arising. Furthermore, as documented under INSERT ... ON DUPLICATE KEY UPDATE Syntax:
If a table contains an AUTO_INCREMENT column and INSERT ... ON DUPLICATE KEY UPDATE inserts or updates a row, the LAST_INSERT_ID() function returns the AUTO_INCREMENT value. Exception: For updates, LAST_INSERT_ID() is not meaningful prior to MySQL 5.1.12. However, you can work around this by using LAST_INSERT_ID(expr). Suppose that id is the AUTO_INCREMENT column. To make LAST_INSERT_ID() meaningful for updates, insert rows as follows:
INSERT INTO table (a,b,c) VALUES (1,2,3)
ON DUPLICATE KEY UPDATE id=LAST_INSERT_ID(id), c=3;
Therefore, you can do:
$db_info->prepare('
INSERT INTO menusize (typesize) VALUES (:typesize)
ON DUPLICATE KEY UPDATE typesize=LAST_INSERT_ID(typesize)
')->execute(array(
':typesize' => $_POST['typesize']
));
$db_info->prepare('
INSERT INTO menucatagorysize
(menusize_ID, menucatagory_ID)
VALUES
(LAST_INSERT_ID(), :catagoryid)
')->execute(array(
':catagoryid' => $_POST['catagoryid']
));
$stmt = $db_info->prepare('
SELECT DISTINCT menuitem_ID
FROM menuprice
WHERE menucatagory_ID = :catagoryid
');
$stmt->execute(array(
':catagoryid' => $_POST['catagoryid']
));
while ($row = $stmt->fetch(PDO::FETCH_ASSOC)) {
// etc.
}
(As an aside, the English word is spelled cat*e*gory, not cat*a*gory.)

Insert data if not exist

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 .

How do I get all the ids of the row created by one multiple row insert statement

I'm new to php. So, please forgive me if this seems like a dumb question.
Say i have a MySQL insert statement insert into table (a,b) values (1,2),(3,4),(5,6). table 'table' has a auto increment field called 'id'.
how can I retrieve all the ids created by the insert statement above?
It will be great if i get an example that uses mysqli.
You can't. I would suggest that you maintain your own ids (using guid or your own auto-increment table) and use it when you insert into the table.
But it's possible to get the auto-increment value for the last inserted using LAST_INSERT_ID():
http://dev.mysql.com/doc/refman/5.0/en/getting-unique-id.html
AngeDeLaMort's answer is almost right. Certainly, the most appropriate way to deal with the problem is to insert one row at a time and poll the insert_id or generate the sequence elsewhere (which has additional benefits in terms of scalability).
I'd advise strongly against trying to determine the last insert_id and comparing this the most recent insert_id after the insert - there's just too may ways this will fail.
But...an alternative approach would be:
....
"INSERT INTO destn (id, data, other, trans_ref)
SELECT id, data, other, connection_id() FROM source";
....
"SELECT id FROM destn WHERE trans_ref=connection_id()";
....
"UPDATE destn SET trans_ref=NULL where trans_ref=connection_id()";
The second query will return the ids generated (note that this assumes that you use the same connection for all 3 queries). The third query is necessary because connection ids to go back into the pool when you disconnect (i.e. are reused).
C.
In some cases, if you have another identifier of sort such as a UserID, you could filter your query by UniqueID's greater than or equal to mysql_insert_id(), limit by the number of affected rows and only display those by the user. This would really only work inside of a transaction.
$SQL = "INSERT INTO Table
(UserID, Data)
VALUES
(1,'Foo'),
(1,'Bar'),
(1,'FooBar')";
$Result = mysql_query($SQL);
$LastID = mysql_insert_id();
$RowsAffected = mysql_affected_rows();
$IDSQL = "SELECT RecordID
FROM Table
WHERE UserID = 1
AND RecordID >= '$LastID'
LIMIT '$RowsAffected'";
$IDResult = mysql_query($IDSQL);
as a follow up to AngeDeLaMort:
You could seperate your inserts and do it something like this:
$data = array (
array(1,2),
array(3,4),
array(5,6)
);
$ids = array();
foreach ($data as $item) {
$sql = 'insert into table (a,b) values ('.$item[0].','.$item[1].')';
mysql_query ($sql);
$id[] = mysql_insert_id();
}
Now all your new id's are in the $id array.
Maybe I can do this
$insert = "insert into table (a,b) values (1,2),(3,4),(5,6)";
$mysqli->query($insert);
$rows_to_be_inserted=3;
$inserted_id = $mysqli->insert_id // gives me the id of the first row in my list
$last_row_id = ($inserted_id+$rows_to_be_inserted)-1;
$mysql->query("select * from table where id between $inserted_id and $last_row_id");
what to you guys say?

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