I have a product info table with more than 130 columns/fields.
I want to write a php script that adds a new product to the table OR updates the existing product if it already exist. The first field is the product key.
The product information is stored in a numbered php array : $product_info[0] to $product_info[130].
Basically something like this :
INSERT INTO table (a,b,c) VALUES ($product_info[0],$product_info[1],$product_info[2])
ON DUPLICATE KEY UPDATE a='$product_info[0]', b='$product_info[1]', c='$product_info[2]'
Is there something more efficient than typing each of the 130 fields twice?
Yes, there is, use the VALUES() function:
INSERT INTO `table` (a, b, c) VALUES (?, ?, ?)
ON DUPLICATE KEY UPDATE a = VALUES(a), b = VALUES (b), c = VALUES(c)
Basically, in the UPDATE part, VALUES(column) will return the specified value for that column for the current row in question. So you can do interesting things like:
ON DUPLICATE KEY UPDATE
a = VALUES(a),
b = VALUES(b) + VALUES(c),
The beauty of that syntax, is it also supports multiple insert rows:
INSERT INTO `table` (a, b, c)
VALUES (?, ?, ?),
VALUES (?, ?, ?),
VALUES (?, ?, ?)
ON DUPLICATE KEY UPDATE a = VALUES(a), b = VALUES (b), c = VALUES(c)
Unfortunately MySQL does not support merging... having an ORM can help ease the pain of coding multiple IF EXISTS UPDATE ... ELSE INSERT code
REPLACE
Related
I don't know how to replace some data if a "user" already exists.
I've tried ON DUPLICATE KEY UPDATE but I came to realize that this will probably not work. Because the only value that isn't updated is 'user' in my code but the other 3 values are constantly updated every 5 minutes.
INSERT INTO online ( `user`, `bot`, `world`, `status` ) VALUES ('$User', '$Name', '$World', '$status')
ON DUPLICATE KEY UPDATE bot = VALUES ('$Name'), world = VALUES ('$World'), status = VALUES ('$status')
The idea is if, for example, user "bob" already exists update his other 3 values bot, world, status, instead of creating a new line and so on.
Edit: this is how I have it setup in Mysql
The argument to VALUES() should be the name of a column, not a string. You put the name of the column that you would have inserted into.
INSERT INTO online ( `user`, `bot`, `world`, `status` ) VALUES ('$User', '$Name', '$World', '$status')
ON DUPLICATE KEY UPDATE bot = VALUES (bot), world = VALUES (world), status = VALUES (status)
I am trying to insert a record to a table with 2 column but I get this error.
My error starts in part of the execute. Anyone that can help me out with this ?
I am using PDO.
My code:
global $conn_kl;
$sql = $conn_kl->prepare("INSERT INTO order_producten VALUES (?,?)");
$sql->execute(array($product_id, $bewerking_id));
The issue is here:
INSERT INTO order_producten VALUES (?,?)
here columns are not defined in this query, in this case it is expected that you have to pass the values for all columns in the table. But you want to insert the values for only 2 columns, so please please specify that columns names like:
INSERT INTO order_producten(column_name1, column_name2) VALUES (?,?)
order_producten will have more or less than two columns and you are setting only two values.
Please specify columns after table name. for example,
INSERT INTO order_producten(id, name) VALUES(?, ?)
For example, code something like this were working for me:
global $conn_kl;
$sql = $conn_kl->prepare("INSERT INTO `order_bewerkingen` VALUES (null, ?, ?, ?)");
$sql->execute(array($order_id, $method, $position));
I need to get either a 1 (the record was inserted) or a 2 (the record was updated). How can this be done using Laravel? I can't find any information about this in their documentation.
DB::insert('insert into ratings
(owner_id, game_id, rating) values (?, ?, ?)
on duplicate key update rating = values(rating)',
[
(int)$body['user'], (int)$gameId, (int)$body['amount']
]);
I know I can get pdo like this:
$pdo = DB::connection()->getPdo();
but I don't think that helps...
You will want to use the function affectingStatement which seems to return the number of affected rows...
$res = DB::affectingStatement('insert into ratings
(owner_id, game_id, rating) values (?, ?, ?)
on duplicate key update rating = values(rating)', [
(int)$body['user'],
(int)$gameId,
(int)$body['amount']
]);
For me, this was returning 2 when a row updates, 1 when a row is inserted, and 0 on failure as expected.
I am inserting values into a database.
If the time the values were created is less than 24 hours then I insert like this,
$stmt = $con->prepare('
INSERT INTO taggedPlaces
(id, created_time, place_id, city, country, latitude, longitude, state, street, zip, name)
VALUES
(?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?)
');
foreach($graphObject['tagged_places']->data as $data) {
if (time() - strtotime($data->created_time) < 86400) {
$stmt->execute(array(
$data->id,
$data->created_time,
$data->place->id,
$data->place->location->city,
$data->place->location->country,
$data->place->location->latitude,
$data->place->location->longitude,
$data->place->location->state,
$data->place->location->street,
$data->place->location->zip,
$data->place->name
));
}
}
Everytime I return to the page it takes the same entries and continuously adds them to the database.
I would like to say something like
if $data->created_time == any created_time value in the DB then don't add this value,
as well as currently I am doing
if (time() - strtotime($data->created_time) < 86400)
to make sure it is not older then 24 hours.
How can I add this condition?
Option 1: (Recommeded)
Make id the primary key for your table, taggedPlaces:
ALTER TABLE taggedPlaces ADD PRIMARY KEY(id)
Then change your insert statement to use INSERT IGNORE which will skip duplicate inserts.
Option 2:
Make created_time a unique field in taggedPlaces:
ALTER TABLE taggedPlaces ADD UNIQUE(created_time);
Then, again, use INSERT IGNORE to skip duplicates.
Option 3: (Not recommeded, but will work)
Prior to running your insert, perform another query to check if $data->created_time is already in the table:
$check = $con->prepare('
SELECT id FROM taggedPlaces
WHERE created_time = ?
');
$check->execute(array($data->created_time));
if (count($check->fetchAll()) == 0) {
// No duplicates found. Proceed...
}
I have a table with 3 columns - id (pk), pageId (fk), name. I have a php script which dumps about 5000 records into the table, with about half being duplicates, with same pageId and name. Combination of pageId and name should be unique. What is the best way to prevent duplicates being saved to the table as I loop through the script in php?
First step would be to set a unique key on the table:
ALTER TABLE thetable ADD UNIQUE INDEX(pageid, name);
Then you have to decide what you want to do when there's a duplicate. Should you:
ignore it?
INSERT IGNORE INTO thetable (pageid, name) VALUES (1, "foo"), (1, "foo");
Overwrite the previously entered record?
INSERT INTO thetable (pageid, name, somefield)
VALUES (1, "foo", "first")
ON DUPLICATE KEY UPDATE (somefield = 'first')
INSERT INTO thetable (pageid, name, somefield)
VALUES (1, "foo", "second")
ON DUPLICATE KEY UPDATE (somefield = 'second')
Update some counter?
INSERT INTO thetable (pageid, name)
VALUES (1, "foo"), (1, "foo")
ON DUPLICATE KEY UPDATE (pagecount = pagecount + 1)
You can also ignore the error with mysql: INSERT IGNORE INTO TABLE ... it will ignore the key error, skip over that insert and move on to the next.
From a mysql point you can do
alter table YOURTABLE add unique index(pageId, name);
If your wording is correct and you want to do it from php you can do
$already_done = array();
foreach ($records as $record)
{
$unique_hash = md5($record['name'].$record['pageId']);
if (!in_array($unique_hash, $already_done))
{
$already_done[] = $unique_hash;
// sql insert here
}
}
either way those should do you just fine.
You can set the PageID and Name to a Unique index in the MySQL database. This way when you insert the rows, it will cause an error, which can be ignored by PHP, and you can just go to the next row.
This assumes you are inserting rows individually. AKA:
foreach($large_data as $fields)
{
mysql_query("INSERT INTO TABLE (`Something`) VALUES('".$fields['something']."');
}