In my database there is a table, that contains only one row: id. It's something like a facade for another table:
Now I want to add an entry into whatever_set, in order to be able to create entries in the dependent tables. In SQL I would write:
INSERT INTO whatever_set VALUES ();
But when I try it with Zend\Db
$action = new Insert(whatever_set');
$data = [];
$action->values($data);
$sql = new Sql($this->dbAdapter);
$statement = $sql->prepareStatementForSqlObject($action);
$result = $statement->execute();
I get an exception:
values or select should be present
There are some possible workarounds (see below). But is there though a "Zend way" to save empty with no VALUES? How to do this?
Workaround 1
Custom Insert class.
Workaround 2
Raw SQL like
$sql = 'INSERT INTO whatever_set VALUES ();';
$result = $this->dbAdapter->getDriver()->getConnection()->execute($sql);
Related
We have two databases A and B and we want to move and replace some columns with all values from database A to database B
Database.SourceTable >>> move to >>> Database.TargetTable
A.User1 >>> >>> B.User2
The user table has many columns, but we only want to replace some columns: col1, col2 and col3.
How can i do this?
Thank you
EDIT: I forget to say, that i only want to replace the first 100 columns id's.
I mean each column has more than 100 id's and i want to replace first 100; not all.
As you're not able to move datas from a database to another one with a single connection, you can't do it with a single query, you need to do it manually, or with a third party language.
You'll need to :
get one connection per database
get a PRIMARY KEY on your table user
export datas from databaseA
inject these datas to databaseB
Example with PHP/PDO :
// Create 2 connections
try {
$dbA = new PDO('mysql:dbname=databaseA;host=localhost', 'root', '');
$dbB = new PDO('mysql:dbname=databaseB;host=localhost', 'root', '');
} catch (PDOException $ex) {
echo 'Connection failed: ' . $ex->getMessage();
}
// Get datas from databaseA
$result = $dbA->query("SELECT id, col1, col2, col3 FROM user");
// Prepare REPLACE query for databaseB
$query = "REPLACE INTO user (id, col1, col2, col3) VALUES";
$values = array();
foreach ($result as $row) {
array_push($values, '('.$row['id'].', "'.$row['col1'].'", "'.$row['col2'].'", "'.$row['col3'].'")');
}
$query .= implode(',', $values);
// Execute REPLACE query on databaseB
$stmt = $dbB->prepare($query);
$stmt->execute();
If table1 on db1 database has t1c1, t1c2,t1c3 columns and you want only t1c1, and t1c3 data be copied into table2 of database db2 with columns t2c1 and t2c2 then
insert into db2.table2( t2c1, t2c2 ) select t1c1, t1c3 from db1.table1
should work for you.
And make sure that you have required privileges set to access data from cross database tables or other objects. And you should also check issues with constraints, if any defined, on table2.
Please i have this quetion,
I have here:
$query = $db->prepare('Update survey_content set '.$type.' = '.$type.'+1 where id = '.$where);
$query->execute();
I'm new to PDO and i really don't know how to create a new query, i need to add a record on the database, is it ok like this?
$query1 = $db->prepare('INSERT INTO daily (selected)
VALUES (.$type.)');
$query1->execute();
Use a placeholder and fill the value in the execute method
$query1 = $db->prepare("INSERT INTO daily (selected)
VALUES (?)");
$query1->execute($selected_data);
or bind the parameter seperately
$query1 = $db->prepare("INSERT INTO daily (selected)
VALUES (:sel_dat)");
$query1->bindParam(":sel_dat",$selected_data);
$query1->execute();
I've created a database for one of my classes simulating a hotel reservation form.
my database table name that I am trying to get values from is tblNameRes and the fields are fldName and pkEmail
I have gotten the values from the database and displayed them in this table here:
http://www.uvm.edu/~cchessia/cs148/assign4/strugg.php
I want to add a column to be able to update and delete these records, but I don't really understand how. I have this code:
$updating = $db->prepare('UPDATE `tblNameRes` SET `name` = `name` + ? WHERE `id` = ?');
$updating->execute(array(20, $id));
but I want to be able to click a link that says "update" and it will take me to another page that allows me to edit the data and submit it back to the database. I would also like to be able to delete the data in the same way.
I'm a little confused to what you would like, would you like the PDO code to both update and delete items from your DB? If so I use this, I'm also new to PDO. Also try not to use ' around table/field names.
/** Code to update */
$query = "UPDATE tblNameRes SET name ='$name' WHERE id = :id";
$stmt = $db->prepare($query);
$stmt->BindValue(':id', $id, PDO::PARAM_INT);
$stmt->execute();
/** Code to delete */
$query = "DELETE FROM tblNameRes WHERE id = :id";
$stmt = $db->prepare($query);
$stmt->BindValue(':id', $id, PDO::PARAM_INT);
$query->execute();
Let me know if this helps, or is not what you wanted. I'm new here so I can't post 'comments'.
/* code for update and remove this 'at the beginning an at the end*/
$query = "UPDATE tblNameRes SET name =:fname WHERE id = :id";
$stmt = $db->prepare($query);
$stmt->BindValue('fname',$fname);
$stmt->execute();
I'm having a problem with running this function. When it runs, it does exactly what I want, except that within my like_requests table the request_id is not the mysql query result linked to the variable $select but Resource Id #22. I thought that resource id's appear when you are trying to echo out a result, but I'm not using echo. What's wrong with the code?
function update_likes($band_requested, $new_likes, $session_user_id) {
$select = mysql_query("SELECT `primary_id` FROM `requests` WHERE
`user_requester_id` = '$session_user_id' AND `person_requested` =
'$band_requested'");
$sql_2 = "INSERT INTO `like_requests` (user_id, request_id) VALUES
('$session_user_id', '$select')";
mysql_query($sql_2);
}
$band_requested = 'rally done';
$new_likes = 239;
$the_session_user_id = 3;
update_likes($band_requested, $new_likes, $the_session_user_id);
UPDATE WITH CORRECTED ANSWER
Here is the code corrected with help from David.
function update_likes($band_requested, $new_likes, $session_user_id)
{
$select = mysql_query("SELECT `primary_id` FROM `requests` WHERE `user_requester_id` =
'$session_user_id' AND `person_requested` = '$band_requested'");
$row = mysql_fetch_row($select);
$request_id = $row[0];
$sql_2 = "INSERT INTO `like_requests` (user_id, request_id) VALUES ('$session_user_id',
'$request_id')";
mysql_query($sql_2);
}
mysql_query returns a resource (http://php.net/manual/en/function.mysql-query.php) not just a scalar value. You'd need to use a function like mysql_fetch_row() to get the, presumably, one row you want, assign that row to a variable $row, then retrieve the primary_id with array syntax like $row['primary_id']. By the way, apparently mysql_query is being eased out and we should use the MySQLi API with the mysqli_query() method.
I'm inserting a row into a MySQL database table. On the first insertion I want a new row to be added, but after that I just want that row to be updated. Here's how I'm doing it. An Ajax request calls the following php file:
<?php
include "base.php";
$bookID = $_POST['bookID'];
$shelfID = $_POST['shelfID'];
$userID = $_SESSION['user_id'];
$query = mysql_query("SELECT shelfID FROM shelves WHERE userID = '$userID' AND shelfID = '$shelfID' AND bookID = '$bookID'");
if (mysql_num_rows($query) == 0) {
$insert = "INSERT INTO shelves (bookID,shelfID,userID) VALUES ('$bookID','$shelfID','$userID')";
mysql_query($insert) or die(mysql_error());
} elseif (mysql_num_rows($query) == 1) { //ie row already exists
$update = "UPDATE shelves SET shelfID = '$shelfID' WHERE userID = '$userID' AND bookID = '$bookID'";
mysql_query($update) or die(mysql_error());
}
?>
As it stands it adds a new row every time.
You should consider using PDO for data access. There is a discussion on what you need to do here: PDO Insert on Duplicate Key Update
I'd flag this as duplicate, but that question is specifically discussing PDO.
You can use the INSERT ... ON DUPLICATE KEY UPDATE syntax. As long as you have a unique index on the data set (i.e. userid + shelfid + bookid) you are inserting, it will do an update instead.
http://dev.mysql.com/doc/refman/5.5/en/insert-on-duplicate.html