How to check with PHP does a SQL database already have - php

I've tried to find the answer to this question but none of the answers fit.
I have two databases, one has 15.000.000 entries and I want to extract the necessary data and store it in a much smaller database with around 33.000 entries. Both databases are open at the same time. Or at least they should be.
While having the big database open and extracting the entries from it, is it possible to check whether the value already exists in a certain table in the smaller database? I just need some generic way which checks that.
In my code I'm first opening both databases (big one is oddsnavi_push, small one is oddsnavi_baby):
$database = "oddsnavi_push";
$db_handle = mysql_connect($server, $user_name, $password);
$db_found = mysql_select_db($database, $db_handle);
$database_baby = "oddsnavi_baby";
$db_handle_baby = mysql_connect($server, $user_name, $password);
$db_found_baby = mysql_select_db($database_baby, $db_handle_baby);
And then I'm starting to read and calculate data from oddsnavi_push:
$SQL_SELECT_ALL = "...giant query...";
$result_select_all = mysql_query( $SQL_SELECT_ALL );
while($db_field_all = mysql_fetch_assoc( $result_select_all ) ) {
$SQL_INSERT="INSERT INTO oddsnavi_baby.calc (id, one, two) VALUES ('$id', '$one', '$two')";
It works up until that point. It takes the data which was read (id, one, two) and inserts them in proper columns in oddsnavi_baby table named calc. It does that properly when the oddsnavi_baby is completely empty.
However, I need it to only update the database IF an entry (based on whether a certain 'id' exists or not) doesn't exist.
EDIT: I will rephrase my question. From the query results (big database) I'm getting strings, for every row. For example $string. How do I open the second database and check if oddsnavi_baby.calc table has the $string value in column Events?

Skip the check and try with just INSERT IGNORE assuming the Id is a unique key.
http://dev.mysql.com/doc/refman/5.5/en/insert.html

Man, I remember coming up against a problem like this once. I think we actually went the slow, expensive route and wrote a script to pull one entry at a time, compare it, and insert it if it didn't already exist.
However, I found this post, which sounds like it might be of some help to you:
http://www.mysqlfaqs.net/mysql-faqs/Tricky-Select-Queries/How-to-compare-data-of-two-tables-of-two-different-databases-in-MySQL
Good luck :)

Why do you really need multiple database? Amount does not matter tables are just fine, no need to split.
//Multiple links to different databases
$dblink1 = mysqli_connect($localhost, $user, $pass, $db1);
$dblink2 = mysqli_connect($localhost, $user, $pass, $db2);
$id = '1'; // Id to check
$query = "SELECT COUNT(*) FROM `table` WHERE id = '1' LIMIT 1";
$result = mysqli_query($dblink1, $query); //query db 1
if(mysql_num_rows($result)) {
$query = "INSERT INTO `table` VALUES(.....)";
$result = mysqli_query($dblink1, $query); //query db 2
}

Related

Delete one column from table, Update another with PHP MYSQL

I have two tables. One table is the matches table (e2wedstrijden) and another table is my scoring table with the points earned etc. (e2teams).
Now I have that I can delete a match from the e2wedstrijden table. And this is working fine.
But I want that if I delete a match from that table. It also add or decrease points to the table ("e2teams"). I tried to compare the tables but this is not working.
So I want for example:
If($row['thuisscore'] == $row['uitscore']) what are to row names in my e2wedstrijden table. So if these two are the same (like 0-0 or 1-1 or something) Than it needs to decrease 1 point from the table e2teams. But only by the teams that are the same as the rows "Thuisteam" and "Uitteam" in my e2wedstrijden table. So the Row Thuisteam (in "e2wedstrijden") Needs to find the same result in ("e2teams") row Team. And this needs to be done the same with the Row Uitteam (in "e2wedstrijden") Needs to find the same result in ("e2teams")
Thuisteam and Uitteam = Dutch for hometeam and awayteam. I think my fault is that the system can't link the 'Thuisteam' from e2wedstrijden to the Team in e2teams but don't know how to solve it
This is my deletematches.php, It deletes the match but doesn't decrease or adds points:
<?php
if(!isset($_COOKIE['E2ingelogd'])) {
header("location:../../index.php");
}
include "../../connect.php";
$dbhandle = mysql_connect($hostname, $username, $password) or die("Could not connect to database");
$selected = mysql_select_db("login", $dbhandle);
$result = mysql_query("SELECT * FROM e2wedstrijden WHERE ID = ".$_GET['del']."");
while($row = mysql_fetch_assoc($result)){
if( $row['thuisscore'] == $row['uitscore']){
echo $row['thuisscore'];
mysql_query("UPDATE e2teams SET Punten = Punten-1 WHERE Team ='".$row['Thuisteam']."'");
mysql_query("UPDATE e2teams SET Gespeeld = Gespeeld-1 WHERE Team = ('".$row['Thuisteam']."'");
mysql_query("UPDATE e2teams SET Verloren = Gelijk-1 WHERE Team ='".$row['Uitteam']."'");
echo "Team is deleted";
}else{
echo 'Update Error!';
}
}
$table_1_delete = mysql_query("DELETE FROM e2wedstrijden WHERE ID = ".$_GET['del']."");
?>
This is my e2teams table:
And this is my E2wedstrijden table:
So i need something like:
UPDATE e2teams SET Punten = Punten-1 WHERE Team = Look in table ("e2wedstrijden) deleted Thuisteam and deleted Uitteam
Hope you can help
You've placed an extra parentheses in the 2nd query for "gespeeld" right after the equal sign:
mysql_query("UPDATE e2teams SET Gespeeld = Gespeeld-1
WHERE Team = ('".$row['Thuisteam']."'");
Is this what isn't updating?
Without being 100% sure on how your data model works, it might make sense at refactoring what you have. Something that might be useful would be to create a view of the summary table and just update the data from the child/master table.... aggregating in the view layer. Views in mysql can be seen here.
If you are stuck with the data model you have (legacy application, etc.) you can possibly look at triggers if you have to modify data in two tables you might want to consider stored procedures or triggers, discussed here and here.
The third thing that comes to mind, is around correlated sub-queries and how you could reference the another table in a sort of update-from. However, you're ID's aren't surrogate keys in this situation.
Also, have a look at sql injection; I haven't looked at PHP in a while but those sql statements kind of look like they are created with sting composition
Good luck,

Avoid multiple Insertion, when insert query called multiple times at same time

$Link= mysql_connect("localhost", "root", "", "");
query = mysql_query($Link,"SELECT col_name FROM table_name WHERE col_name='$val'");
$num=mysql_num_rows($query);
if($num==0){
$query1 = mysql_query($Link,"INSERT INTO table_name(col_name)VALUES('$val')"); )
}
There will be multiple calls to query at same time. I used this method to avoid multiple insertion, but sometimes multiple rows are inserted.Please help me. Thanks in advance.
thanks for your help..
I m not calling this code in loop but multiple calls at same time from different users are made.
mysql_query syntax: resource mysql_query ( string $query [, resource $link_identifier = NULL ] )
$query1 = mysql_query("INSERT INTO table_name (col_name)VALUES('".$val."')");
Full code:
$Link= mysql_connect("localhost", "root", "", "");
$query = mysql_query($Link,"SELECT col_name FROM table_name WHERE col_name='".$val."'");
$num=mysql_num_rows($query);
if($num==0){
$query1 = mysql_query("INSERT INTO table_name(col_name)VALUES('".$val."')");
}
If there are multiple simultaneous requests to the web server(s) that host the PHP, the select could return no results for more than one of these requests, and then there may be multiple inserts due to each such request performing the insert.
If you are not running windows and have only one web server, you could synchronize access to this code using PHP's semaphores.
But even better, can you put a unique constraint on the value of this column in the table in mysql?
Though your code is correct and it should not insert again, but
You should add a unique key to your field and then use Insert Ignore
"INSERT IGNORE INTO table_name(col_name)VALUES('$val')"

sqlite3 - efficient way to count rows returned from SELECT statement in PHP without using COUNT()

I'm an SQL noob and learning how to use PDO. I'm doing a course which introduces basic user login functions. In an example of a login page, they check the username/password against a MySQL database. I edited their code slightly to be able to simultaneously check whether the user/pass combo exists and also grab the user's first name:
$sql = sprintf("SELECT firstname FROM users WHERE username='%s' AND password='%s'",
mysql_real_escape_string($_POST["username"]),
mysql_real_escape_string($_POST["password"]));
// execute query
$result = mysql_query($sql);
if (mysql_num_rows($result) == 1) {
$_SESSION["authenticated"] = true;
// get contents of "firstname" field from row 0 (our only row)
$firstname = mysql_result($result,0,"firstname");
if ($firstname != '')
$_SESSION["user"] = $firstname;
}
What I want to do is use SQLite instead and do the same thing. Searching around has only resulted in people saying you should use a SELECT COUNT(*) statement, but I don't want to have to use an extra query if it's possible. Since I'm SELECTing the firstname field, I should only get 1 row returned if the user exists and 0 if they don't. I want to be able to use that number to check if the login is correct.
So far I've got this:
$dsn = 'sqlite:../database/cs75.db';
$dbh = new PDO($dsn);
$sql = sprintf("SELECT firstname FROM users WHERE username='%s' AND password='%s'",
$_POST["username"],
$_POST["password"]);
// query the database and save the result in $result
$result = $dbh->query($sql);
// count number of rows
$rows = sqlite_num_rows($result);
if ($rows == 1) { ...
But this is returning Warning: sqlite_num_rows() expects parameter 1 to be resource, object given.
Is there a way I can do this efficiently like in MySQL, or do I have to use a second query?
EDIT:
I found this, not sure if it's the best way but it seems to work: How to get the number of rows grouped by column?
This code let me do it without the second query:
// query the database and save the result in $result
$result = $dbh->query($sql);
// count number of rows
$rows = $result->fetch(PDO::FETCH_NUM);
echo 'Found: ' . $rows[0];
$rows is an array so I can just count that to check if it's > 0.
Thanks to everyone who commented. I didn't know until now that there were 2 different approaches (procedural & object oriented) so that helped a lot.
Normally, you can use PDOStatement::rowCount(), however, SQLite v3 does not appear to provide rowcounts for queries.
You would need to seperately query the count(*), or create your own counting-query-function.
The documentation comments have an example of this
A bit late, but i tried this with SQLite3 successful:
$result = $db->query('SELECT * FROM table_xy');
$rows = $result->fetchAll();
echo count($rows);

MySqli not rolling back appropriately

I have a database listed as $db under mysqli. This database is contains into two tables, I listed them below as table and table2 (just for this example). Table2's rows requires an id from table. This is fine, but there might be a problem adding the columns into table2 thus requiring a rollback routine. However, it doesn't seem to be working.
I started with turning off the auto-commit. I then tried to put in the rollback command even though I am using the die command to signal a failure. As far as I am concerned the transaction could be blasted into oblivion in mid operation and the database should still be stable. So I am not sure what is going on here unless the database is completely ignoring the fact that I am trying to turn off auto-commit.
The basic structure of my code is listed below:
function problem($str)
{
global $db;
mysqli_rollback($db);
die($str);
}
mysqli_autocommit($db,false);
//Basic check if exists
$sqlstr = "SELECT * FROM table WHERE name = '$name';";
$r = mysqli_query($db,$sqlstr);
if (mysqli_num_rows($r)>0){problem("A row already exists under that id");}
//Insert the row
$sqlstr = "INSERT INTO table (name,v1,v2,v3) VALUES ('$name','$v1','$v2','$v3');";
$r = mysqli_query($db,$sqlstr);
if (!$r){problem("Could not insert into the table. $sqlstr");}
//Get the generated id part 1
$sqlstr = "SELECT id FROM table WHERE name = '$name';";
$r = mysqli_query($db,$sqlstr);
if (!$r){problem("Could not add into the table. $sqlstr");}
//Get the generated id part 2
$row = mysqli_fetch_assoc($r);
$eid = $row['id'];
//A simple loop
$count = count($questions);
for ($i=1;i<=$count;$i++)
{
//This is where it typically could die.
$r = mysqli_query($db,"INSERT INTO table2 VALUES (...);");
if (!$r){problem("Could not add to the table2. $sqlstr");}
}
mysqli_commit($db);
Is there something I am missing? I tried to follow the examples I found for the auto-commit as closely as I could.
Transactions only work if the table engine supports them, e.g. InnoDB.

How to connect to 2 databases at the same time in PHP

I am trying to connect to 2 databases on the same instance of MySQL from 1 PHP script.
At the moment the only way I've figured out is to connect to both databases with a different user for each.
I am using this in a migration script where I am grabbing data from the original database and inserting it into the new one, so I am looping through large lists of results.
Connecting to 1 database and then trying to initiate a second connection with the same user just changes the current database to the new one.
Any other ideas?
You'll need to pass a boolean true as the optional fourth argument to mysql_connect(). See PHP's mysql_connect() documentation for more info.
If your database user has access to both databases and they are on the same server, you can use one connection and just specify the database you want to work with before the table name. Example:
SELECT column
FROM database.table
Depending on what you need to do, you might be able to do an INSERT INTO and save a bunch of processing time.
INSERT INTO database1.table (column)
SELECT database2.table.column
FROM database2.table
Lucas is correct. I assume that both the databases are hosted on the same host.
Alternatively, you can create only 1 db connection and keep swapping the databases as required. Here is pseudo code.
$db_conn = connect_db(host, user, pwd);
mysql_select_db('existing_db', $db_conn);
-- do selects and scrub data --
mysql_select_db('new_db', $db_conn);
-- insert the required data --
I would suggest using two connection handlers
$old = mysql_connect('old.database.com', 'user', 'pass);
mysql_select_db('old_db', $old);
$new = mysql_connect('new.database.com','user','pass);
mysql_select_db('new_db', $new)
// run select query on $old
// run matching insert query on $new
If it's an option, use PDO: you can have as many database connections open as you like.
Plus, assuming your executing the same queries over and over, you can use prepared statements.
You can easily use 2 databases in same time with following Codes:
<?php
define('HOST', "YOURHOSTNAME");
define('USER', "YOURHOSTNAME");
define('PASS', "YOURHOSTNAME");
define('DATABASE1', "NAMEOFDATABASE1");
define('DATABASE2', "NAMEOFDATABASE2");
$DATABASE1 = mysqli_connect(HOST, USER, PASS, DATABASE1);
$DATABASE2 = mysqli_connect(HOST, USER, PASS, DATABASE2);
if(!$DATABASE1){
die("DATABASE1 CONNECTION ERROR: ".mysqli_connect_error());
}
if(!$DATABASE2){
die("DATABASE2 CONNECTION ERROR: ".mysqli_connect_error());
}
$sql = "SELECT * FROM TABLE"; /* You can use your own query */
$DATABASE1_QUERY = mysqli_query($DATABASE1, $sql);
$DATABASE2_QUERY = mysqli_query($DATABASE2, $sql);
$DATABASE1_RESULT = mysqli_fetch_assoc($DATABASE1_QUERY);
$DATABASE2_RESULT = mysqli_fetch_assoc($DATABASE2_QUERY);
/* SHOW YOUR RESULT HERE WHICH DATABASE YOU WANT FROM */
echo $DATABASE1_RESULT['id'];
echo $DATABASE2_RESULT['id'];
/*After complete your all work don't forgot about close database connections*/
mysqli_close($DATABASE1);
mysqli_close($DATABASE2);
?>
First Connect Two Database
$database1 = mysql_connect("localhost","root","password");
$database2 = mysql_connect("localhost","root","password");
Now Select The Database
$database1_select = mysql_select_db("db_name_1") or die("Can't Connect To Database",$database1);
$database_select = mysql_select_db("db_name_2") or die("Can't Connect To Database",$database2);
Now if we want to run query then specify database Name at the end like,
$select = mysql_query("SELECT * FROM table_name",$database1);

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