i want to connect two mysql databases one is located in localhost and other one is located in a server
here is what i have done so far and i am not getting either error or data
<?php
$con=mysql_connect('120.247.201.8:3306','root','root');
$con1=mysql_connect('localhost','root','');
//mysql_connect('localhost','root','');
if(!$con){
die('Try Again');
}
if(!$con1){
die('Try Again');
}
mysql_select_db("iot",$con1);
mysql_select_db("lora_gateway",$con);
$result =mysql_query("SELECT lora_gateway.`server_log`.`created_at`, lora_gateway.`server_log`.`temperature`FROM iot.`device` inner JOIN `lora_gateway`.`server_log` on `lora_gateway`.`server_log`.`gateway_Id` = `iot`.`device`.`gatewayId` where iot.`device`.`deviceId`='23' ORDER BY lora_gateway.`server_log`.`created_at` desc");
$num= mysql_num_rows($result);
print_r($num);
?>
This is a solution for multiple servers, connections and DBs.
Two or more MySQL 5 db servers
Two or more locations (local and/or anything else)
Two or more different databases
Two or more tables and fields
Tested and Works fine :-)
//Define your database connections and select your database want to use. In this example I use two connections and two DBs. But you can use more than two.
<?php
//MySQL Server 1
$dbhost1 = "127.0.0.1";
$dbuser1 = "dbuser1";
$dbpassword1 = "dbpass1";
$db1 = "database1";
$connection1 = mysql_connect($dbhost1,$dbuser1,$dbpassword1) or die (mysql_error());
mysql_select_db($db1,$connection1);
//MySQL Server 2
$dbhost2 = "xxx.xxx.xxx.xxx";
$dbuser2 = "dbuser2";
$dbpassword2 = "dbpass2";
$db2 = "database2";
$connection2 = mysql_connect($dbhost2,$dbuser2,$dbpassword2) or die (mysql_error());
mysql_select_db($db2,$connection2);
//The SQL statement
$sql =" SELECT database1.tablename1.fieldname1 AS field1, database2.tablename2.fieldname2 AS field2 FROM database1.tablename1,database2.tablename2";
//Execute query and collect results in $results
$results = mysql_query($sql);
//Print result until end of records
while($rows = mysql_fetch_array($results)){
print $rows["field1"]." | ".$rows["field2"]."<br>";
}
?>
First of all, try to accustom yourself to using PDO or at least the mysqli_ functions. It's the future. :)
mysql_query's second parameter, the connection link, is optional. If omitted, it uses the last connection opened with mysql_connect. (See php.net Documentation)
Ergo, always use $con or $con1 as 2nd parameter in order to use the correct connection.
Then, provided that your queries are correct, it should work as expected.
what about to add connection to mysql_query() ?
$result = mysql_query("SELECT lora_gateway.`server_log`.... desc", $con);
Related
When i recently wanted to list all tables in a database in php i made the simple MySQL query:
$tables_query = mysql_query('show tables');
while ($test = mysql_fetch_array($tables_query)) {
echo "Table: {$test[0]}<br />";
}
The first result is
TABLES 105
address_book
I don't have a table called "TABLES 105" but the mysql_num_rows also shows, that there is 105 results, even that my database only contains 104 table
If i try to request "show tables" directly on the MySql server, it works fine and i get 104 rows as result. It also worked before and i can't seem to find anything about this, so im hoping someone can help me in here.
It also affect when i call directly to the mysql server. I got access with an other user login for an other database, on the same server and here is no issues at all.
Its questionable how that 105 got there in the first place, most likely this is caused by that mysql_num_rows function that you mentioned as fetch_array actually fetches the rows, but here's one on MySQLi, stop using MySQL anymore:
$db = 'test'; // database name
$con = mysqli_connect('localhost', 'username', 'password', $db);
$tables_query = mysqli_query($con, "SHOW TABLES FROM {$db}");
while($table = mysqli_fetch_assoc($tables_query)) {
echo $table["Tables_in_{$db}"], '<br/>';
}
An alternative way of course is to delve into information_schema:
$db = 'test'; // database name
$con = mysqli_connect('localhost', 'root', '', $db);
$tables_query = mysqli_query($con, "SELECT * FROM INFORMATION_SCHEMA.TABLES WHERE TABLE_SCHEMA = '{$db}'");
while($table = mysqli_fetch_array($tables_query)) {
echo $table['TABLE_NAME'], '<br/>';
}
I have information spread out across a few databases and want to put all the information onto one webpage using PHP. I was wondering how I can connect to multiple databases on a single PHP webpage.
I know how to connect to a single database using:
$dbh = mysql_connect($hostname, $username, $password)
or die("Unable to connect to MySQL");
However, can I just use multiple "mysql_connect" commands to open the other databases, and how would PHP know what database I want the information pulled from if I do have multiple databases connected.
Warning : mysql_xx functions are deprecated since php 5.5 and removed since php 7.0 (see http://php.net/manual/intro.mysql.php), use mysqli_xx functions or see the answer below from #Troelskn
You can make multiple calls to mysql_connect(), but if the parameters are the same you need to pass true for the '$new_link' (fourth) parameter, otherwise the same connection is reused. For example:
$dbh1 = mysql_connect($hostname, $username, $password);
$dbh2 = mysql_connect($hostname, $username, $password, true);
mysql_select_db('database1', $dbh1);
mysql_select_db('database2', $dbh2);
Then to query database 1 pass the first link identifier:
mysql_query('select * from tablename', $dbh1);
and for database 2 pass the second:
mysql_query('select * from tablename', $dbh2);
If you do not pass a link identifier then the last connection created is used (in this case the one represented by $dbh2) e.g.:
mysql_query('select * from tablename');
Other options
If the MySQL user has access to both databases and they are on the same host (i.e. both DBs are accessible from the same connection) you could:
Keep one connection open and call mysql_select_db() to swap between as necessary. I am not sure this is a clean solution and you could end up querying the wrong database.
Specify the database name when you reference tables within your queries (e.g. SELECT * FROM database2.tablename). This is likely to be a pain to implement.
Also please read troelskn's answer because that is a better approach if you are able to use PDO rather than the older extensions.
If you use PHP5 (And you should, given that PHP4 has been deprecated), you should use PDO, since this is slowly becoming the new standard. One (very) important benefit of PDO, is that it supports bound parameters, which makes for much more secure code.
You would connect through PDO, like this:
try {
$db = new PDO('mysql:dbname=databasename;host=127.0.0.1', 'username', 'password');
} catch (PDOException $ex) {
echo 'Connection failed: ' . $ex->getMessage();
}
(Of course replace databasename, username and password above)
You can then query the database like this:
$result = $db->query("select * from tablename");
foreach ($result as $row) {
echo $row['foo'] . "\n";
}
Or, if you have variables:
$stmt = $db->prepare("select * from tablename where id = :id");
$stmt->execute(array(':id' => 42));
$row = $stmt->fetch();
If you need multiple connections open at once, you can simply create multiple instances of PDO:
try {
$db1 = new PDO('mysql:dbname=databas1;host=127.0.0.1', 'username', 'password');
$db2 = new PDO('mysql:dbname=databas2;host=127.0.0.1', 'username', 'password');
} catch (PDOException $ex) {
echo 'Connection failed: ' . $ex->getMessage();
}
I just made my life simple:
CREATE VIEW another_table AS SELECT * FROM another_database.another_table;
hope it is helpful... cheers...
Instead of mysql_connect use mysqli_connect.
mysqli is provide a functionality for connect multiple database at a time.
$Db1 = new mysqli($hostname,$username,$password,$db_name1);
// this is connection 1 for DB 1
$Db2 = new mysqli($hostname,$username,$password,$db_name2);
// this is connection 2 for DB 2
Try below code:
$conn = mysql_connect("hostname","username","password");
mysql_select_db("db1",$conn);
mysql_select_db("db2",$conn);
$query1 = "SELECT * FROM db1.table";
$query2 = "SELECT * FROM db2.table";
You can fetch data of above query from both database as below
$rs = mysql_query($query1);
while($row = mysql_fetch_assoc($rs)) {
$data1[] = $row;
}
$rs = mysql_query($query2);
while($row = mysql_fetch_assoc($rs)) {
$data2[] = $row;
}
print_r($data1);
print_r($data2);
$dbh1 = mysql_connect($hostname, $username, $password);
$dbh2 = mysql_connect($hostname, $username, $password, true);
mysql_select_db('database1', $dbh1);
mysql_select_db('database2',$dbh2);
mysql_query('select * from tablename', $dbh1);
mysql_query('select * from tablename', $dbh2);
This is the most obvious solution that I use but just remember, if the username / password for both the database is exactly same in the same host, this solution will always be using the first connection. So don't be confused that this is not working in such case. What you need to do is, create 2 different users for the 2 databases and it will work.
Unless you really need to have more than one instance of a PDO object in play, consider the following:
$con = new PDO('mysql:host=localhost', $username, $password,
array(PDO::ATTR_PERSISTENT => true));
Notice the absence of dbname= in the construction arguments.
When you connect to MySQL via a terminal or other tool, the database name is not needed off the bat. You can switch between databases by using the USE dbname statement via the PDO::exec() method.
$con->exec("USE someDatabase");
$con->exec("USE anotherDatabase");
Of course you may want to wrap this in a catch try statement.
You might be able to use MySQLi syntax, which would allow you to handle it better.
Define the database connections, then whenever you want to query one of the database, specify the right connection.
E.g.:
$Db1 = new mysqli('$DB_HOST','USERNAME','PASSWORD'); // 1st database connection
$Db2 = new mysqli('$DB_HOST','USERNAME','PASSWORD'); // 2nd database connection
Then to query them on the same page, use something like:
$query = $Db1->query("select * from tablename")
$query2 = $Db2->query("select * from tablename")
die("$Db1->error");
Changing to MySQLi in this way will help you.
You don't actually need select_db. You can send a query to two databases at the same time. First, give a grant to DB1 to select from DB2 by GRANT select ON DB2.* TO DB1#localhost;. Then, FLUSH PRIVILEGES;. Finally, you are able to do 'multiple-database query' like SELECT DB1.TABLE1.id, DB2.TABLE1.username FROM DB1,DB2 etc. (Don't forget that you need 'root' access to use grant command)
if you are using mysqli and have two db_connection file. like
first one is
define('HOST','localhost');
define('USER','user');
define('PASS','passs');
define('**DB1**','database_name1');
$connMitra = new mysqli(HOST, USER, PASS, **DB1**);
second one is
define('HOST','localhost');
define('USER','user');
define('PASS','passs');
define(**'DB2**','database_name1');
$connMitra = new mysqli(HOST, USER, PASS, **DB2**);
SO just change the name of parameter pass in mysqli like DB1 and DB2.
if you pass same parameter in mysqli suppose DB1 in both file then second database will no connect any more. So remember when you use two or more connection pass different parameter name in mysqli function
<?php
// Sapan Mohanty
// Skype:sapan.mohannty
//***********************************
$oldData = mysql_connect('localhost', 'DBUSER', 'DBPASS');
echo mysql_error();
$NewData = mysql_connect('localhost', 'DBUSER', 'DBPASS');
echo mysql_error();
mysql_select_db('OLDDBNAME', $oldData );
mysql_select_db('NEWDBNAME', $NewData );
$getAllTablesName = "SELECT table_name FROM information_schema.tables WHERE table_type = 'base table'";
$getAllTablesNameExe = mysql_query($getAllTablesName);
//echo mysql_error();
while ($dataTableName = mysql_fetch_object($getAllTablesNameExe)) {
$oldDataCount = mysql_query('select count(*) as noOfRecord from ' . $dataTableName->table_name, $oldData);
$oldDataCountResult = mysql_fetch_object($oldDataCount);
$newDataCount = mysql_query('select count(*) as noOfRecord from ' . $dataTableName->table_name, $NewData);
$newDataCountResult = mysql_fetch_object($newDataCount);
if ( $oldDataCountResult->noOfRecord != $newDataCountResult->noOfRecord ) {
echo "<br/><b>" . $dataTableName->table_name . "</b>";
echo " | Old: " . $oldDataCountResult->noOfRecord;
echo " | New: " . $newDataCountResult->noOfRecord;
if ($oldDataCountResult->noOfRecord < $newDataCountResult->noOfRecord) {
echo " | <font color='green'>*</font>";
} else {
echo " | <font color='red'>*</font>";
}
echo "<br/>----------------------------------------";
}
}
?>
I have two databases, one online (mysql) and one in my office (SQL Server) which I would like to compare and update where a value is different.
I am using php to connect to the SQL Server database and run a query to retrieve the information, then connecting to the Mysql database running a query. Then I need to compare the two queries and update where necessary.
Is there somewhere I can look for tips on how to do this, I am sketchy on PHP and struggling really.
This is as far as I have got-:
<?php
$Server = "**server**";
$User = "**user**";
$Pass = "**password**";
$DB = "**DB**";
//connection to the database
$dbhandle = mssql_connect($Server, $User, $Pass)
or die("Couldn't connect to SQL Server on $Server");
//select a database to work with
$selected = mssql_select_db($DB, $dbhandle)
or die("Couldn't open database $DB");
//declare the SQL statement that will query the database
$query = "SELECT p.id, p.code, ps.onhand";
$query .= "FROM products p with(nolock)";
$query .= "INNER JOIN productstockonhanditems ps with(nolock)";
$query .= "ON ps.ProductID = p.ID";
$query .= "WHERE ps.StockLocationID = 1";
//execute the SQL query and return records
$get_offlineproduct2 = mssql_query($query);
mysql_connect("**Host**", "**username**", "**password**") or die(mysql_error());
mysql_select_db("Database_Name") or die(mysql_error());
$get_onlineproducts = mysql_query(SELECT w.ob_sku, w.quantity
FROM product_option_value AS w
ORDER BY ob_sku)
or die(mysql_error());
//close the connection
mssql_close($dbhandle);
?>
I am looking to compare the value p.code to w.ob_sku and whenever they match copy the value of ps.onhand to w.quantity so the online database has the correct quantities from the office database.
My question I guess is how close am I to getting this right? Also am I doing this the right way, I don't want to get so far and realise that i am just wasting my time...
Thanks!
You do not need to fetch any record from MySQL, since you actually want to update it.
I would do something like this:
$query = 'SELECT p.code, ps.onhand FROM (...)';
// execute the SQL query and return a result set
// mssql_query() actually returns a resource
// that you must iterate with (e.g.) mssql_fetch_array()
$mssqlResult = mssql_query($query);
// connect to the MySQL database
mysql_connect("**Host**", "**username**", "**password**") or die(mysql_error());
mysql_select_db("Database_Name") or die(mysql_error());
while ( $mssqlRow = mssql_fetch_array($mssqlResult) ) {
$mssqlCode = $mssqlRow['code'];
$mssqlOnHand = $mssqlRow['onhand'];
mysql_query(
"UPDATE product_option_value SET quantity = $mssqlOnHand WHERE ob_sku = $mssqlCode"
// extra quotes may be required around $mssqlCode depending on the column type
);
}
I need to perform a simply query.
Literally, all I need to perform is:
SELECT price, sqft, zipcode FROM homes WHERE home_id = X
When I use PHP PDO, which I've read is the recommended way to connect to a MySQL database, simply creating the connection takes a measured 610ms.
My code is below:
try {
$conn_str = DB . ':host=' . DB_HOST . ';dbname=' . DB_NAME;
$dbh = new PDO($conn_str, DB_USERNAME, DB_PASSWORD);
$params = array();
$sql = 'SELECT price, sqft, zipcode FROM homes WHERE home_id = :home_id';
$params[':home_id'] = X;
$stmt = $dbh->prepare($sql);
$stmt->execute($params);
$result_set = $stmt->fetchAll(PDO::FETCH_ASSOC);
// json output
ob_start("ob_gzhandler");
header('Content-type: text/javascript; charset=utf-8');
print "{'homes' : ";
print json_encode( $result_set );
print '}';
ob_end_flush();
$dbh = null;
} catch (PDOException $e) {
die('Unable to connect');
}
Question: What's the fastest way for me to connect to my MySQL database to perform the query above?
If the slowness is due to having to reach over the network for each connection, and mysql having to do a reverse DNS lookup to check through its GRANTs table, then that overhead could very well account for a large chunk of the latency. Switching to persistent connections would make it a one-time cost for the life of the connection.
However, this does lead to othe problems. Since transactions are rolled back and locks released when the connection holding them is closed, going persitent means they'll stay active. Without taking great care in your code to not leave the connection in an inconsistent state, you could very well create a deadlock or at least lock out all other connections until you go in manually and clean up.
Fastest possible :
mysqli_connect("servername", "user", "pass") or die("can't connect");
mysqli_select_db("dbname") or die("can't select database");
list($price, $sqft, $zipcode) = mysqli_fetch_array(mysqli_query("SELECT price, sqft, zipcode FROM homes WHERE home_id = ".mysqli_real_escape_string($home_id)));
[EDIT]: Now using mysqli instead of mysql.
Guess PDO is as fast as MYSQLI.
I think your problem is how you connect with PDO.
Propably your connectionstring looks like:
:host=localhost;:dbname=foo
And there is the problem... PDO tries to connect to localhost but PDO uses the DNS to turn localhost into 127.0.0.1 and this is what costs time.
If you use 127.0.0.1 directly you wont have these problems anymore :)
So the connectionstring must look like
:host=127.0.0.1;:dbname=bar
as of version php 5.3.0 the fastest and most lightweight way of calling into the db from php is as follows:
This example uses the mysql/ext (not mysqli) and calls stored procedures
$conn = mysql_connect("localhost", "user", "pass");
mysql_select_db("db");
$sql = sprintf("call get_user(%d)", 1);
$result = mysql_query($sql);
mysql_free_result($result);
mysql_close($conn);
The stored procedure:
delimiter #
create procedure get_user
(
in p_user_id int unsigned
)
begin
select
u.user_id, u.username, u.status_id, s.name as status_name, ...
from
users u
inner join user_status s on u.status_id = s.status_id
...
where
u.user_id = p_user_id;
end #
delimiter ;
I have information spread out across a few databases and want to put all the information onto one webpage using PHP. I was wondering how I can connect to multiple databases on a single PHP webpage.
I know how to connect to a single database using:
$dbh = mysql_connect($hostname, $username, $password)
or die("Unable to connect to MySQL");
However, can I just use multiple "mysql_connect" commands to open the other databases, and how would PHP know what database I want the information pulled from if I do have multiple databases connected.
Warning : mysql_xx functions are deprecated since php 5.5 and removed since php 7.0 (see http://php.net/manual/intro.mysql.php), use mysqli_xx functions or see the answer below from #Troelskn
You can make multiple calls to mysql_connect(), but if the parameters are the same you need to pass true for the '$new_link' (fourth) parameter, otherwise the same connection is reused. For example:
$dbh1 = mysql_connect($hostname, $username, $password);
$dbh2 = mysql_connect($hostname, $username, $password, true);
mysql_select_db('database1', $dbh1);
mysql_select_db('database2', $dbh2);
Then to query database 1 pass the first link identifier:
mysql_query('select * from tablename', $dbh1);
and for database 2 pass the second:
mysql_query('select * from tablename', $dbh2);
If you do not pass a link identifier then the last connection created is used (in this case the one represented by $dbh2) e.g.:
mysql_query('select * from tablename');
Other options
If the MySQL user has access to both databases and they are on the same host (i.e. both DBs are accessible from the same connection) you could:
Keep one connection open and call mysql_select_db() to swap between as necessary. I am not sure this is a clean solution and you could end up querying the wrong database.
Specify the database name when you reference tables within your queries (e.g. SELECT * FROM database2.tablename). This is likely to be a pain to implement.
Also please read troelskn's answer because that is a better approach if you are able to use PDO rather than the older extensions.
If you use PHP5 (And you should, given that PHP4 has been deprecated), you should use PDO, since this is slowly becoming the new standard. One (very) important benefit of PDO, is that it supports bound parameters, which makes for much more secure code.
You would connect through PDO, like this:
try {
$db = new PDO('mysql:dbname=databasename;host=127.0.0.1', 'username', 'password');
} catch (PDOException $ex) {
echo 'Connection failed: ' . $ex->getMessage();
}
(Of course replace databasename, username and password above)
You can then query the database like this:
$result = $db->query("select * from tablename");
foreach ($result as $row) {
echo $row['foo'] . "\n";
}
Or, if you have variables:
$stmt = $db->prepare("select * from tablename where id = :id");
$stmt->execute(array(':id' => 42));
$row = $stmt->fetch();
If you need multiple connections open at once, you can simply create multiple instances of PDO:
try {
$db1 = new PDO('mysql:dbname=databas1;host=127.0.0.1', 'username', 'password');
$db2 = new PDO('mysql:dbname=databas2;host=127.0.0.1', 'username', 'password');
} catch (PDOException $ex) {
echo 'Connection failed: ' . $ex->getMessage();
}
I just made my life simple:
CREATE VIEW another_table AS SELECT * FROM another_database.another_table;
hope it is helpful... cheers...
Instead of mysql_connect use mysqli_connect.
mysqli is provide a functionality for connect multiple database at a time.
$Db1 = new mysqli($hostname,$username,$password,$db_name1);
// this is connection 1 for DB 1
$Db2 = new mysqli($hostname,$username,$password,$db_name2);
// this is connection 2 for DB 2
Try below code:
$conn = mysql_connect("hostname","username","password");
mysql_select_db("db1",$conn);
mysql_select_db("db2",$conn);
$query1 = "SELECT * FROM db1.table";
$query2 = "SELECT * FROM db2.table";
You can fetch data of above query from both database as below
$rs = mysql_query($query1);
while($row = mysql_fetch_assoc($rs)) {
$data1[] = $row;
}
$rs = mysql_query($query2);
while($row = mysql_fetch_assoc($rs)) {
$data2[] = $row;
}
print_r($data1);
print_r($data2);
$dbh1 = mysql_connect($hostname, $username, $password);
$dbh2 = mysql_connect($hostname, $username, $password, true);
mysql_select_db('database1', $dbh1);
mysql_select_db('database2',$dbh2);
mysql_query('select * from tablename', $dbh1);
mysql_query('select * from tablename', $dbh2);
This is the most obvious solution that I use but just remember, if the username / password for both the database is exactly same in the same host, this solution will always be using the first connection. So don't be confused that this is not working in such case. What you need to do is, create 2 different users for the 2 databases and it will work.
Unless you really need to have more than one instance of a PDO object in play, consider the following:
$con = new PDO('mysql:host=localhost', $username, $password,
array(PDO::ATTR_PERSISTENT => true));
Notice the absence of dbname= in the construction arguments.
When you connect to MySQL via a terminal or other tool, the database name is not needed off the bat. You can switch between databases by using the USE dbname statement via the PDO::exec() method.
$con->exec("USE someDatabase");
$con->exec("USE anotherDatabase");
Of course you may want to wrap this in a catch try statement.
You might be able to use MySQLi syntax, which would allow you to handle it better.
Define the database connections, then whenever you want to query one of the database, specify the right connection.
E.g.:
$Db1 = new mysqli('$DB_HOST','USERNAME','PASSWORD'); // 1st database connection
$Db2 = new mysqli('$DB_HOST','USERNAME','PASSWORD'); // 2nd database connection
Then to query them on the same page, use something like:
$query = $Db1->query("select * from tablename")
$query2 = $Db2->query("select * from tablename")
die("$Db1->error");
Changing to MySQLi in this way will help you.
You don't actually need select_db. You can send a query to two databases at the same time. First, give a grant to DB1 to select from DB2 by GRANT select ON DB2.* TO DB1#localhost;. Then, FLUSH PRIVILEGES;. Finally, you are able to do 'multiple-database query' like SELECT DB1.TABLE1.id, DB2.TABLE1.username FROM DB1,DB2 etc. (Don't forget that you need 'root' access to use grant command)
if you are using mysqli and have two db_connection file. like
first one is
define('HOST','localhost');
define('USER','user');
define('PASS','passs');
define('**DB1**','database_name1');
$connMitra = new mysqli(HOST, USER, PASS, **DB1**);
second one is
define('HOST','localhost');
define('USER','user');
define('PASS','passs');
define(**'DB2**','database_name1');
$connMitra = new mysqli(HOST, USER, PASS, **DB2**);
SO just change the name of parameter pass in mysqli like DB1 and DB2.
if you pass same parameter in mysqli suppose DB1 in both file then second database will no connect any more. So remember when you use two or more connection pass different parameter name in mysqli function
<?php
// Sapan Mohanty
// Skype:sapan.mohannty
//***********************************
$oldData = mysql_connect('localhost', 'DBUSER', 'DBPASS');
echo mysql_error();
$NewData = mysql_connect('localhost', 'DBUSER', 'DBPASS');
echo mysql_error();
mysql_select_db('OLDDBNAME', $oldData );
mysql_select_db('NEWDBNAME', $NewData );
$getAllTablesName = "SELECT table_name FROM information_schema.tables WHERE table_type = 'base table'";
$getAllTablesNameExe = mysql_query($getAllTablesName);
//echo mysql_error();
while ($dataTableName = mysql_fetch_object($getAllTablesNameExe)) {
$oldDataCount = mysql_query('select count(*) as noOfRecord from ' . $dataTableName->table_name, $oldData);
$oldDataCountResult = mysql_fetch_object($oldDataCount);
$newDataCount = mysql_query('select count(*) as noOfRecord from ' . $dataTableName->table_name, $NewData);
$newDataCountResult = mysql_fetch_object($newDataCount);
if ( $oldDataCountResult->noOfRecord != $newDataCountResult->noOfRecord ) {
echo "<br/><b>" . $dataTableName->table_name . "</b>";
echo " | Old: " . $oldDataCountResult->noOfRecord;
echo " | New: " . $newDataCountResult->noOfRecord;
if ($oldDataCountResult->noOfRecord < $newDataCountResult->noOfRecord) {
echo " | <font color='green'>*</font>";
} else {
echo " | <font color='red'>*</font>";
}
echo "<br/>----------------------------------------";
}
}
?>