Inserting data in multiple databases

Question

I need to enter data from a single html form in two tables which are created in two different databases, and I need to do this on single submit any suggestions what can be done for achieving this.

I have the following code:

<?php

echo "Entering";

$one= mt_rand(1000000000,9999999999);
$two= mt_rand(1000,9999);


echo "<br><br>getting values";

$user= mt_rand(1000000000,9999999999);
$useralias = $one.$two;
$first= $_POST['first_name'];
$last= $_POST['last_name'];
$email=$_POST['email'];
$country=$_POST['country'];
$city = $_POST['city'];
$zipcode = $_POST['zipcode'];
$address = $_POST['address'];
$phone = $_POST['phone'];
$fax = $_POST['fax'];
$website= $_POST['website'];
$company= $_POST['company'];

echo $first;
echo "<br>".$last;

echo "<br><br>setting database etc one";

$host = "localhost";
$database = "mya2billing";
$table = "cc_card";
$username = "root";
$password = "mehusnain";

echo "<br><br>executing query one";

$con = mysql_connect($host , $username, $password );
if(!$con){
echo "Connection failed";
}
else{
mysql_select_db($database);
$query = "INSERT INTO $table (username, useralias, firstname, lastname, email, country,   city, zipcode, address, phone, fax, company_name, company_website) VALUES   ('$user','$useralias','$first','$last','$email','$country','$city','$zipcode','$address','$ phone','$fax','$website','$company')";

echo $query;

if(mysql_query($query)){
echo "<br><br>Insertion done in $table";
$con.close();
}
else{
echo "<br><br>Failed in $table";
$con.close();
}
}

echo "<br><br>setting databse 2 etc";


$host = "localhost";
$database = "voixe";
$table = "hak_users";
$username = "root";
$password = "mehusnain";

echo "<br><br>executing query 2";

$con = mysql_connect($host , $username, $password );
if(!$con){
echo "Connection failed";
}
else{
mysql_select_db($database);
$query = "INSERT INTO $table (user_login, user_pass, user_nicename, user_email,     display_name) VALUES ('$user','password','$first." ".$last','$email','$first')";

echo $query;

if(mysql_query($query)){
echo "<br><br>Insertion done in $table";
$con.close();
}
else{

echo "

Failed in $table"; $con.close(); } }

?>

One thing more that is not a single echo statement is working.....

just a heads up, I'd move away from using mysql_ as it's no longer supported. Aim to use the MySQLi extension instead - stackoverflow.com/questions/8891443/… — Sam Holguin
– Sam Holguin, Commented Apr 23, 2014 at 10:31

Vandana Pareek · Accepted Answer · 2014-04-23 11:14:23Z

2

Please check,following code may help you.

$con1 = mysql_connect('localhost', 'user1', 'pass1');
$rv1 = mysql_select_db('db1', $con1);
if(!$con1){
  echo "Connection failed";
}
else{
  mysql_query("INSERT INTO test (name) VALUES('ABC')");
  mysql_close($con1);
}


$con2 = mysql_connect('localhost', 'user2', 'pass2');
$rv2 = mysql_select_db('db2', $con2);
if(!$con2){
  echo "Connection failed";
}
else{
  mysql_query("INSERT INTO test (name) VALUES('ABC')");
  mysql_close($con2);
}

answered Apr 23, 2014 at 11:14

Vandana Pareek

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1 Comment

wild Over a year ago

if i want to create connection only once from both database and execute quries several time with both of them then?

SagarPPanchal · Accepted Answer · 2014-04-23 11:22:09Z

1

Use dbname.tblname with insert query like

$con1 = mysql_connect('localhost', 'user1', 'pass1');
$rv1 = mysql_select_db('db1', $con1);
if(!$con1){
  echo "Connection failed";
}
else{
  mysql_query("INSERT INTO db1.test (name) VALUES('ABC')");
  mysql_close($con1);
}


$con2 = mysql_connect('localhost', 'user2', 'pass2');
$rv2 = mysql_select_db('db2', $con2);
if(!$con2){
  echo "Connection failed";
}
else{
  mysql_query("INSERT INTO db2.test (name) VALUES('ABC')");
  mysql_close($con2);
}

answered Apr 23, 2014 at 11:22

SagarPPanchal

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Comments

Muhammad Bilal · Accepted Answer · 2015-05-29 10:51:13Z

1

  enter code here`$conn1 = mysql_select_db('db1');
  mysql_open($conn1);
 insert query ->>>
 mysql_close($con1);



 mysql_close($conn2);
$conn2 = mysql_select_db('db2');
insert query ->>>
 mysql_close($conn2);

answered May 29, 2015 at 10:51

Muhammad Bilal

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Comments

mickmackusa · Accepted Answer · 2021-07-28 11:27:00Z

I guess this should help. But not tested.

ADVICE: Avoid using mysql_* statement as they are deprecated in recent PHP versions. Learn mysqli_* prepared or PDO and start implementing.

<?php
if (isset($_POST['submit'])) {

// FIRST DB

$con1 = new mysqli('localhost', 'user', 'password', 'db1');

/* check connection */
if (mysqli_connect_errno()) {
   printf("Connect failed: %s\n", mysqli_connect_error());
   exit();
}

$stmt = $mysqli->prepare("INSERT INTO SampleTable VALUES (?)");
$stmt->bind_param('s', $sample);   // bind $sample to the parameter

// escape the POST data for added protection
$sample = isset($_POST['sample'])
      ? $mysqli->real_escape_string($_POST['sample'])
      : '';

/* execute prepared statement */
$stmt->execute();

printf("%d Row inserted.\n", $stmt->affected_rows);

/* close statement and connection */
$stmt->close();

/* close connection */
$mysqli->close();



//  SECOND DB

$con2 = new mysqli('localhost', 'user', 'password', 'db2');

/* check connection */
if (mysqli_connect_errno()) {
   printf("Connect failed: %s\n", mysqli_connect_error());
   exit();
}

$stmt = $mysqli->prepare("INSERT INTO SampleTable VALUES (?)");
$stmt->bind_param('s', $sample);   // bind $sample to the parameter

// escape the POST data for added protection
$sample = isset($_POST['sample'])
      ? $mysqli->real_escape_string($_POST['sample'])
      : '';

/* execute prepared statement */
$stmt->execute();

printf("%d Row inserted.\n", $stmt->affected_rows);

/* close statement and connection */
$stmt->close();

/* close connection */
$mysqli->close();


  }
?>

Muhammad H Tahir · Accepted Answer · 2014-04-23 12:37:50Z

0

Thank you all for your answers but i figured it out by myself. U just removed $con.close() from 1st insertion query and removed some concatination from 2nd query and it worked for me atleast.

edited Apr 23, 2014 at 12:37

user1907906

answered Apr 23, 2014 at 11:49

Muhammad H Tahir

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1 Comment

Vandana Pareek Over a year ago

yes i've figured it out too($con.close())..but forgot to mention ;)

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