connect to sql database with php - php

Question

Im trying to connect to sql database with PHP.For some reason when i run the code i am getting below error.
Parse error: syntax error, unexpected '$result' (T_VARIABLE) in C:\xampp2\htdocs
\tutorials\abb.php on line 13
My Code is this...
$user = 'root' ;
$pass = 'xcvsdffd' ;
$db = 'testdb' ;
$con = new mysqli('localhost', $user , $pass , $db) or die("UNABLE TO CONNECT");
$selected = mysql_select_db($db,con)
$result = mysql_query("SELECT * FROM test");
while ($row = mysql_fetch_array($result)) {
echo "ID:".$row{'id'}." Name:".$row{'name33'}."
".$row{'year'}."<br>";
}
//close the connection and recordset objects freeing up resources
$result->Close();
$con->Close();
$result = null;
$con = null;
?>

Answer 1

There are two reasons which was the reason for your errors..
1.
You missed a semicolon ; and a $ sign. Below is valid one.
$selected = mysql_select_db($db,$con); //Replace this with your existing line.
2.
You are mixing mysql_* and mysqli_*
Sidenote:
This(mysql_*) extension is deprecated as of PHP 5.5.0, and will be removed in the future. Instead, the MySQLi or PDO_MySQL extension should be used. Switching to PreparedStatements is even more better to ward off SQL Injection attacks !
The PDO way...
<?php
$dsn = 'mysql:dbname=testdb;host=localhost';
$user = 'root';
$password = 'xcvsdffd';
try
{
$dbh = new PDO($dsn, $user, $password ,array(PDO::MYSQL_ATTR_INIT_COMMAND => 'SET NAMES utf8'));
$dbh->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
}
catch (PDOException $e)
{
echo 'Connection failed: ' . $e->getMessage();
}
For more information.. read the PHP Manual. or use PreparedStatements in MySQLi

Answer 2

Mysqli works with MySQL version 4.1.13 or newer; if your version is old then u should connect your db the old way like this:
$con=mysql_connect("localhost","root","pass");
mysql_select_db("database_name",$con);

Answer 3

They always tell me that MySQL command is now depreciated or something. That's why it's a good practice to use PDO or MySQLI. It's easy to use, if you'll appreciate it's use.
Now going to your code:
$connection = new mysqli("localhost","user","password","database name");
$selectDatabase = mysqli_select_db($connection,"database name");
$connection->query ("SELECT * FROM test");
Hope this helps

Answer 4

Try this below , if you are creating mysqli connection then you dont have to use mysql_select_db($db,con) for again select database. or use semicolon after this line
<?php
$mysqli = new mysqli("localhost", "my_user", "my_password", "test");
/* check connection */
if (mysqli_connect_errno()) {
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}
/* return name of current default database */
if ($result = $mysqli->query("SELECT DATABASE()")) {
$row = $result->fetch_row();
printf("Default database is %s.\n", $row[0]);
$result->close();
}
/* change db to world db */
$mysqli->select_db("world");
/* return name of current default database */
if ($result = $mysqli->query("SELECT DATABASE()")) {
$row = $result->fetch_row();
printf("Default database is %s.\n", $row[0]);
$result->close();
}
$mysqli->close();
?>
Reference

Answer 5

there might be a semicolon missing in a line .
mysql_select_db($db,$con);

Answer 6

Try this
<?php
$username = 'root';
$pass = 'pass';
$db = 'db';
$link = mysqli_connect('localhost',$username,$pass,$db);
if(!link){
echo "Not Connected to DB" . mysqli_connect_error();
$result = mysqli_query($link, "SELECT * FROM test");
while($row = mysqli_fetch_array($result)){
echo "ID: " . $row['id'];
echo "Name: " . $row['name33'];
echo "Year: " . $row['year'];
}
?>
Querying in PHP must be consistent, meaning if you use mysqli you should use it throughout the PHP file, and if you use mysqli you must also use it throughout the PHP file. mysqli and mysql also has different code constructions. So search on the web for their proper construction. Hope this helps