Inserting values into database using object oriented programming [duplicate]

Question

Possible Duplicate:
How to successfully rewrite old mysql-php code with deprecated mysql_* functions?

I have not been able to insert the values into the databse . However I do not get any error message as well .

<html>
<body>
<form action="database.php" method="post">
Name : <input type ="text" name = "name"/>
Number  :<input type ="text" name = "number"/>
<input type ="submit" value = "submit"/>
</form>
</body>
</html>

database.php

<?php
class Database
{
    var $host;
    var $user;
    var $pass;
    var $data;
    var $con;
    var $table;
    var $db;

    public function controls()
    {
        $this->host="localhost";
        $this->user="cgiadmin";
        $this->pass="cgi";
        $this->data="j2";
    }

    public function connection()
    {
        $this->con="mysql_connect($this->host,$this->user,$this->pass)";
    }
    public function tablename()
    {
        $this->table="Insert into employee(name,number) values('$_POST[name]','$_POST[number]')";
    }
    public function databaseconnection()
    {
        $this->db="mysql_select_db($this->data,$this->con)";
    }

}
$name=new Database;
$name->connection();
if(!($name->con))
{
    echo "'Error: ' . mysql_error()";
}

$name->databaseconnection();
$name->tablename();

echo "thanks for taking the survey";

?>

@Blender I am new to this language . So finding it difficult to deal with it :( — Jathin
– Jathin, Commented Jun 14, 2012 at 6:34

Vaishu · Accepted Answer · 2012-06-14 07:03:01Z

2

Try this...

Modify tablename() function

public function tablename($nam,$num)
    {
        $this->table=mysql_query("INSERT INTO employee(name,number) VALUES ('$nam','$num')");
    }

Get values and call tablename() function

$name=new Database;
$name->connection();
if(!($name->con))
{
    echo "'Error: ' . mysql_error()";
}

$name->databaseconnection();

$nam=$_POST[name];
$num=$_POST[number];
$name->tablename($nam,$num);

echo "thanks for taking the survey";

answered Jun 14, 2012 at 7:03

Vaishu

2,3733 gold badges23 silver badges25 bronze badges

Add a comment |

hakre · Accepted Answer · 2012-06-14 21:48:21Z

-3

You need some magic :) Read about quotes in php

<?php
class Database
{
    var $host;
    var $user;
    var $pass;
    var $data;
    var $con;
    var $table;
    var $db;

    public function controls()
    {
        $this->host="localhost";
        $this->user="cgiadmin";
        $this->pass="cgi";
        $this->data="j2";
    }

    public function connection()
    {
        $this->con = mysql_connect($this->host,$this->user,$this->pass);
    }
    public function tablename()
    {
        $this->table=mysql_query("INSERT INTO employee(name,number) VALUES ('".$_POST[name]."','".$_POST[number]."')");
    }
    public function databaseconnection()
    {
        $this->db=mysql_select_db($this->data,$this->con);
    }

}
$name=new Database();
$name->controls();
$name->connection();
if(!($name->con))
{
    echo 'Error: ' . mysql_error();
}

$name->databaseconnection();
$name->tablename();

echo "thanks for taking the survey";

?>

edited Jun 14, 2012 at 21:48

hakre

199k55 gold badges453 silver badges865 bronze badges

answered Jun 14, 2012 at 6:16

Ruben Nagoga

2,2182 gold badges20 silver badges32 bronze badges

5

Holy SQL injection vulnerability, Batman!

Michael Borgwardt
– Michael Borgwardt

06/14/2012 06:21:19
Commented Jun 14, 2012 at 6:21
@Ruben Nagoga I am getting an error message . Is it a problem with my database ? Error message is : Warning: mysql_connect() [function.mysql-connect]: Access denied for user 'www-data'@'localhost' (using password: NO) in /home/jnagesh/public_html/SampleSurvey/MVC/database.php on line 20 Error: Access denied for user 'www-data'@'localhost' (using password: NO) Warning: mysql_select_db(): supplied argument is not a valid MySQL-Link resource in /home/jnagesh/public_html/SampleSurvey/MVC/database.php on line 28

Jathin
– Jathin

06/14/2012 06:29:59
Commented Jun 14, 2012 at 6:29
Warning: mysql_query() [function.mysql-query]: Access denied for user 'www-data'@'localhost' (using password: NO) in /home/jnagesh/public_html/SampleSurvey/MVC/database.php on line 24 Warning: mysql_query() [function.mysql-query]: A link to the server could not be established in /home/jnagesh/public_html/SampleSurvey/MVC/database.php on line 24 thanks for taking the survey

Jathin
– Jathin

06/14/2012 06:30:21
Commented Jun 14, 2012 at 6:30
Is this a problem in program or an error in the database ?

Jathin
– Jathin

06/14/2012 06:30:44
Commented Jun 14, 2012 at 6:30
@MichaelBorgwardt Could you please help me correct the error

Jathin
– Jathin

06/14/2012 06:54:14
Commented Jun 14, 2012 at 6:54

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