Salary update

**dcr33** · 12-14-2011, 04:57 AM

I have this program -
To Develop a php application to modify salary of an employee,your application should provide facility to choose an employee from a drop-down list.

I tried with the following code but it cannot change the record in the table "emp" inside the "db1" database. There are 3 pages in total. The table "emp" has 3 fields -emp_id,ename,salary.
Here is the code-
1)emp_select.php-

Code:

<html>
<head>
<title></title>
</head>
<body>
<form action="emp_mod.php" method="post">
Choose an employee <select name="emp">
<?php
$host="localhost";
$user="root";
$passwd="";
$con=mysql_connect($host,$user,$passwd);
mysql_select_db('db1',$con);
$sql="select emp_id,ename from emp";
$result=mysql_query($sql,$con);

?> 
<input type="submit" value="show">
</form>
</body>
</html>

2)emp_mod.php -

Code:

<html>
<head>
<title>Untitled Document</title>
</head>

<body>
<?php
$eid=$_POST['emp'];
$con=mysql_connect('localhost','root','');
mysql_select_db('db1',$con);
$sql="select ename,salary from emp where emp_id='$eid'";
$result=mysql_query($sql,$con);
$rec=mysql_fetch_row($result);
 $enm=$rec[0];
$sal=$rec[1];
?>
<form action="emp_upd.php" mehtod="post">
Emp. ID <input type="text" name="eid" value="<?php echo $eid; ?>" readonly="readonly"><br />
Emp. Name: <input type="text" readonly="readonly" name="enm" value="<?php echo $enm;?>"><br />
Mthly Salary: <input type="text"  name="sal" value="<?php echo $sal;?>"><br />	
<br />
<input type="submit" value="Save">
</form>
</body>
</html>

3)emp_upd.php-

Code:

<?php
$eid=$_POST['eid'];
$sal=$_POST['sal'];
$con=mysql_connect('localhost','root','');
mysql_select_db('db1',$con);
$sql="update emp set salary='$sal' where emp_id='$eid'";
if(mysql_query($sql,$con))
{
echo "Record changed<br>";
}
else
{
 die('Unable to change salary<br>');
 }
 ?>

Where can be the problem in the above code?
Pls help
Thanks in advance

**fobos** · 12-14-2011, 04:12 PM

emp_id,ename,salary

emp_select.php

PHP Code:


<html>

<head>

<title></title>

</head>

<body>

<form action="emp_mod.php" method="post">

Choose an employee 

<select name="emp">

<?php

$con = mysql_connect("localhost","root","");

mysql_select_db("db1", $con);

$result = mysql_query("SELECT * FROM emp");

while($row = mysql_fetch_array($result))  {

  echo "<option value='".$row['emp_id']."'>".$row['ename']."</option>";

 }

mysql_close($con);

?> 

</select>

<input type="submit" value="show">

emp_mod.php

PHP Code:


<html>

<head>

<title>Untitled Document</title>

</head>



<body>

<?php

$emp_id = $_POST['emp'];

echo $emp_id; // Test to see if the emp_id is passed to next page. if so just delete this.

$con = mysql_connect("localhost","root","");

mysql_select_db("db1", $con);

$result = mysql_query("SELECT * FROM emp WHERE emp_id = '$emp_id'");

$row = mysql_fetch_array($result);

?>

<form action="emp_upd.php" mehtod="post">

Emp. ID <input type="text" name="id" value="<?php echo $row['emp_id']; ?>" readonly="readonly"><br />

Emp. Name: <input type="text" readonly="readonly" name="name" value="<?php echo $row['ename']; ?>"><br />

Mthly Salary: <input type="text"  name="salary" value="<?php echo $row['salary']; ?>">    

<br />

<input type="submit" value="Save">

</form>

</body>

</html>

mysql_close($con);

emp_update.php

PHP Code:


$id = $_POST['id'];

$name = $_POST['name'];

$salary = $_POST['salary'];

$con = mysql_connect("localhost","root","");

mysql_select_db("db1", $con);

mysql_query("UPDATE emp SET salary = '$salary' WHERE emp_id = '$id'");

echo "Sucessfuly updated ".$name." 's salary to ".$salary;



mysql_close($con);

?>

I hope this works for you. Im doing this at work, so i apologize if there might be an error. just point it out and i will help you.

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