Error in inserting record

hi all i have been inserting records in to database using php using the below program which is saved as “insert.php”.
but it is displaying the following error
Parse error: syntax error, unexpected ‘<’ in C:\xampp\htdocs\insert1.php on line 18

dont no what went wrong.
can u tell me how to solve it…

<?php
$host="localhost"; // Host name 
$username="root"; // Mysql username 
$password=""; // Mysql password 
$db_name="test"; // Database name 
$tbl_name="emp"; // Table name

// Connect to server and select databse.
mysql_connect("$host", "$username", "$password")or die("cannot connect"); 
mysql_select_db("$db_name")or die("cannot select DB");

$sql="SELECT * FROM $tbl_name";
$result=mysql_query($sql);

// Count table rows 
$count=mysql_num_rows($result);

<?php
while($rows=mysql_fetch_array($result))
{
?>
<tr>
<td align="center" bgcolor="#FFFFFF"><input name="checkbox[]" type="checkbox" id="checkbox[]" value="y" /></td>
<td align="center"><input name="empno[]" type="text" id="name" value="<? echo $rows['empno'];?>"></td>
<td align="center"><input name="empname[]" type="text" id="empname" value="<? echo $rows['empname'];?>"></td>
<td align="center"><input name="desig[]" type="text" id="desig" value="<? echo $rows['desig'];?>"></td>
</tr>

<?php
}
?>
<input type="submit" name="Submit" value="Submit">

<?php

// Get values from form 
$no=$_POST['empno'];
$name=$_POST['empname'];
$desig=$_POST['desig'];

// Check if button name "Submit" is active, do this 
if(array_key_exists('Submit', $_POST))
{
for($i=0;$i<count($count);$i++)
{
     //protect form sql injection
    $a = (int) $_POST['empno'][$i]; 
    $b = mysql_real_escape_string( $_POST['empname'][$i] ); 
    $c = mysql_real_escape_string( $_POST['desig'][$i] ); 
 //read the query
 $sql="INSERT INTO '$tbl_name' (empno, empname, desig) VALUES('{$a}', '{$b}', '{$c}')";
 mysql_query($sql) or die(mysql_error());
}
}

You have an opening PHP tag where it shouldn’t be, find the following code

<?php
while($rows=mysql_fetch_array($result))

and replace it with

while($rows=mysql_fetch_array($result))

i have modified the code like u said above but it is also not executing…
below is the modified code…


<?php
$host="localhost"; // Host name 
$username="root"; // Mysql username 
$password=""; // Mysql password 
$db_name="test"; // Database name 
$tbl_name="emp"; // Table name

// Connect to server and select databse.
mysql_connect("$host", "$username", "$password")or die("cannot connect"); 
mysql_select_db("$db_name")or die("cannot select DB");

$sql="SELECT * FROM $tbl_name";
$result=mysql_query($sql);

// Count table rows 
$count=mysql_num_rows($result);
while($rows=mysql_fetch_array($result))
{
?>
<tr>
<td align="center" bgcolor="#FFFFFF"><input name="checkbox[]" type="checkbox" id="checkbox[]" value="y" /></td>
<td align="center"><input name="empno[]" type="text" id="name" value="<? echo $rows['empno'];?>"></td>
<td align="center"><input name="empname[]" type="text" id="empname" value="<? echo $rows['empname'];?>"></td>
<td align="center"><input name="desig[]" type="text" id="desig" value="<? echo $rows['desig'];?>"></td>
</tr>

<?php
}
?>
<input type="submit" name="Submit" value="Submit">

<?php

// Get values from form 
$no=$_POST['empno'];
$name=$_POST['empname'];
$desig=$_POST['desig'];

// Check if button name "Submit" is active, do this 
if(array_key_exists('Submit', $_POST))
{
for($i=0;$i<count($count);$i++)
{
     //protect form sql injection
    $a = (int) $_POST['empno'][$i]; 
    $b = mysql_real_escape_string( $_POST['empname'][$i] ); 
    $c = mysql_real_escape_string( $_POST['desig'][$i] ); 
 //read the query
 $sql="INSERT INTO '$tbl_name' (empno, empname, desig) VALUES('{$a}', '{$b}', '{$c}')";
 mysql_query($sql) or die(mysql_error());
}
}


The syntax seems OK now, what’s the error that you’re getting?

my database name is “test”.my table name is “emp”.
emp table contains 3 fields as shown below…

empno empname desig

1111 raju pilot
1112 ram chef
1113 ramu doctor
1114 paul engineer
1115 ajay player

now i need to add values in to database using php.after entering values for 3 fields and click on the submit button it should be updated on the table…
tell me how to do it…i dont need check box for inserting the values…
below is the code…
what to modify in the below code for getting the above requirements…


<?php
$host="localhost"; // Host name 
$username="root"; // Mysql username 
$password=""; // Mysql password 
$db_name="test"; // Database name 
$tbl_name="emp"; // Table name

// Connect to server and select databse.
mysql_connect("$host", "$username", "$password")or die("cannot connect"); 
mysql_select_db("$db_name")or die("cannot select DB");

$sql="SELECT * FROM $tbl_name";
$result=mysql_query($sql);

// Count table rows 
$count=mysql_num_rows($result);
while($rows=mysql_fetch_array($result))
{
?>
<tr>
<td align="center" bgcolor="#FFFFFF"></td>
<td align="center"><input name="empno[]" type="text" id="name" value="<? echo $rows['empno'];?>"></td>
<td align="center"><input name="empname[]" type="text" id="empname" value="<? echo $rows['empname'];?>"></td>
<td align="center"><input name="desig[]" type="text" id="desig" value="<? echo $rows['desig'];?>"></td>
</tr>

<?php
}
?>
<input type="submit" name="Submit" value="Submit">

<?php

// Get values from form 
$no=$_POST['empno'];
$name=$_POST['empname'];
$desig=$_POST['desig'];

// Check if button name "Submit" is active, do this 
if(array_key_exists('Submit', $_POST))
{
for($i=0;$i<count($count);$i++)
{
     //protect form sql injection
    $a = (int) $_POST['empno'][$i]; 
    $b = mysql_real_escape_string( $_POST['empname'][$i] ); 
    $c = mysql_real_escape_string( $_POST['desig'][$i] ); 
 //read the query
 $sql="INSERT INTO '$tbl_name' (empno, empname, desig) VALUES('{$a}', '{$b}', '{$c}')";
 mysql_query($sql) or die(mysql_error());
}
}