error in inserting record

ravi951's Avatar, Join Date: Aug 2011
Go4Expert Member
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 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);
<?
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());
}
}

Last edited by shabbir; 17Aug2011 at 13:42.. Reason: Code blocks
0
Webdeveloper's Avatar, Join Date: Jun 2011
Go4Expert Member
Hi,

This is because of the syntax error in the SQL query being generated. The best way to identify the error is to echo the query generated on the screen and you will be able to see the error.

Hope this helps.

Cheers,

~Maneet
0
pein87's Avatar
Ambitious contributor
I think you need to get the basics of php before you even continue on any further with this project. Your doing so many things wrong its not even funny. Try http://www.w3schools.com/php/default.asp and once you understand the basics come back and do this basic project. Your programming foundation needs to be sturdy before you can attempt to create programs in any language.