1. We have moved from vBulletin to XenForo and you are viewing the site in the middle of the move. Though the functional aspect of everything is working fine, we are still working on other changes including the new design on Xenforo.
    Dismiss Notice

error in inserting record

Discussion in 'PHP' started by ravi951, Aug 17, 2011.

  1. ravi951

    ravi951 New 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.....
    ="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";
    // Count table rows 
    <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>
    <input type="submit" name="Submit" value="Submit">
    // Get values from form 
    // Check if button name "Submit" is active, do this 
    //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 a moderator: Aug 17, 2011
  2. Webdeveloper

    Webdeveloper New Member


    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.


  3. pein87

    pein87 New Member

    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.

Share This Page