error in inserting record

**ravi951** · 08-17-2011, 05:06 AM

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());
}
}

**JShor** · 08-17-2011, 05:21 AM

Fixed. You put in a <?php in the middle of your PHP execution, throwing an error.

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);

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());
}
}

**ravi951** · 08-17-2011, 05:30 AM

even now also it is displaying error.
unable to insert values in textbox......

**traq** · 08-17-2011, 06:08 AM

Originally Posted by ravi951

even now also it is displaying error.
unable to insert values in textbox......

your statement is unclear. Are you still receiving the same error? same error, different location? different error?

( saying "it doesn't work", without elaborating, is very, very un-useful. do your absolute best to figure out what's going on, and contribute to the solution, before throwing it back as "still not working." )