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());
}
}
5 replies to this topic
#1
Posted 16 August 2011 - 09:09 PM
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#2
Posted 16 August 2011 - 09:16 PM
You appear to be opening a PHP tag (<?php) within a PHP script, and that is not a valid keyword.
You must close the tag before you open another one:
You must close the tag before you open another one:
?> <?php
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#3
Posted 16 August 2011 - 09:32 PM
here is the modified version. i remove <?php but also not executing.....
<?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());
}
}
#5
Posted 18 August 2011 - 06:47 PM
I don't see any <form> tag in your code...
The data won't be sended if it is not inside a form.
The data won't be sended if it is not inside a form.
#6
Posted 21 August 2011 - 10:56 AM
I think it's better to make two pages, simple one to put your form in and give it action to e.g(insert.php) this page will treat your data and insert them into your database.
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