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Parse error: syntax error, unexpected '$_POST' (T_VARIABLE) in C:\xampp\htdocs\DBINSERT.php on line 11

ksweety2012

By ksweety2012
in PHP Coding Help

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Go to solution Solved by Ch0cu3r,

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ksweety2012 Newbie

ksweety2012

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Posted (edited)
<?PHP

 

$user_name = "ganga";

$password = "gangamma";

$database = "ganga";

$server = "localhost";

 

$conn = mysql_connect($server, $user_name, $password);

 

print "Connection to the Server is now opened";

$sql = 'INSERT INTO `ganga`(`eno`, `ename`) VALUES ('$_POST[eno]','$_POST[ename]')';

mysql_select_db('ganga',$conn);

$retval = mysql_query( $sql, $conn );

echo "Entered data successfully\n";

mysql_close($conn);

?>

 

Getting php error: Parse error: syntax error, unexpected '$_POST' (T_VARIABLE) in C:\xampp\htdocs\DBINSERT.php on line 11

help this please...gangadhara

Edited by ksweety2012
  • Solution
Ch0cu3r Newbie

Ch0cu3r

Posted
  • Solution
Posted

Never use raw $_POST variables in your query. You should sanitize them before using them.

$eno = mysql_real_escape_string($_POST['eno']);
$ename= mysql_real_escape_string($_POST['ename']);
$sql = "INSERT INTO `ganga`(`eno`, `ename`) VALUES ('eno','$ename')";

Another thing you should be doing is to validate them. You should check they exists and are in the format your expect. 

 

NOTE: the mysql_* functions are deprecated, meaning they are not longer supported and could soon be removed complete from future version of PHP. I recommend you to use either PDO or MySQLi instead.

ksweety2012 Newbie

ksweety2012

Posted
  • Author
Posted

Thank you soooo much...... 

 

 

Record updated in mysql db, but eno is saved as 0 instead of 18 (which i enered actual eno.).

 

 

my form.html page code is like this:

 

html>
<body>
A small example page to insert some data in to the MySQL database using PHP
<form action="DBINSERT.php" method="post">
Eno<input type="number" name="eno" /><br><br>
Ename: <input type="text" name="ename" /><br><br>
 
<input type="submit" />
</form>
</body>
</html>
 
 
Please look into the code where i used input type="number" name='eno"
 
and I passed eno as 18. but database took as 0, 
 
regards,
Ch0cu3r Newbie

Ch0cu3r

Posted
Posted

My bad 'eno' in the query should off been '$eno'. In fact, if eno should be a number then validate that it is number first

 

This is an example of how you should be sanitize/validate your data before using it in your query

<?php
 
$user_name = "ganga";
$password = "gangamma";
$database = "ganga";
$server = "localhost";
 
$conn = mysql_connect($server, $user_name, $password);
mysql_select_db('ganga',$conn);

// check a post request has been made before using $_POST values
if($_SERVER['REQUEST_METHOD'] == 'POST')
{
    $errors = array();

    // check eno isset and is has a numeric value (is a number)
    if(isset($_POST['eno']) && is_numeric($_POST['eno']))
    {
        // cast value to an interger (whole number)
        $eno = intval($_POST['eno']);
    }
    // eno is not set, or is not a numberic value
    else
    {
        // set error message
        $errors[] = 'eno must be a number';
    }

    // check ename isset and is not an empty value
    if(isset($_POST['ename']) && !empty($_POST['ename']))
    {
        // sanitize
        $ename = mysql_real_escape_string($_POST['ename']);
    }
    // ename is not set, or is empty
    else
    {
        // set error message
        $errors[] = 'ename cannot be empty.';
    }

    // if there are no errors after validation, then insert data into database
    if(empty($errors))
    {
        $sql = "INSERT INTO `ganga`(`eno`, `ename`) VALUES ($eno, '$ename')";
        $retval = mysql_query( $sql, $conn );
        if($retval)
            echo "Entered data successfully\n";
    }
    // there are are errors
    else
    {
        // display error message(s)
        echo "Invalid data provided:<br />" . implode('<br />', $errors);
    }
}
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