update query with php mysql is not updating the database table entries

Question

<?php
include 'dbConfig.php';
$address=mysql_real_escape_string($_POST['address']);
$pincode=mysql_real_escape_string($_POST['pincode']);
$phone=mysql_real_escape_string($_POST['phone']);
$email=mysql_real_escape_string($_POST['contactemail']);
$id=1;
$sql="UPDATE contactus SET address = '$address' , pincode = '$pincode' , phone = '$phone' , email = '$email' WEHER id='$id'";
$result=mysql_query($sql);
if (!$result)
{
    die('Error: '. mysqli_error());

}

it gives answer like this:-

DB initiated
Error:

What is the problem..?

You can use the built in mysql error function to get details if it was a query error php.net/manual/en/function.mysql-error.php — Scott Helme
– Scott Helme, Commented Aug 6, 2013 at 9:11

Conrad Lotz · Accepted Answer · 2013-08-06 09:10:23Z

1

The WHERE is spelt incorrectly as WEHER:

Try this

$sql="UPDATE contactus SET address = '$address' , pincode = '$pincode' , phone = '$phone' , email = '$email' WHERE id='$id'";

answered Aug 6, 2013 at 9:10

Conrad Lotz

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Comments

score 1 · Accepted Answer · 2013-08-06 09:20:18Z

1

Use a correct mysqli syntax. here an example

<?php
$con=mysqli_connect("example.com","peter","abc123","my_db");
// Check connection
if (mysqli_connect_errno())
  {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
  }

mysqli_query($con,"UPDATE Persons SET Age=36
WHERE FirstName='Peter' AND LastName='Griffin'");

mysqli_close($con);
?>

source: http://www.w3schools.com/php/php_mysql_update.asp

And fix you query

 WEHER id='$id'";

by

WHERE id='$id'"

edited Aug 6, 2013 at 9:20

answered Aug 6, 2013 at 9:09

user1409909

Comments

Goutam Pal · Accepted Answer · 2013-08-06 10:02:03Z

1

$sql="UPDATE contactus SET address = '$address' , pincode = '$pincode' , phone = '$phone' , email = '$email' WEHER id='$id'";
$result=mysql_query($sql);
if (!$result)
{
    die('Error: '. mysqli_error());

}

should be

$sql="UPDATE contactus SET address = '$address' , pincode = '$pincode' , phone = '$phone' , email = '$email' WHERE id='$id'";
$result=mysql_query($sql);
if (!$result)
{
    die('Error: '. mysql_error());

}

answered Aug 6, 2013 at 10:02

Goutam Pal

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Comments

AndreaBogazzi · Accepted Answer · 2013-08-06 09:18:42Z

0

Are you passing the "link identifier" parameter in mysql_query function?

resource mysql_query ( string $query [, resource $link_identifier = NULL ] )

i know is not mandatory, but you are executing query with mysql_query, getting the error with mysqli, dbconfig we cannot see it. So is difficult to guess.

answered Aug 6, 2013 at 9:18

AndreaBogazzi

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