Loading

taking information from database and display on textbox

taking information from database and display on textbox

Abdurrahman Yaşar

Abdurrahman Yaşar

10y1kPHP

I am trying to fix it for 2 hours,but cannot see the informations on textbox. Can you tell me what is wrong ?

$con = mysqli_connect("localhost","root","root");

if (!$con)

{

die('Could not connect: ' . mysqli_error());

}

mysqli_select_db("banka", $con);

$con = mysqli_connect("localhost","root","root");

if (!$con)

{

die('Could not connect: ' . mysqli_error());

}

mysqli_select_db("banka", $con);

$result = mysqli_query( "SELECT * FROM customer WHERE cusid='$_POST[name]'" );

while( $rowval= mysqli_fetch_array( $result ) ){

$cusid= $rowval['cusid'];

$cusname= $rowval['cusname'];

$cussurname= $rowval['cussurname'];

$cususername= $rowval['cususername'];

$cuspassword= $rowval['cuspassword'];

$adress= $rowval['adress'];

$tel= $rowval['tel'];

$trmoney= $rowval['trmoney'];

$dolarmoney= $rowval['dolarmoney'];

}

mysqli_close($con);

?>

Enter Account no

1 Reply

Know the answer? Post it — somebody with the same question will find it here.

Gokulraj Sankar10y ago
hi
in your query where condition you used thecusidin the query but your post value is a name.
kindly check your where a condition
i hope it will help you.
SELECT * FROM customer WHERE cusid='$_POST[name]'


Id:
Name
Surname:
Username:
Password:
Address:
Telephone:
TL amount:
Dolar amount: