updating database

[franknu]

i want to update a databse using a password and a username in the row i am just having some trouble here is my code

PHP: Select all

<?
 
 $host = "localhost";
$username = "localhost";
$password = "abc123";
$database = "contacts";
$db = mysql_connect($host, $username, $password);
mysql_select_db($database);
$BusinessName = (isset($_POST['BusinessName']) ? $_POST['BusinessName'] : ""); 
 $Slogan = (isset($_POST['Slogan']));
 $Business_Address = (isset($_POST['Business_Address']));
 $Tel = (isset($_POST['Tel']));
 $Website = (isset($_POST['Website']));
 $Email = (isset($_POST['Email']));
 $Member_Status = (isset($_POST['Member_Status']));
 $Fax =(isset($_POST['Fax']));
 $type = (isset($_POST['type']));
 $make = (isset($_POST['make']));
 $Categories = (isset($_POST['Categories']));
 $Keyword = (isset($_POST['Keyword']));
 $Picture1 =  (isset($_POST['Picture1']));
 $Headline =  (isset($_POST['Headline']));
 $Slogan2 = (isset($_POST['Slogan2']));
 $Description1 = (isset($_POST['Description1']));
 $Description2 = (isset($_POST['Description2']));
 $Description3= (isset($_POST['Description3']));
 $Contact2 = (isset($_POST['Contact2']));
 $Picture2 =  (isset($_POST['Picture2']));
 $Picture3 = (isset($_POST['Picture3']));
 $Picture4 =  (isset($_POST['Picture4']));
 $User_Name = (isset($_POST['User_Name']));
 $Password = (isset($_POST['Password']));
 
 
$User_Name = mysql_query("SELECT `Password` FROM `Business_Info` WHERE `User_Name` = '$User_Name'"); 
 
if($Password == $Password) 
 
 
{ 
$query = "UPDATE `Business_Info` SET `BusinessName`= '$BusinessName', `Slogan`='$Slogan', 
 `Business_Address`='$Business_Address', 
`Tel`='$Tel', `Website`='$Website', `Email`='$Email', `Member_Status`='$Member_Status', 
`Fax`='$Fax', `type`='$type', `make`='$make', `Categories`='$Categories', `Keyword`='$Keyword', 
`picture1`='$Picture1', `Headline`='$Headline', `Slogan2`='$Slogan2', `Description1`='$Description1', 
`Description2`='$Description2', `Description3`= '$Description3', `Contact2`='$Contact2', `Picture2`='$Picture2',
 `Picture3`='$Picture3', `Password`='$Password' WHERE `User_Name`='$User_Name'";
$result = mysql_query($query) or die(mysql_error()); 
 
  } 
 
    else 
    { 
 
  echo "Incorrect Password or User Name Try again "; 
  exit; 
} 
echo "$query" ;
 
?>

this is what i am displaying

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UPDATE `Business_Info` SET `BusinessName`= '', `Slogan`='', `Business_Address`='', `Tel`='', `Website`='', `Email`='', `Member_Status`='', `Fax`='', `type`='', `make`='', `Categories`='', `Keyword`='', `picture1`='', `Headline`='', `Slogan2`='', `Description1`='', `Description2`='', `Description3`= '', `Contact2`='', `Picture2`='', `Picture3`='', `Password`='1' WHERE `User_Name`='Resource id #3' 

[flann]

is there an error?

[franknu]

Originally Posted by flann

is there an error?

it is no lettting me update all i am getting is the samething is on the update query

[djme]

I think it will work if you take the `s out like below, ive never used those for this but they may not make a difference. The only other reason I can see is if you have some of your fields wrongly labelled (you do have picture1 and the others as Picture2 and Picture3).

PHP: Select all

$query = "UPDATE Business_Info SET BusinessName='$BusinessName', Slogan='$Slogan', 
 Business_Address='$Business_Address', 
Tel='$Tel', Website='$Website', Email='$Email', Member_Status='$Member_Status', 
Fax='$Fax', type='$type', make='$make', Categories='$Categories', Keyword='$Keyword', 
picture1='$Picture1', Headline='$Headline', Slogan2='$Slogan2', Description1='$Description1', 
Description2='$Description2', Description3= '$Description3', Contact2='$Contact2', Picture2='$Picture2',
 Picture3='$Picture3', Password='$Password' WHERE User_Name='$User_Name'";
$result = mysql_query($query) or die(mysql_error()); 

[ukgeoff]

All you are actually doing is setting all your variables, $slogan, etc, to 'true' or 'false'. The result of your test - isset($_POST['slogan']);, etc.

What you should be doing is something like:

Code: Select all

if(isset($_POST['slogan'])){
  $slogan = $_POST['slogan'];
}