PHP - Moved: Updating Query

why is it not updating?


<?php
$_POST['email'];
$_POST['submit'];
if (isset($_POST['submit']))
{ 
/*$sql =" UPDATE class_members SET email ='".$_POST['email']."', WHERE S_UID ='".$_POST['mem_id']."' ";*/
$result = mysql_query("UPDATE class_members SET email ='".$_POST['email']."', WHERE mem_id='".$_POST['mem_id']."'") 
or die(mysql_error());  
} 

?>

Hi all, for my website ive got a fuel station which ive coded so users can fill up there there car fuel tanks, but the code all works part from one part of it.

The code:

The part of the code which isnt working:

<?php
$usermoney = $fetch->money;
$costs2 = $fetch_fuelstation->price * $fueldiff;
if ($costs2 > $usermoney){
echo "You dont have enouth money.";
} elseif ($costs2 <= $usermoney){
$nmoney = $usermoney - $costs2;

mysql_query("UPDATE users SET money='$nmoney' WHERE username='$username'") or die (mysql_error());
}
?>

Ive got a varibal called $usermoney which selects the users current money from the database, but what im trying todo is take the fuel price away from there current money, but its not subtracting the money as it should.

Anyone see why its not taking the money?
Thanks.

Hey Guys,

Ive been searching for a while, but havent really come up with any answers.

At the minute i have a page where i can add a new topic to a database, but i want to be able to update the query on another page if that makes sense 

So as an example say i have this below thats just been inserted into the database with an ID of 2.

<?php

// Details
$id = $_POST['id'];
$title = $_POST['title'];
$detail = $_POST['detail'];

mysql_connect("localhost", "web148", "123") or die(mysql_error());
mysql_select_db("web148") or die(mysql_error());

mysql_query("SET names 'utf8'");
mysql_query("INSERT into sunnyargues(ID, title, body) VALUES ('2', ''&#36229;&#32423;&#39532;&#37324;&#22885;&#24847;&#22823;&#21033;&#20877;&#36973;&#31181;&#26063;&#27495;&#35270;', '&#24403;&#34987;&#38382;&#21450;&#22312;&#21451;&#35850;&#36187;&#19978;&#30340;&#19981;&#21451;&#22909;&#19968;&#24149;&#20250;&#19981;&#20250;&#37325;&#28436;&#26102;&#65292;&#26364;&#22855;&#23612;&#35828;&#65292;&#8220;&#26377;&#36825;&#31181;&#21487;&#33021;&#24615;&#12290;&#8221; &#8220;&#25105;&#24403;&#28982;&#19981;&#24076;&#26395;&#36825;&#26679;&#30340;&#20107;&#24773;&#21457;&#29983;&#65292;&#20294;&#26159;&#35841;&#20063;&#19981;&#33021;&#20445;&#35777;&#12290;&#8221;'");
echo "<h1>Article Updated!</h1>";

?>

On the sunny argues page i currently have, where it only selects ID 1:

<?php

mysql_connect("localhost", "web148", "123") or die(mysql_error());
mysql_select_db("web148") or die(mysql_error());
mysql_query("SET names 'utf8'");
$result = mysql_query("SELECT body FROM sunnyargues WHERE ID='1'");
$row = mysql_fetch_array( $result );
echo nl2br($row['body']);
?>

Is there a way i can update this query with the new ID of the 2nd article once its been inserted into the database.

I hope that makes sense 

Hi there.  I'm totally new (about a week!) with php and mysql and am encountering a problem that perhaps someone can help me with? I've looked through to see if a similar problem has appeared or been solved, but without success, so apologies if I am repeating something.    In php I am trying to update 7 fields from a form from which a user has edited/modified any of the fields in a chosen record (except id).   Here is the code:   $id=$_GET['id'];

I have a form that allows my client to update some products. Now the products are simple just basic info and 1 picture.

I have set this up so they can edit the products and change the information, having done this many times in the past, but now hit a puzzling block that I am baffled.

The client when editing is presented with the form with the information pulled from the database and the form fields loaded with that data ready to edit.

The image can either be left alone or they can choose to upload a new image. They are shown the image they currently have stored in the database.

The problem I have is EVEN if they decide not to upload an image and change other information, when the submit the form it must be sending a blank value for the image somewhere as it is updating the database and removing the image reference as if it has been removed.

I have an if/else statement based on the form to perform 2 different queries for the update in mysql.

Here is the code for the form update, as you can see the image should not update?? Please help??


 if ($_SERVER['REQUEST_METHOD'] =='POST') {

//This stops SQL Injection in POST vars 
 foreach ($_POST as $key => $value) { 
  $_POST[$key] = mysql_real_escape_string($value); 
 }

// **************************** THIS IS FOR NO NEW IMAGE ********************************
 
 if ($_SERVER['REQUEST_METHOD'] =='POST' && empty($_FILES['product_image']['name'])) {

   # setup SQL statement for no new image
  $SQL = " UPDATE products SET product_title = '{$_POST['product_title']}', product_description = '{$_POST['product_description']}', standard_price = '{$_POST['standard_price']}', deluxe_price = '{$_POST['deluxe_price']}' WHERE product_id = '{$_REQUEST['product_id']}' ";
}

// **************************** THIS IS FOR A NEW IMAGE ********************************

 else {

// Check the image type is a jpeg or gif for the image.
 if (($_FILES['product_image']['type'] != "image/gif") && ($_FILES['product_image']['type'] != "image/jpeg") && ($_FILES['product_image']['type'] != "image/pjpeg")) {
  echo "<FONT FACE=\"Verdana\"><SPAN CLASS=\"content\">You have chosen not to upload a <b>Product Image</b>.<BR></SPAN>" ;
 } elseif ($_FILES['product_image']['size'] > 100000) {
  echo "<FONT FACE=\"Verdana\"><SPAN CLASS=\"content\">The file size is bigger than 300kb.<BR></SPAN>" ;
 } else {
  move_uploaded_file($_FILES['product_image']['tmp_name'], "/httpdocs/product_images/".$_FILES['product_image']['name']) ;
 echo "<FONT FACE=\"Verdana\"><SPAN CLASS=\"content\"><B>Your front image has successfully uploaded.</B><BR></SPAN>" ;
  }
}

  # setup SQL statement for update
  $SQL = " UPDATE products SET product_title = '{$_POST['product_title']}', product_description = '{$_POST['product_description']}', standard_price = '{$_POST['standard_price']}', deluxe_price = '{$_POST['deluxe_price']}', product_image = '{$_FILES['product_image']['name']}' WHERE product_id = '{$_REQUEST['product_id']}' ";
}
  
  #execute SQL statement
  $result = mysql_db_query( *****,"$SQL",$connection );

  # check for error
  if (!$result) { echo("ERROR: " . mysql_error() . "\n$SQL\n");    }