Failing to alter data in database

Question

The purpose of this script is to allow the user to add "money" to their account and then withdraw the money. This whole task is a fictitious one for school so i'm not worried about security or using real money or anything like that.

so far the user has the ability to add money. But I cannot get the withdraw to work.

    $funds = $_POST['funds'];
    $withdraw_or_add = $_POST['list'];


if($withdraw_or_add == "add"){
    $sql = "UPDATE users ". "SET userFunds = $funds ". "WHERE userId = 1";
} else {
    $info = mysql_fetch_array(mysql_query("SELECT * FROM users WHERE userId = '1'"));
    $new_fund == $info['userFunds'] - $funds;
    $sql = "UPDATE users ". "SET userFunds = $new_fund ". "WHERE userId = 1";
}

This is the code I have written for this section. When i used it on the server i get this message in return.

Could not update data: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'WHERE userId = 1' at line 1

My knowledge of PHP is not fantastic. So any help would be much appreciated.

`users ` in your query is any constant or table name itself? — Milan Gupta, Nov 15 '16 at 11:16
Try writing query like this: `"UPDATE users SET userFunds = ' ".$funds." ' WHERE userId = 1";` — Milan Gupta, Nov 15 '16 at 11:18
@MilanGupta Okay i did that and now when i withdraw "money" it takes the number that should be used for subtraction and just uses that as the new figure. So when i want to take away 5 from 20 it should be 15. But instead it just changes to 5. — Craig Harkins, Nov 15 '16 at 11:22
Every time you use [the `mysql_`](http://stackoverflow.com/questions/12859942/why-shouldnt-i-use-mysql-functions-in-php) database extension in new code **[a Kitten is strangled somewhere in the world](http://2.bp.blogspot.com/-zCT6jizimfI/UjJ5UTb_BeI/AAAAAAAACgg/AS6XCd6aNdg/s1600/luna_getting_strangled.jpg)** it is deprecated and has been for years and is gone for ever in PHP7. If you are just learning PHP, spend your energies learning the `PDO` or `mysqli` database extensions. [Start here](http://php.net/manual/en/book.pdo.php) — RiggsFolly, Nov 16 '16 at 11:07

Milan Gupta · Answer 1 · 2016-11-15T12:30:37.617

1

Try this:

$funds = $_POST['funds'];
$withdraw_or_add = $_POST['list'];


if($withdraw_or_add == "add")  
{
  $sql = "UPDATE users SET userFunds = '".$funds."' WHERE userId = 1";  
  $sql =  mysql_query($sql) or die(mysql_error());
}   
else  
{
  $info = mysql_query("SELECT * FROM users WHERE userId = '1'");
  $info = mysql_fetch_assoc($info);
  $new_fund = $info['userFunds'] - $funds;
  $sql = "UPDATE users SET userFunds = '".$new_fund."' WHERE userId = 1";  
  $sql =  mysql_query($sql) or die(mysql_error());
}

edited Nov 15 '16 at 12:30

answered Nov 15 '16 at 11:30

Milan Gupta

1,181
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21

I'm still getting the same problem. It is still changing the value to what it should be subtracting. – Craig Harkins Nov 15 '16 at 12:19
1

To confirm: `$funds = 5` and `$info['userFunds'] = 20` are the values and you are getting incorrect value in `$new_fund`. Is it so? – Milan Gupta Nov 15 '16 at 12:23
yes. So in this example `$new_fund` should = 15. But for me it's equaling 5. – Craig Harkins Nov 15 '16 at 12:26
1

Did your print value of `$new_fund` or you are checking updated record record in DB? – Milan Gupta Nov 15 '16 at 12:28
checking in the database. – Craig Harkins Nov 15 '16 at 12:28
1

I just added `$sql = mysql_query($sql) or die(mysql_error());` in answer. Looks like your update query was not being fired at all. Can you check with edited answer? – Milan Gupta Nov 15 '16 at 12:31
Now it's returning `No database selected`. – Craig Harkins Nov 15 '16 at 12:35
Let us [continue this discussion in chat](http://chat.stackoverflow.com/rooms/128163/discussion-between-craig-harkins-and-milan-gupta). – Craig Harkins Nov 15 '16 at 12:45

Failing to alter data in database

1 Answers1