Updating query

[dean7]

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.

[Pikachu2000]

My best guess would be that one or more of the variables has no value. Have you echoed them to see if the values are what you'd expect?

[dean7]

To add to whats happening, ive echoed the $nmoney varibal out which does contain a value of the users new money, just in the mysql query it isnt updating the database   

 

Is there something im doing wrong in that part of the code?

[Pikachu2000]

There are no errors generated, right?

 

What is the data type of the of the `money` field in the database? Is it DECIMAL, FLOAT, VARCHAR, or what?

 

If it's a numeric field type, try the query without the quotes around $nmoney, i.e. SET money = $nmoney

[dean7]

  Quote

There are no errors generated, right?

 

What is the data type of the of the `money` field in the database? Is it DECIMAL, FLOAT, VARCHAR, or what?

 

If it's a numeric field type, try the query without the quotes around $nmoney, i.e. SET money = $nmoney

Sorry about late reply but its just a VARCHAR field in the database.

[Pikachu2000]

Change your query execution to not include the query string. Store it in a variable, temporarily comment out the execution and echo the variable to see if it looks right.

 

$query = "UPDATE users SET money='$nmoney' WHERE username='$username'";
//mysql_query($query) or die (mysql_error());
echo $query;

[dean7]

  Quote

Change your query execution to not include the query string. Store it in a variable, temporarily comment out the execution and echo the variable to see if it looks right.

 

$query = "UPDATE users SET money='$nmoney' WHERE username='$username'";
//mysql_query($query) or die (mysql_error());
echo $query;

Ah thanks , for doing that its manged to give me an error which could be good:

 

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 'Resource id #11' at line 1

 

But how would I get around this error?

 

 

[Pikachu2000]

Either $username or $nmoney contains a query result resource instead of a valid value. You'll need to figure out which one, and why.

[thomashw]

Change your query string to this:

$query = "UPDATE users SET money='".$nmoney."' WHERE username='".$username."'";

 

[gizmola]

  Quote

Change your query string to this:

$query = "UPDATE users SET money='".$nmoney."' WHERE username='".$username."'";

 

There is no difference between this query and the one he already had.  This has nothing to do with his problem.  Pikachu2000 already identified the issue.

[thomashw]

  Quote

  Quote

Change your query string to this:

$query = "UPDATE users SET money='".$nmoney."' WHERE username='".$username."'";

 

There is no difference between this query and the one he already had.  This has nothing to do with his problem.  Pikachu2000 already identified the issue.

I've had problems using the syntax he's using, so that is why I suggested changing the string to what I posted.

[gizmola]

  Quote

  Quote

  Quote

Change your query string to this:

$query = "UPDATE users SET money='".$nmoney."' WHERE username='".$username."'";

 

There is no difference between this query and the one he already had.  This has nothing to do with his problem.  Pikachu2000 already identified the issue.

I've had problems using the syntax he's using, so that is why I suggested changing the string to what I posted.

 

I'm sorry to be harsh, but that's poor advice.  I can only suggest that you really study PHP "interpolation".  The only issue with interpolation is that you can not include array variables using single quotes around the key, but even this has a workaround.

 

 

$r['day'] = 'monday';

echo "today is $r['day']";
// syntax error

echo "today is $r[day]";
// works but requires constant lookup

echo "today is {$r['day']}";
// works 

 

Speaking personally, it is both easier to read and easier to maintain interpolated strings in my opinion.

 

Having to concatenate every single variable is a lot of work, and a major headache when you have to alter queries. 

 

 

[dean7]

  Quote

Either $username or $nmoney contains a query result resource instead of a valid value. You'll need to figure out which one, and why.

I blive that it is $nmoney which contains the resource instead of a valid value, but how would I sort that to contain a valid value?