trouble with function not returning the correct value

benwestgarth

Hi all,
I have not used functions very often in PHP, and I'm having a hard time getting what I would have thought to be an easy function working. I have included the code below.

I have a page which needs to make repeated calls to a MySQL database. Inside the function the SQL does a simple count(*) on a range of tables and at the end of the function should return the $result. But it doesn't.

And there are a few oddities. When the page outputs it shows that $query and $execQuery exist (from the debug comments), but there is no indication that $result has been initiated. And if I change the debug to die on finding no $result, it does. It is as though the function is ignoring the line commencing with $result. Please help!!

$result;
function listings($table,&$result)
{
  $query = "select count(*) from $table where customer_id='$check_id'";
  if($query) echo "There is a query.";
  $execQuery = mysql_query($query);
  if($execQuery) echo "There is an execQuery.";
  $result = mysql_result($execQuery, 0, 0);
  if($result) echo("There is a result ");
  return $result;
}

listings(accommodation,$result);    //call function to see if listing exists
if ($result == 0)
{
  echo "$result";
  echo "<a href='add_accommodation.php'>Add an accommodation listing to the database.</a>";
  echo "<br><br>";
}

bernouli

try this

$result= listings(accommodation,$result);    //call function to see if listing exists
if ($result == 0)
{
  echo "$result";
  echo "<a href='add_accommodation.php'>Add an accommodation listing to the database.</a>";
  echo "<br><br>";
}

benwestgarth

Thanks for your reply, but it still doesn't work. It just won't return $result. I have stared at this for a long time now, and I can't understand why the function won't give $result a value.

Any other suggestions?

Thanks , Ben

bernouli

and what did youget after you run the script?
you must have get some string in your browser right?

maybe $result != 0, so that it wont enter the if clause

bobthedog

Hmmmm - could be lots of different things, but without your output, I can't really tell.

The first place to look would probably be at this line

$result= listings(accommodation,$result);

Note the missing quotes around the table name (accomodation)?

Well - that's my first guess anyhow. My second one would be that your result set is returning a number greater than 0. Therefore you will never get into the if block and your code will never be executed.

As an additional point, I notice you are passing in $result as a call by ref, and then returning it as well. TBH, you are only returning one value - so there is no need to pass result in as well (since its original value is not being used in the functon).

i.e.

function listings($table)
{
  $query = "select count(*) from $table where customer_id='$check_id'";
  if($query) echo "There is a query.";
  $execQuery = mysql_query($query);
  if($execQuery) echo "There is an execQuery.";
  $result = mysql_result($execQuery, 0, 0);
  if($result) echo("There is a result ");
  return $result;
}

$result = listings('accommodation');    //call function to see if listing exists

Good luck

benwestgarth

Hi again, thanks for the ideas, bernouli and bobthedog, but the problem is not that I can't get into the loop. In fact, in this particular instance it shouldn't enter the loop. The problem is that the function will not alter $result.

Here is my code snip...

$result;
function listings($table,&$result)
{
  $query = "select count(*) from $table where customer_id='$check_id'";
  if($query) echo "QUERY<br />";    //debug line
  $execQuery = mysql_query($query);
  if($execQuery) echo "EXECQUERY<br />";    //debug line
  $result = mysql_result($execQuery, 0, 0);
  if($result) echo "RESULT<br />";    //debug line
  return $result;
}

$result = listings('accommodation',$result);    //call function to see if listing exists
if ($result == 0)
{
  echo "$result";
  echo "<a href='add_accommodation.php'>Add an accommodation listing to the database.</a>";
  echo "<br><br>";
}

Here is the output...

QUERY
EXECQUERY
0Add an accommodation listing to the database.

QUERY
EXECQUERY
Add a sightseeing listing to the database

Note that the debug lines confirm that both $query and $execQuery have been set, but it denies that $result exists! When the function returns, $result should == 1, but it == 0. This is because I have declared $result outside of the function and it is just using the default 0 value for an uninitialised variable. It is not using the return value from the function. The most peculiar thing is that if I set $result to explicitly == some other value inside the function ie. $result == 9, it returns the correct value!!!! It just doesn't like the line:

  $result = mysql_result($execQuery, 0, 0);

although, if I use this code outside the function it works perfectly.

Thanks for your help, and I really hope someone can see what I'm doing wrong.

Cheers, Ben

[deleted]

I tested you function on a simple state_id table where AK = 1, etc. and it worked fine. Your function call will only work in cases where there are Zero results. Try calling your function using <? echo listings(accomodation,$results) ?> and see if you have a different outcome than the function call you are currently using.

bobthedog

Ok - so - why isn't it producing output when you do this...

if($result) echo("There is a result ");

Could is be that $result has been set to 0??? I think so - otherwise we would get output (unless of course it has returned 'false' because you have specified invalid offsets). So - it seems to be running and then entering the loop as required.

If it is doing the wrong thing, then I would suggest checking that the field with the specified id actually does exist within your database. Try outputting the $query line to find out what the generated query comes out as and run it directly through mysql via the command line to see if it matches a row.

Can I also suggest tidying up your function a bit. Don't call it with $result passed in by ref AND return the result, that's just confusing.

Try this:

function listings($table)
{
  $query = "select count(*) from $table where customer_id='$check_id'";
  if($query) echo "QUERY<br />";    //debug line
  $execQuery = mysql_query($query);
  if($execQuery) echo "EXECQUERY<br />";    //debug line
  $result = mysql_result($execQuery, 0, 0);
  if($result) echo "RESULT<br />";    //debug line
  return $result;
}

$result = listings('accommodation');

benwestgarth

Thanks, Bobthedog,

Even if $result was set to zero, the script should still echo the debug line shouldn't it? Besides which, $result definitely isn't zero. If I run the mysql_query outside of the function, $result ==1 and the debug line is echoed. It's only when it is inside the function that it doesn't work which makes absolutely no sense to me.

Like I said, I haven't used functions much, so I reckon there's something I'm missing. Also, the code I posted was the end of a series of fruitless attempts to get this working. I was blindly stabbing at ideas when I started trying to pass $result to the function as a ref. Thanks for getting me back on track.

ANy ideas or suggestions?

Cheers, Ben

bobthedog

curses wildly

I cannot believe I didn't see this in the first place... ah well... code blindness must have set in in my previous mails. 🙂

Anyhow - to go over your first point....

"0" does actually mean the same thing as "false" when you do an if condition as you do in your code. i.e.

if (0) { ... }

is logically the same as

if (false) { ... }

If you want to check specifically for a false value then you need to do a value AND type check represented by three consecutive = signs.

i.e. if ($result === false) { ... }

Anyhow - to get back to the problem you are having....$check_id is never declared within the function (since the scope of this variable is only within the function). Therefore the select is "select count(*) from accomodation where customer_id=''", which obviously is not matching any rows in your table.

To rectify this you need to also pass $check_id into the function as well as $table.

So - modified code would read like this...

function listings($table, $check_id)
{
  $query = "select count(*) from $table where customer_id='$check_id'";
  if($query) echo "QUERY<br />";    //debug line
  $execQuery = mysql_query($query);
  if($execQuery) echo "EXECQUERY<br />";    //debug line
  $result = mysql_result($execQuery, 0, 0);
  if($result) echo "RESULT<br />";    //debug line
  return $result;
}

$result = listings('accommodation', $check_id);

If you don't understand about scoping of variables then go read up on it at http://www.php.net/.

Anyhow - sorry for not seeing this in the first place 🙂

benwestgarth

Outstanding! Thanks bobthedog! I had declared $check_id as a session variable and for some reason in my head that meant its scope extended the whole way through the program.

The script works exactly as it was intended now. I can't thank you enough for your patience. Good karma to you.

Cheers, Happy Ben 😃