Using Old Scripts On Php5

[talmik]

My old scripts suddenly dont work now that my server upgraded to php 5 for example

 

This error-

Warning: mysql_result() expects parameter 2 to be long, string given in /home/plegend/public_html/entry.php on line 165

This code-

$roomname = mysql_result($currentroom, '$roomnumber', "name");

 

What is differant now?

[Pikachu2000]

Clearly, the second parameter is expected to be long, but you've given it a string value. Have you echoed it to see what exactly it is?

 

http://us3.php.net/manual/en/function.mysql-result.php

[Barand]

Doesn't matter what it contains - as it is in single quotes it is just the string '$roomnumber'.

 

Remove the quotes.

 

And I can't believe it worked in older versions either.

[Pikachu2000]

Didn't even notice those. So, yeah, it's a string. And I agree, that code must never have worked.

[talmik]

Removing the quotes sadly does nothing. It did work before, have been using the code fine for about 4 years

[ManiacDan]

The second argument to mysql_result is a number. It must be a number. You originally had the string '$roomnumber'. Are you saying that once you actually use the variable $roomnumber it's giving the same error? Try intval($roomnumber).

[PFMaBiSmAd]

I'm going to guess it 'worked' because a non-numerical string, treated as a number, is zero.

 

What exactly is your query and code leading up to that point? Are you actually trying to fetch something other than a row 0 value?

[ManiacDan]

I think even a non-numeric string will throw an error, since these lower-level c functions tend to be strict.

[talmik]

Thanks for your comments guys. This is the full section of code

 

$roomnumber = mysql_result($currentuser, 0, "currentroom");

$currentroom = mysql_query("SELECT * FROM rooms WHERE number = $roomnumber LIMIT 1");

$roomname = mysql_result($currentroom, '$roomnumber', "name");

$roomname = explode(" ", $roomname);

$roomname = $roomname[0];

$roomdescription = mysql_result($currentroom, 0, "description");

 

If I delete the line in question, the site works but it does not work correctly. It is basically calling from a list of rooms which are each named and number

[Barand]

Even if roomnumber were numeric, as you only ever select a single row in that query, any roomnumber other than 0 would give an error.

 

We can't see the previous query but it looks as though you have a query to get a roomnumber and then another query using that to get the room details. You should be doing all that in a single query using a JOIN.

[PFMaBiSmAd]

Also, mysql_result is the slowest way of retrieving data because it performs a data seek, then a fetch. You should just use a mysql_fetch_assoc or mysql_fetch_row statement.

[talmik]

Sadly I am not a programmer, I had these scripts custom made for me years ago. I cant even find a host who still uses the older php lol

[ManiacDan]

These scripts were poorly written, even "back then." This line, for example, is wrong and has always been wrong. It was only through a fluke of multiple overlapping layers of wrong that it managed to work in the first place.

[PFMaBiSmAd]

What is differant now?

 

Aside from a Warning message, the lines of code you have posted should be producing the same result as before, Either php5 introduced that Warning message for the parameter type or the code was always producing the error, but it was hidden due to the error_reporting or display_errors/log_errors settings.

 

Ignoring the Warning message, what else isn't working? What output are you getting v.s. what you expect?

[PFMaBiSmAd]

I just tried your code for one of my tables/columns, and along with that Warning about the parameter type, it doesn't fetch any data.

 

If you change that errant parameter from '$roomnumber' to a 0 (zero), your code will fetch the value from the result of the query.

[PFMaBiSmAd]

Here's the likely php change that caused the preexisting error in your code to stop returning any data (in my test above, mysql_result returned a null value when using code with the incorrect '$roomnumber' 2nd parameter) -

 

The newer internal parameter parsing API has been applied across all the extensions bundled with PHP 5.3.x. This parameter parsing API causes functions to return NULL when passed incompatible parameters.