Returning Array From PHP Function

Question

What is wrong with this code? I am trying to return an array and use it in the caller function.

function my_subjects()
        {   
            $all_my_teams=my_teams();
            $k=0;
            $_subjects=array();
            while($team=mysql_fetch_assoc($all_my_teams))
            {
                $q="SELECT $this->user_field FROM $team";
                $r=mysql_query($q);

                while($i=mysql_fetch_assoc($r))
                {
                    $_subjects[k++]=$i;
                }
            }                   

            return $_subjects;
        }

Note: the function my_teams() returns the value similar to the $r variable which is used through all_my_teams variable. it contains all the names of teams.

Aside from using the deprecated mysql_* functions I can't see anything odd in a quick look. What doesn't happen that should? — lc.
– lc., Commented Jan 22, 2013 at 15:58
why are you running two queries? this should be done as a single JOINed query. — Marc B
– Marc B, Commented Jan 22, 2013 at 15:59
What happens when you run this? What is returned - are you just getting an empty array? — Arthur Richards
– Arthur Richards, Commented Jan 22, 2013 at 16:05

Community · Accepted Answer · 2023-11-17 19:22:09Z

1

Turn up error_reporting to see if your code is generating errors.
Check if your query is successful if( ! $r=mysql_query($q) ) { die(mysql_error()); }
var_dump($_subjects); to see if the data is what you expect it to be.
Maybe actually tell us what's going wrong? You've just posted a block of code and told us "it doesn't work." which isn't terribly indicative of a problem.
$k is irrelevant, just use $_subjects[]=$i;. [wouldn't cause an error, just easier]
Stop using mysql_* functions and port your code to PDO or MySQLi to:
- Benefit from parameterized queries which can help protect against SQL injection.
- Stop everyone on SO starting an unrelated argument in the comments about it.

edited Nov 17, 2023 at 19:22

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answered Jan 22, 2013 at 16:18

Sammitch

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Comments

Hanky Panky · Accepted Answer · 2013-01-22 15:57:36Z

0

while($i=mysql_fetch_assoc($r))
                {
                    $_subjects[k++]=$i;
                }

Here you also have to provide a field name for $i. Something like

while($i=mysql_fetch_assoc($r))
                {
                    $_subjects[$k++]=$i["field_name"];
                }

Array returning part is just fine.

Edit: Your variable k was missing a $ sign as well.

answered Jan 22, 2013 at 15:57

Hanky Panky

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1 Comment

Arthur Richards Over a year ago

You don't need to include a field name since $i is just an associative array. But you are correct that 'k' was missing a $ sign - that is likely what's causing the problem. I totally missed that when reading his code!

Community · Accepted Answer · 2023-11-20 22:45:59Z

0

$_subjects[$k++] = $i;

should be fine, since you're using mysql_fetch_assoc() $i will contain an associative array of the result set.

If this is returning an empty array (is that the correct problem?), you should double check that your sql is correct and actually returning data/data that you expect.

Edit: Like Hanky Panky mentioned, 'k' is missing a $ sign in your code - that is likely to be your problem - PHP should be throwing a parse error for that one, so make sure you have error_reporting() enabled.

edited Nov 20, 2023 at 22:45

CommunityBot

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answered Jan 22, 2013 at 16:02

Arthur Richards

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2 Comments

Sammitch Over a year ago

Are you sure you mean $i++?

Arthur Richards Over a year ago

haha, no. thanks @Sammitch - I shouldn't do this stuff before I drink my coffee :p

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