select random variable from sql and assign it to a function

Question

I have this function:

function userPicBySkill($mainPassion){
$query  = mysql_query("SELECT id, username, imagename FROM users WHERE mainpassion =    '$mainPassion' ORDER BY RAND() LIMIT 5"); 

while ($row = mysql_fetch_array($query)) { 

$uid = $row["id"];
$username = $row["username"];   
$imagename = $row["imagename"];

echo "<a href='/$username'>  <img src='image/$imagename' width='40' height='41' alt =    '$username'></a>"; 
 } 
   }

it works if I assign:

 $mainpassion = 'some skill';

What I would like to do is randomly choose 5 'skills' from my table 'skills':

+----------+-----------+
|skill_id  | skill_name|
+----------+-----------+
|        1 |   guitar  |  
|        2 |   cooking |   
|        3 |   math    |    
|        4 |   plumbing|
|        5 |   piano   |
+----------+-----------+

and then get 5 users pictures for each skill. I tried this but it doesn't show anything:

function findRandomSkill(){

$skill_list = mysql_query("SELECT skill_name FROM skills ORDER BY RAND() LIMIT 5");
while($row = mysql_fetch_array($skill_list)){
$skill = $row['skill_name'];
echo '<div class="userBySkillDiv">
<h5>'.$skill.'</h5>';
userPicBySkill($skill);
echo'</div>';
}

yes, if I do $mainpassion = 'guitar' and then userPicBySkill($mainPassion)...it works — Mat
– Mat, Commented Oct 8, 2012 at 6:22
You say "it doesnt show anything" - if you view source, can you see maybe the empty img tag? Or is that not being shown either? — Christian
– Christian, Commented Oct 8, 2012 at 6:23
i think the problem is because of same $row in both functions. Try to change them. — Yogesh Suthar
– Yogesh Suthar, Commented Oct 8, 2012 at 6:25
these functions are both in same class or just two separate functions?? — Yogesh Suthar
– Yogesh Suthar, Commented Oct 8, 2012 at 6:26

Ja͢ck · Accepted Answer · 2012-10-08 06:29:12Z

3

It's better to apply joins instead of doing 25 queries:

SELECT id, username, imagename 
FROM users 
INNER JOIN (SELECT skill_name
    FROM skills
    ORDER BY RAND() 
    LIMIT 5
) userskills ON users.mainpassion = userskills.skill_name
ORDER BY RAND()
LIMIT 5

answered Oct 8, 2012 at 6:29

Ja͢ck

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1 Comment

Mat Over a year ago

you're right. i have to use join in this case! +1 for your advice. thanks

Yogesh Suthar · Accepted Answer · 2012-10-08 06:35:04Z

0

use different $row variable in function like this, this is causing problem.

function userPicBySkill($mainPassion){
$query  = mysql_query("SELECT id, username, imagename FROM users WHERE mainpassion =    '$mainPassion' ORDER BY RAND() LIMIT 5"); 

while ($row1 = mysql_fetch_array($query)) { 

$uid = $row1["id"];
$username = $row1["username"];   
$imagename = $row1["imagename"];

echo "<a href='/$username'>  <img src='image/$imagename' width='40' height='41' alt =    '$username'></a>"; 
 } 
}

answered Oct 8, 2012 at 6:35

Yogesh Suthar

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5 Comments

Ja͢ck Over a year ago

-1. $row is only used once within the function scope and according to the question is called from another function .. the two scopes having nothing in common.

Mat Over a year ago

@Jack I just changed that and now it works. I agree with you that the $row should be used only within the function scope...that's why i didn't think it could be the problem

Ja͢ck Over a year ago

@mattia You actually had both blocks of code in the same scope then?

Mat Over a year ago

@Jack yes,that was the problem

Ja͢ck Over a year ago

@mattia That's completely contrary to your question then, because it shows the top function being called by the below; and as such, it's impossible that such a local variable name has any effect.

Osiris · Accepted Answer · 2012-10-08 06:23:14Z

0

What do you mean by 'doesn't show anything'? Is the list of 5 skill names being returned?

To get 5 random skills, try $skill_list = mysql_query("SELECT DISTINCT(skill_name) FROM skills ORDER BY RAND() LIMIT 5");

answered Oct 8, 2012 at 6:23

Osiris

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