PHP Random from MYSQL

[plznty]

How would I make it so that

for every set of data where a fieldname is 1+ it picks a random one out.

For example

User            Hit

Luke            0

Peter            1

Alex              3

Peter            1

 

For every value where Hit is 1 or over it selects out of the query results a random username.

[joesaddigh]

SELECT column FROM table

WHERE Hit >0

ORDER BY RAND()

LIMIT 1

 

[Pikachu2000]

Before you do that, Google 'mysql why order by rand() is bad'.

[joesaddigh]

Is there a good alternative?

[wannabephpdude]

I would definitely have a unique id for any table [Edit] when comes to storing users for sure. If you can work that out then the following will be helpful.

 

Take a look at solution 2 as opposed to using rand() within your query.

[plznty]

so how could I gather users who have 1<= and then select a random id from the results?

[Pikachu2000]

Before you do that, Google 'mysql why order by rand() is bad'.

 

Did you do this ^^^?

 

Reading some of the results will help you make the decision whether you want to use ORDER BY RAND() or not. There are also better solutions in each explanation of the downfalls.

[plznty]

I've got about 1000 results to pick from

maybe 1/2 of which will needed to be called.

Is this amount of data okay.

Yes i did read

Thanks

[Pikachu2000]

I'd have to say with that small of a record set, you'll be OK. Just remember that you used ORDER BY RAND() there, and if the query starts to bog down as the table grows, change to a more efficient method.

[plznty]

Alright thanks!

[plznty]

Reopened.

How do i write this out I keep getting Resource id #2 output.

Thanks

[plznty]

How would I write this.

(bump) (was off first page)

[Pikachu2000]

Post the code you're using to query the DB and display the result.

[plznty]

I keep getting Resource id #2 output.

[litebearer]

As Pika said...

Post the code you're using to query the DB

[plznty]

oops 2x post.

[plznty]

    $query = mysql_query("SELECT * FROM users WHERE clicks >0 ORDER BY RAND() LIMIT 1;");
    echo $query;
echo $row['email'];

[Pikachu2000]

There's an error in the query syntax (semicolon inside the quotes), and you need to actually do something with the query result before you can use the values it returns. You also had nothing in place to check to see if the query succeeded.

 

$query = "SELECT * FROM users WHERE clicks > 0 ORDER BY RAND() LIMIT 1";
if( $result = mysql_query($query) ) {
     $array = mysql_fetch_assoc($result);
     echo $row['email'];
} else {
     echo '<br>Query: ' . $query . '<br>Produced error: ' . mysql_error() . '<br>';
}

[plznty]

it displays nothing.

This is doing my head in.

Thanks for helping though

<?php
    $connect = mysql_connect("localhost", "dbname", "pass") or die(mysql_error());
    mysql_select_db("db") or die(mysql_error());
$query = "SELECT * FROM users WHERE clicks > 0 ORDER BY RAND() LIMIT 1";
if( $result = mysql_query($query) ) {
     $array = mysql_fetch_assoc($result);
     echo $row['id'];
} else {
     echo '<br>Query: ' . $query . '<br>Produced error: ' . mysql_error() . '<br>';
}
?>

 

Produces nothing

[mentalist]

hmmm

     $array = mysql_fetch_assoc($result);
     echo $row['id'];

i'm sure your issue lies here, something to do with the names of variables...

[Pikachu2000]

Change

echo $row['id'];

 

to

echo $row['id'] . ' id should be here';

 

To see if if the code is at least entering that part of the conditional.

[plznty]

<?php
    $connect = mysql_connect("localhost", "", "") or die(mysql_error());
    mysql_select_db("") or die(mysql_error());
$query = "SELECT * FROM users WHERE clicks > 0 ORDER BY RAND() LIMIT 1";
if( $result = mysql_query($query) ) {
     $array = mysql_fetch_assoc($result);
     echo "hello".$row['id'];
} else {
     echo '<br>Query: ' . $query . '<br>Produced error: ' . mysql_error() . '<br>';
}
?>

 

Does display "hello"

so something to do with $row variable etc

[Pikachu2000]

D'oh! Change $array = mysql_fetch . . . to $row = mysql_fetch . . .  Sorry about that.

[plznty]

D'oh! Change $array = mysql_fetch . . . to $row = mysql_fetch . . .  Sorry about that.

Your a god. Thanks!

[Pikachu2000]

If you put this code on a live production server, you'll probably want to change the error reporting that's in the else{} to something generic like "Sorry, there was a database error." and log the actual error in a logfile instead, but that's another subject.