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I am using this code:

 

<?php

$q = mysql_query("SELECT * FROM `table`");

$r = mysq_fetch_array($q);

$r = array_rand($r);

 

echo $r['column'];

 

?>

 

But that doesnt work.

 

What I am trying to do, is select a random row from the db, then I can display the columns. How would I do that? Thanks

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https://forums.phpfreaks.com/topic/117798-solved-rand/
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You need to realize what mysql_fetch_array is doing

mysql_fetch_array produces the "next" row in the query resource to be displayed.

 

using array_rand($row) will only grab a random field from that query row.

 

 

http://www.phpfreaks.com/forums/index.php/topic,125759.0.html

 

that shows exactly how to do a proper query for a random row.

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https://forums.phpfreaks.com/topic/117798-solved-rand/#findComment-605873
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obviously more than you have since I don't just give them  a code snippet without an explanation of why their methods are invalid or inappropriate for their goals and what your method does.

 

You really shouldn't dabble in the php boards if you aren't willing to give complete and proper answers to people.

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