4 random pictures (thumbs)

Anon

How do i randomly select 4 diferent pictures?

I'm using the function below to select 1 random picture... BUT when i use this function 4 times i get 4 pictures with sometimes 2 the same... (logical offcourse) How can i avoid this?

I want 4 diferent random pictures or ID's to include in an img tag.

Thanks,

Gerry

function random_picture() {
global $CFG;

$qid = mysql_query("SELECT * from download_image");

$max = mysql_num_rows($qid)-1;

mt_srand((double)microtime()*1000000);
if ($max > 0) {
$random = mt_rand(0,$max);
} else {
$random = 0;
}

for ($x=0;$x<=$random;$x++) {
$myrow = mysql_fetch_array($qid);
}

$id = $myrow["id"];

echo "<img src='./admin/images/idc/" . $id ."_t.jpg' border=0>";

}

Anon

try using this for your sql_statement
$sql = "select * from download_image order by rand() limit 4";

just my 2 cents...if works for me

Anon

Using something I got from here I think, partly, anyways, syntax is probably dodgy, normally is 🙂

<?php
srand((double)microtime()*1000000);
$randstring = "";
for ($i=0;$i<1;$i++) {
$randstring = $randstring.rand(1,4);
}
if ($randstring == "1") {echo "<img src=\"http://www.site.com/images/img1.gif\">"; }
elseif ($randstring == "2") {echo "<img src=\"http://www.site.com/images/img2.gif\">"; }
elseif ($randstring == "3") {echo "<img src=\"http://www.site.com/images/img3.gif\">"; }
elseif ($randstring == "4") {echo "<img src=\"http://www.site.com/images/img4.gif\">"; }
?>

It should work, thought it may not be the best way of doing it.
(No, I don't fully understand it, just that it can generate me a random number of 30 digits and so with a little editing and some knowledge, it should work, heck, I've been wanting something similar for my own site anyways)

Anon

lol, no wonder it was faster than my reply! Of course, if you don't want to go to the trouble of using a SQL db for it, but then this method is a lot faster with a much smaller php file, and allows for easy editing of which pics are to be used!

prophetize

This may help:

<?
include("dbconnect.inc");
$id=rand(1,8);
$sql="SELECT * FROM randimg WHERE id=$id";
$result=@($sql);
$row=mysql_fetch_array($result);
echo "<img src=\"images/" . $row["src"] . "\" alt=\"" . $row["name"] ."\">"
?>

With this in mind, just have a DB set up with the following fields:

id
name
src

In addition, at the top ... "$id=rand(1,8)" ... just change the integers according to the number of images you want to choose from.

Hope this helps,
Prophetize