Checking if the value has changed in a while loop help

[TEENFRONT]

Hey

 

Il jump straight in with my question. I have a while() loop outputting info from the db but im having trouble telling if a certain value has changed from the last loop, and if it is different i want to show an image. Showing an image is not the problem, just the telling if the value is different to the last loop. OK heres my code

 

$date = explode(" ", $date);     // This splits the date and time (EG 22/01/08 22:15) field into just the date giving me $date[0] in this format DD/MM/YYYY
$jobday = explode("/", $date[0]);   // This splits the $date[0] to just the first set of numbers so it outputs DD (EG 22) as $jobday[0]
echo $jobday[0];

 

so it loops through the results and $jobday[0] will output the following results in 6 loops along with an image EG

 

22  img

22  img

22  img

21  img

21  img

20  img

 

how do i tell if the number has changed between loops? So i only insert an img if the number is different from the last....sooo like this

 

22  img

22

22

21  img

21

20  img

 

 

I tried things like if($jobday[0] != $jobday[0])  but as $jobday[0] is set inside my loop, $jobday[0] with always equal $jobday[0]. And if i try setting $jobday[0] outside of the while loop then obviosly its not set right for each loop...

 

Any help????? Im going mad

[kenrbnsn]

You need to use another variable to store the previous value:

<?php
$temp = array();
for ($i=0;$i<6;$i++)
$temp[] = rand(20,24);
sort($temp);
$prev = '';
foreach ($temp as $v) {
echo $v;
echo ($v != $prev)?' something':'';
echo '<br>';
$prev = $v;
}
?>

 

Ken

[TEENFRONT]

Hey

 

Thanks for that. Wooooooosh straight over my head.

 

Any additional help would be appreciated. This is a cut down version of my code, how would i implement kenrbnsn's example above?

 

while($r=mysql_fetch_array($result))
{

$jobday = explode("/", $date[0]);
echo $jobday[0];
echo $image;

}

 

So the above will output this (say there is 6 records to loop through from the db)

 

22  img

22  img

21  img

21  img

20  img

20  img

 

i only want to show "img" if the date number if different to the previous date. kenrbnsn's example looks like what i need when i run it in isolation but i dont know how to fit that into my while loop to work?

 

[kenrbnsn]

Try:

<?php
$prev = '';
while($r=mysql_fetch_array($result))
{
    $jobday = explode("/", $date[0]);
    echo $jobday[0];
    if ($jobday[0] != $prev) echo ' ' . $image . '<br>';
    $prev = $jobday[0];
}
?>

 

Ken