[SOLVED] constant checking

[Canadiengland]

for the following script, how would i make it display multiple unequipped items but not display the same ones more than once?

 

//unequipped items -------------------------------------------------------------------
    $uneq="SELECT * from km_items where owner='$player' and status='u'";
    $uneq2=mysql_query($uneq) or die("Could not get user stats");
    $uneq3=mysql_fetch_array($uneq2);
if (!empty ($uneq3[id])) {
echo "<a href=equip.php?itemid=$uneq3[id]><img src='$uneq3[itempic]'";
} else {
}

 

thanks!

[JasonLewis]

change this:

 

$uneq="SELECT * from km_items where owner='$player' and status='u'";

 

to this, i think:

 

$uneq="SELECT DISTINCT(*) from km_items where owner='$player' and status='u'";

[Canadiengland]

hmm.. didnt work

[JasonLewis]

oh wait, sorry you dont have a while-loop.

 

try this then.

 

//unequipped items -------------------------------------------------------------------
$uneq="SELECT DISTINCT(*) from `km_items` where `owner`='$player' and `status`='u'";
$uneq2=mysql_query($uneq) or die("Could not get user stats");

$shown_items = ""; //this is just another checker, incase my mysql is gay.
while($uneq3=mysql_fetch_array($uneq2)){
if (!empty($uneq3['id']) && !in_array({$uneq3['itempic']}, $shown_items)){
echo "<a href='equip.php?itemid={$uneq3['id']}'><img src='{$uneq3['itempic']}'></a><br>";
$shown_items[] = $uneq3['itempic'];
}
}

 

i hope i did it right that time.

[Canadiengland]

still doesnt work... i think its cause im using an older version of php or mysql or something.. using whatever comes with wampserver.. but its getting late thanks for trying, ill mess with it more tomorrow.

[JasonLewis]

what is the exact problem now?

 

not showing all results or showing the same result more then once or are there errors being produced?

[Canadiengland]

//unequipped items -------------------------------------------------------------------

$uneq="SELECT DISTINCT * from km_items where owner='$player' and status='u'";

$uneq2=mysql_query($uneq) or die("Could not get user stats");

$uneq3=mysql_fetch_array($uneq2);

 

$shown_items = ""; //this is just another checker, incase my mysql is gay.

while($uneq3=mysql_fetch_array($uneq2))

{

if (!empty($uneq3['id']) && !in_array($uneq3['itempic']))

{

echo "<a href=equip.php?itemid=$uneq3[id]><img src='$uneq3[itempic]'<br>";

$shown_items[] = $uneq3['itempic'];

}

}

print "</table><br><br>";

    }

  }

 

this code gives me the effect but it also says (on the page)

 

Warning: Wrong parameter count for in_array() in C:\wamp\www\killmonster\index.php on line 242

 

 

any ideas?

 

242 is if (!empty($uneq3['id']) && !in_array($uneq3['itempic']))

[Canadiengland]

buuuuuuuuuuuuuump

[per1os]

Just a note on the DISTINCT modifier, it generally does not work on * it needs to be a specific column, because doing a distinct on * makes it so every single column must match to not display it.

 

IE:

SELECT DISTINCT(lname), userid, pass ...  would work like it should.

 

On that note:

 

 

http://us2.php.net/manual/en/function.in-array.php

 

!in_array($uneq3['itempic']) ---- What are you checking for here? You need at least 2 parameters, an array and a needle.

 

//unequipped items -------------------------------------------------------------------
$uneq="SELECT DISTINCT * from km_items where owner='$player' and status='u'";
$uneq2=mysql_query($uneq) or die("Could not get user stats");
$uneq3=mysql_fetch_array($uneq2);

$shown_items = array(); 
while($uneq3=mysql_fetch_array($uneq2))
{
if (!empty($uneq3['id']) && !in_array($uneq3['itempic'], $shown_items))
{
echo "<a href=equip.php?itemid=$uneq3[id]><img src='$uneq3[itempic]'
";
$shown_items[] = $uneq3['itempic'];
}
}
print "</table>

";
    }
  }

 

See if that works.

[Canadiengland]

wow that worked - thanks for helping AND explaining what you did, so i won't make future errors now!

[Canadiengland]

may have spoke too soon - that displays a max of 2 seperate pics and both pics are linked to the same url... hmm

[per1os]

<?php
//unequipped items -------------------------------------------------------------------
$uneq="SELECT * from km_items where owner='$player' and status='u'";
$uneq2=mysql_query($uneq) or die("Could not get user stats");
//$uneq3=mysql_fetch_array($uneq2); -- I do not know why this was there?

$shown_items = array(); 
while($uneq3=mysql_fetch_array($uneq2))
{
if (!empty($uneq3['id']) && !in_array($uneq3['itempic'], $shown_items))
{
echo "<a href=equip.php?itemid=$uneq3[id]><img src='$uneq3[itempic]'
";
$shown_items[] = $uneq3['itempic'];
}
}
print "</table>

";
    }
  }
?>

 

If you can give me a database structure, that can help, because chances are the problem is in the query.

[Canadiengland]

database structure of km_items is

 

name

power

type

expperturn

rageperturn

itempic

owner

id

status

cost

rarity

raritymax

itemfind

[Canadiengland]

well i kinda fixed it... the way you put it up there made it so that each item would be displayed according to pic... but i wanted it to be each item.. so i changed

 

!in_array($uneq3['itempic'], $shown_items))

 

to

 

!in_array($uneq3['id], $shown_items))

 

 

but ive still got the problem of each pic has the same url when they are 2 different ids

[Canadiengland]

SOLVED! added a <td> after each echo and that solved it. thanks for all the help project and frost!