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FT3

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Posts posted by FT3

  1. slect rand() nestrada ?

    in Vispārēji

    Posted · Edited by FT3

    un ka to izlabot ?

     

    pilns kods

    <?php
$con = mysql_connect("localhost","aaaa","aaaa");
if (!$con){die('Could not connect: ' . mysql_error());}
mysql_select_db("quix_base", $con);
$now = time();
if($_GET['type'] == 'banner'){
$result = mysql_query("SELECT * FROM `banner_ad` WHERE `expire` > '$now' ORDER BY rand() LIMIT 1")or die(mysql_error());
while($row = mysql_fetch_array($result)){
echo "
<a href='".$row['adress']."' target='_blank'><img src='".$row['banner']."' title='".$row['title']."' width='468' height='60'></a>
";
}
}else{
$result = mysql_query("SELECT * FROM `text_ad` WHERE `expire` > '$now' ORDER BY rand() LIMIT 2");
while($row = mysql_fetch_array($result)){
echo "
<a target='_blank' href='".$row['url_line']."' >
<div id='text_advert'>
<B>".$row['bold_line']."</B>
<BR>
".$row['text_line1']."
<BR>
".$row['text_line2']."
<BR>
<b>".$row['url_line']."</b>
</div>
</a>
";
}
}
mysql_close($con);
?>

     

    text_ad izvelk viss okey , bet to banner_ad neka ,

  2. slect rand() nestrada ?

    in Vispārēji

    Posted · Edited by FT3

    Sveiski , man te atgadijusies ķibele

     

    $result = mysql_query("SELECT * FROM `banner_ad` WHERE `expire` > '$now' ORDER BY rand() LIMIT 2");
while($row = mysql_fetch_array($result)){
echo "
<a href='".$row['adress']."' target='_blank'><img src='".$row['banner']."' title='".$row['title']."' width='468' height='60'></a>
";

     

    izvada tikai 1 banneri un visu laiku vienu un to pašu

     

    bet ja noņemu

    WHERE `expire` > '$now'

     

    tad rand() strada ;/

  4. replace

    in Vispārēji

    Posted

    Sveiki ir problema neka nevaru atrisinat ..

     

    iekš input ir šadi value

    value="0.15 Ls"
value="0.30 Ls"
value="0.45 Ls"
value="0.60 Ls"
value="0.75 Ls"
value="0.95 Ls"
value="1.25 Ls"
value="1.50 Ls"
value="1.75 Ls"
value="2.00 Ls"

     

    ka izdarit ta lai pie $_POST izvada 15 , 30 , 125 u. c.

     

    Iepriekš paldies :)

  6. jquery .html().replace

    in Javascript

    Posted

    Sveiki , vajadziga palidziba :)

     

    es lietoju šo jquery replace kodu , bet ir viena problema kods aizvieto tikai vienu [test]

    ka lai panakt lai viņš aizvietotu visus [test] kuri atrodas majalapa ?!

     

    $('#ipbwrapper').html($('#ipbwrapper').html().replace('[test]','<span class="att">test</span> '));

     

    Iepriekš paldies, un sorry par garumzimem ;D

  9. js iframe

    in Javascript

    Posted · Edited by FT3

    jo gala rezultata man vajadzes panakt ta

     

    <script>var rs_id=UNIQ ID;</script>
<script src="http://manalapa.lv/krutajs.js"></script>

     

     

    pieliku to width un height lkm nepreizi , tik un ta neko nerada .

     

    <script>
var vote_hd = document.getElementsByTagName('head')[0];
var vote_tag = document.createElement('iframe');
vote_tag.width = '300';
vote_tag.height = '300';
vote_tag.src = 'http://manalapa.lv?p=' + rs_id;
vote_hd.appendChild(vote_tag);
</script>

  10. js iframe

    in Javascript

    Posted

    Sveiki taka no js vispar neko nejedzu , prasu jusu palidzibu :)

     

    Ka šim kodam pielikt width un height lai rada to iframe ?

    width un height ir nepieciešams lai parbauditu scripta darbibu , jo tgd viņš itka ielade bet neko nerada .

     

    <script>var rs_id=MANS ID;</script>
<script>
var vote_hd = document.getElementsByTagName('head')[0];
var vote_tag = document.createElement('iframe');
vote_tag.src = 'http://manalapa.lv?p=' + rs_id;
vote_hd.appendChild(vote_tag);
</script>

     

     

    iepriekš tnq

  12. mysql select problema

    in Vispārēji

    Posted

    Sveiki.

     

    Man te tada lieta, man datubaze ir izveidotas 3 kategorijas un 3 forumi , bet izvada tikai vienu ! why ?

     

    <?
include('connect.php');
$result = mysql_query("SELECT * FROM `rc_forum_category` ORDER BY `id` ", $connect) or die(mysql_error());
 if(mysql_num_rows($result) == 0){echo 'forum DOWN !';}
 for ($i = 0; $i < mysql_num_rows($result); $i++)
 {
 	$c_id = mysql_result($result, $i, "id");
 	$c_title = mysql_result($result, $i, "title");

$result = mysql_query("SELECT * FROM `rc_forum_forum` WHERE `category` = '$c_id'", $connect) or die(mysql_error());
 for ($i = 0; $i < mysql_num_rows($result); $i++)
 {
 	$f_id = mysql_result($result, $i, "id");
 	$f_title = mysql_result($result, $i, "title");  

echo '
<b>'.$c_title.'</b><BR>
'.$f_title.'
';
}
}
?>

  22. from php to cmd

    in Vispārēji

    Posted 

    <?php
$text = $_POST['text'];
if($_POST['go']){
exec("/say $text");
}
?>
<form method="POST">
<input name="text" >
<input type="submit" name="go">
</form>

     

    mans kods izskatas šadi , bet nekadu komandu vins neaizsuta uz cmd ;[[

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