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homers

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Posts posted by homers

  4. Pievieno citā tabulā

    in Iesācējiem

    Posted

    Pamainīju uz $_POST['join']

    <?php
if(isset($_POST['join'])){
$tmt = mysql_query("SELECT * FROM komanda WHERE playera_id=$id");
if(mysql_num_rows($tmt)  > 0){
$resultsas = mysql_query("SELECT * FROM komanda WHERE playera_id = $id") or die(mysql_error());
$ras = mysql_fetch_array($resultsas);
$usera_team = $ras['nosaukums'];
$komanda = $usera_team;
$turnirs = $s['nos'];
$t_idd = isset($_GET['id']);
$statuss = "2";
mysql_query("INSERT INTO komandas_turnira(komanda, turnirs, turnira_id, status) VALUES('$komanda', '$turnirs', '$t_idd', '$statuss')");
echo "<script>alert('Tagad gaidi kamēr tavu komandu apstiprinās!');</script>";
echo '<meta http-equiv="refresh" content="0;url=http://localhost/?p=cup&id='.$_GET['id'].'">';
}else{
echo "Tev nav komandas!";
}
}else{
echo "";
}
?>

  5. Pievieno citā tabulā

    in Iesācējiem

    Posted · Edited by homers

    Tatad piemēram es atrodas http://*.lv/?p=cup&id=1 spiežu Pievienoties turnirā viss notiek labi, bet kad atrodos http://*.lv/?p=cup&id=2 spiežu pievienoties un tabulā pievienojas visa informācija tikai turnira_id parādas 1

    Kas ir janomaina lai tabulā pievienotos pareiza turnira id ?

     

    Kods

    <?php
if($_GET['p'] == 'cup' and isset($_GET['id']) and $_GET['id'] == $_GET['id'] and isset($_GET['go']) and $_GET['go'] == 'join'){
$tmt = mysql_query("SELECT * FROM komanda WHERE playera_id=$id");
if(mysql_num_rows($tmt)  > 0){
$resultsas = mysql_query("SELECT * FROM komanda WHERE playera_id = $id") or die(mysql_error());
$ras = mysql_fetch_array($resultsas);
$usera_team = $ras['nosaukums'];
$komanda = $usera_team;
$turnirs = $s['nos'];
$t_idd = isset($_GET['id']);
$statuss = "2";
mysql_query("INSERT INTO komandas_turnira(komanda, turnirs, turnira_id, status) VALUES('$komanda', '$turnirs', '$t_idd', '$statuss')");
echo "<script>alert('Tagad gaidi kamēr tavu komandu apstiprinās!');</script>";
echo '<meta http-equiv="refresh" content="0;url=http://lapa.lv/?p=cup&id='.$_GET['id'].'">';
}else{
echo "Tev nav komandas!";
}
}else{
echo "";
}
?>

  8. mysql fetch array errors

    in Iesācējiem

    Posted · Edited by homers

    Errors:

    Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in C:\xampp\xampp\htdocs\index.php on line 188

     

    188 rinda:

    while ($row = mysql_fetch_array($results, MYSQL_ASSOC)) {

     

    Viss kods:

    <?php
$k = mysql_result(mysql_query("SELECT COUNT(*) as Num FROM komandas_turnira"),0);
$skaits = "8";
$s_rez = $skaits-$k;
$open = "o";
$results = mysql_query("SELECT id,nos,x FROM turniri WHERE status=".$open." ORDER BY id DESC");
while ($row = mysql_fetch_array($results, MYSQL_ASSOC)) {
printf("<img src='/style/icons/arrow_right.gif'> <a href=?p=cup&id=%s>%s Cup</a> [<font color=green>%s</font>/<font color=red>8</font>]<br/>", $row["id"], $row["nosaukums"], $e);
}
?>

  13. Rada tikai 1 nepareizo tekstu

    in Iesācējiem

    Posted

    Kods:

    <?
$epasts = isset($_POST['epasts']);
$check = mysql_query("SELECT epasts FROM lietotaji WHERE epasts=$epasts");
$epastss = $check[0];
$niks = isset($_POST['uzvards']);
$check1 = mysql_query("SELECT niks FROM lietotaji WHERE niks=$niks");
$nikss = $check1[0];
if($epasts == $epastss){
if($niks == $nikss){
if($_POST['epasts'] == $_POST['epasts_conf']){
if($_POST['parole'] == $_POST['parole_conf']){
if(!empty($_POST['vards']) and !empty($_POST['uzvards']) and !empty($_POST['epasts']) and !empty($_POST['parole']))
{
// insert

	'Tagad vari ielogoties!';
}
else
{
	 "Aizpildi visus laukus!<br />";
}
}else{
echo "Paroles nesakrīt!<br />";
}
}else{
echo "E-pasti nesakrīt!<br />";
}
}else{
echo "Niks jau aizņemts!<br />";
}
}else{
echo "E-pasts jau ir aizņemts!<br />";
}
?>

    Bet rada tikai šo tekstu E-pasts jau ir aizņemts!

    Esmu meiģinājis ievadīt daudzus citus variantus kuru nav datubāzē, bet rada tikai to tekstu..

    Kas par problēmu?

  18. izvēlk tikai vienu..

    in Iesācējiem

    Posted · Edited by homers

    Tatad man vajag izvilkt komandas sastavu, bet skripts izvēlk tikai vienu spēletaju..

    kods:

    <?php
{
$results=mysql_query("SELECT * FROM komandas_sastavs WHERE id=".$_GET['id']."") or die(mysql_error());
$s = mysql_fetch_array($results);
echo "» <a href='?p=user&id={$s['playera_id']}'>{$s['playera_n']}</a><br>";
}
?>

  22. help

    in Draza

    Posted

    Vieglākais variants ir šāds:

     

    Izveido lietotāja db tabulu online, kad lietotājs ielogojas tabula updeito kolonu online uz vārdu on.

    Tad pie online izvēlc šadi:

    mysql_query("SELECT * FROM lietotaji WHERE online='on'");

    Bet kad izlogojas tabulā updeitojas kolonu online uz vārdu off

  24. help

    in Draza

    Posted · Edited by homers

    Nu vienkarši nomainot tabulas nosaukumu kodā

     

    <?
//konekts
$delay = "900";
$delays = time() - $delay;
$useri = mysql_query("SELECT * FROM TABULA WHERE PEDEJA AKTIVITATE >= ".$delays."");
$reg = mysql_num_rows($useri);
$viesi = mysql_query("SELECT * FROM TABULA WHERE id = '0' and LAIKS >= ".$delays."");
$viesi = mysql_num_rows($viesi);
echo '
Viesi: '.$viesi.'<br />
Lietotāji: '.$reg.'<br />';
?>

  25. help

    in Draza

    Posted 

    <?
//konekts
$delay = "900";
$delays = time() - $delay;
$useri = mysql_query("SELECT * FROM ibf_members WHERE last_activity >= ".$delays."");
$reg = mysql_num_rows($useri);
$viesi = mysql_query("SELECT * FROM ibf_sessions WHERE member_id = '0' and running_time >= ".$delays."");
$viesi = mysql_num_rows($viesi);
echo '
Viesi: '.$viesi.'<br />
Lietotāji: '.$reg.'<br />';
?>

     

    hz, vai iet

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