FT3
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Posts posted by FT3
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es sapr0ta , ne l0ti labi bet nu pietiekami ;]
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mekleju ari g0ogle.lv ir kaukadi scripti , bet nevaru viņus adopdet ;/
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Labdien te tada lieta , vjg dabut php login scriptu kurš strada uz sessiam
piemers :
Login : input laucinš Passsword : input laucinš
ja tu esi veiksmigi ielogojies
ņem informaciju no mysql datubazes
un rada profila informaciju !
p.s. mekleju foruma nek0 tadu neatrdau ;/
ļoti ļoti vjg šo scriptu , iepriekš paldies ;)
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sk , paldies teu ;]]
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2easy , pld teu ;]] viss iet ;] ka tu kautk0 saprati maneya murga ? :D
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kods nau murgains , man viss perfecti strada ! ;]]
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Labdien te tada lieta kur man jaieliek kodu
echo "Neizdevas"
Lai vinu raditu kad nau aizpilditi input lauki ?
rekur kods !
<?function add(){ ?> <form action="" method="POST" > <table width="100%" cellspacing="3" cellpadding="0"> <tr><td><b> title 1: </b></td><td> <input type="text" name="title1"></td></tr> <tr><td><b> title 2: </b></td><td> <input type="text" name="title2"></td></tr> <tr><td><input type="submit" name="submit" value="Pievienot"></td><td> </table> </form> <? } if ($_POST['submit']) { $title1 = $_POST['title1']; $title2 = $_POST['title2']; if (($title1 != "") && ($title2 != "")) { ?> <b>Done!</b><? echo "<meta http-equiv='refresh' content='2;url=java script: history.go(-1)'>"; mysql_query("INSERT INTO titles (title1, title2) VALUES('$title1', '$title2') ") or die(mysql_error()); }}else{add();} ?>
Visska esmu meginajis bet neka nesanak ;/
Iepriekš paldies ! :)
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nja nezini , nedot tizlus linkus.
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Labdien , ta tada lieta vai jus varetu pateikt kodu kurš parbauda vai ir aizpilditi title, text, autor, date lauki
iepriekš paldies !
<?php require("connect.php"); $title = $_POST[title]; $text= $_POST[text]; $autor = $_POST[autor]; $date = $_POST[date]; mysql_query ("INSERT INTO news (title, text, autor, date) VALUES ('$title','$text','$autor','$date')") or die(mysql_error()); echo "Zina ir veiksmigi pievienota !"; ?>
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:/
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Labdien te tada lieta
vai kads nevaretu iedot kodiņu
lai piem aizpildi formas
un pec ta ka tu uzpied submit pogu
viss aizutas uz MySql
iepriekš paldies ;]
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Aleksejs ko teu vjg lai tu varetu salikt ?
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;((
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vari iedot uzreiz savenotu kodu ? :D
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Labdien .
te tads jautajums vai jus nevaretu parveidot kodu lai varetu celt max 3 buildingus
iepriekš paldies
<?php $buildings=buildings($_SESSION["user"][10]); check_c($_GET["town"], $_SESSION["user"][10]); $cq=get_c($_GET["town"]); ?> <font color="#996633" size="-1"> <b> <? if (count($cq)) ":<br \>"; for ($i=0; $i<count($cq); $i++) { $name=explode("-", $buildings[$cq[$i][1]][2]); if ($cq[$i][2]>-1) { $s=1; $l=$land[$cq[$i][1]][$cq[$i][2]]+1; } else { $s=0; $l=$data[$cq[$i][1]]+1; } echo "[<a class='q_link' href='cancel_c.php?town=".$_GET["town"]."&b=".$cq[$i][1]."&subB=".$cq[$i][2]."'>x</a>] ".$name[$s].", ".$lang['level']." ".$l." - <span id='cq".$i."'>".$cq[$i][0]."</span><script type='text/javascript'> var id=new Array(50); timer('cq".$i."', 'town.php?town=".$_GET["town"]."'); </script><br \>"; } echo ""; ?> </b> </font>
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<?php $result2 = mysql_query("SELECT * FROM users WHERE online='Y' AND id!=$id"); while($row = mysql_fetch_array($result2)) { $img = getimagesize($row['bilde']); ?> <div> <?php if($img[0] > 100) echo "<a href='{$row['bilde']}' target='_blank'><img src='{$row['bilde']}' alt='Profila bilde' width='100' /></a><br />"; else echo "<img src='{$row['bilde']}' alt='Profila bilde' /><br />";?> <?php echo "{$row['vaards']} {$row['uzvaards']}"; ?> </div> <?php } ?>
vjg ustajsti lai nerada to avatar bet gan Users online: skaits :?
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Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in C:\AppServ\www\v2\includes\online.php on line 8 Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in C:\AppServ\www\v2\includes\online.php on line 10 Viesi: Lietotaji:
erors
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query
in Vispārēji
Posted
labdien te tada lieta uzkodeju kodu bet , rada tukšu lapu kas nepareiz, k0 lai dara ?
phpmyadmin screen :
pats kods :
Iepriekš paldies !
Ja Nezini Tad Neposto ! xD