isset + get no db

[IBEX]

isset + get no db

Kāds varētu pastāstīt par šo lietu?

Esmu izveidojis, tadu lietu, ka man aiznes pieream uz lapu video, tur paradas kategorijas kadas ir iespejamas un man ir nepiecieshams lai uzpiezhot uz kategorijas (piem. muzika) aizved ari uz to muzikas sadalu, kur ir tie muzikas klipi un talak uzpiezhot uz ta klipa uz pashu klipu.. cik noprotu ar isset un get tas butu iespejams? kads nevaretu paradit kā?.. pavisam simple tikai :)

[bubu]

Ja godīgi - nekā nesapratu kāds kaut kādam isset un "get no db" (wtf?) ir sakars ar tavu bezsakarīgi garo teikumu..

Kur tieši ir problēma?

[anonīms]

<?php
if($_GET['sadala'] == 'video')
{
include "lapas/video.php";
}
elseif($_GET['sadala'] == 'galerija')
{
include "lapas/galerija";
}
else
{
include "lapas/sakums.php";
}
?>

 

Tad video.php failā

 

$visi_video = mysql_query("SELECT * FROM video");
while($video = mysql_fetch_array($visi_video))
{
echo "<a href="?sadala=video&id=".$video['0'].">Video bilde</a>";
}

 

un kad aizet uz katru noteikto video tad

 

$video = mysql_query("SELECT * FROM video where id = {$_GET['id']}");
$video = mysql_fetch_array($video);

 

un viss pārējais..

Laikam par to domāji.

[IBEX]

nu tie ir tie hiperlinki, jāā, tas jeu man bija, bet man vajag, kipa kad uzpiezh uz to ?sadala=video&id=7 tad ari kkas paradaas nevis balta lapa :) cik noprotu tas ar if (!isset($_GET['cat_id']) || !isnum($_GET['cat_id'])) { kko tadu ir dabunams gatavs, bet kaaa?.. :/

[KarlisBa]

Tev tak tikko to iepriekšējā postā parādīja... Ja url'ī ir

http://kautkas.com/index.php?sadala=video

Tad to arī attiecīgi nolasa:

if (isset($_GET['sadala']) AND $_GET['sadala'] == "kas") { echo "attiecīgā sadaļa - video"; }

Edited October 19, 2008 by KarlisBa

[IBEX]

<?php

$result = mysql_query("select * from category ORDER BY id"); 
while($r=mysql_fetch_array($result)) 
{ 
$id=$r["id"]; 
$name=$r["name"]; 
if (isset($_GET['id']) AND $_GET['id'] == "video") { echo "<a href='?id=video?cat=$id'>$name</a> | "; }

} 

?>

Kods tas.. nu sanak, ka paradas man tas kategorijas, bet uzpiezhot uz linka aiziet uz baltu lapu.. Ka man shaja lapa kko ieksha dabut? :) plz :)))

Edited October 19, 2008 by IBEX

[anonīms]

<?php


if($_GET['id'] == 'video' AND $_GET['cat'])
{
$rez = mysql_query("SELECT * FROM video WHERE cat = {$_GET['cat']}");
while($rr = mysql_fetch_array($rez))
{
echo '<a href="?id=video&cat='.$_GET['cat'].'&video_id='.$rr['id'].'">'.$rr['video_nosaukums'].'</a>';
}
}
else
{
$result = mysql_query("select * from category ORDER BY id");
while($r=mysql_fetch_array($result))
{
$id=$r["id"];
$name=$r["name"];
echo "<a href='?id=video&cat=$id'>$name</a><br />"; }
}
}
?>

 

 

un vispaŗ tev tur sanāca tas links www.tavalapa.lv/?id=video?cat=1, bet vajadzēja www.tavalapa.lv/?id=video&cat=1

[IBEX]

ahaa.. sk.. un tālāk, man sanāk..

<?php

if($_GET['id'] == 'video' AND $_GET['cat'] te vel likt to video_id un get to id ne? )

{

$rez = mysql_query("SELECT * FROM video WHERE cat = {$_GET['cat']}");

while($rr = mysql_fetch_array($rez))

{

echo 'te naks info par video';

}

}

else

{

if($_GET['id'] == 'video' AND $_GET['cat'])

{

$rez = mysql_query("SELECT * FROM video WHERE cat = {$_GET['cat']}");

while($rr = mysql_fetch_array($rez))

{

echo '<a href="?id=video&cat='.$_GET['cat'].'&video_id='.$rr['id'].'">'.$rr['video_nosaukums'].'</a>';

}

}

else

{

$result = mysql_query("select * from category ORDER BY id");

while($r=mysql_fetch_array($result))

{

$id=$r["id"];

$name=$r["name"];

echo "<a href='?id=video&cat=$id'>$name</a><br />"; }

}

}

?>

[anonīms]

<?php
if($_GET['id'] == 'video' AND $_GET['cat'] AND $_GET['video_id']
{
$vidd = mysql_query("SELECT * FROM video WHERE id = {$_GET['video_id']}");
$vidd = mysql_fetch_arraY($vidd);

}
else
{
if($_GET['id'] == 'video' AND $_GET['cat'])
{
$rez = mysql_query("SELECT * FROM video WHERE cat = {$_GET['cat']}");
while($rr = mysql_fetch_array($rez))
{
echo '<a href="?id=video&cat='.$_GET['cat'].'&video_id='.$rr['id'].'">'.$rr['video_nosaukums'].'</a>';
}
}
else
{
$result = mysql_query("select * from category ORDER BY id");
while($r=mysql_fetch_array($result))
{
$id=$r["id"];
$name=$r["name"];
echo "<a href='?id=video&cat=$id'>$name</a><br />"; }
}
}
?>

[IBEX]

mjā, tā jeu domāju, ok, liels paldies, tas tākā būtu, bet viņam kkas nepatīk..

Undefined index: cat in C:\Program Files\EasyPHP 2.0b1\www\own\switch\video.php on line 2

cik sapratu, ka nau linkā tie parametri kas ir if norādīti, tāpēc viņš bļauj.. kā tikt no tā vaļā?.. :/

if($_GET['id'] == 'video' AND $_GET['cat'] AND $_GET['video_id']) <--- otra linija..

Edited October 29, 2008 by IBEX

[Val]

skat. postu #5

[IBEX]

ERROR REPORTING palidzeja :) Tnx :)

[Val]

vispār jau problēma palika; tu viņu tikai "paslēpi".

[IBEX]

nuuu, jā, bet ka savādāk.. nesaprotu.. :/

<?php
if($_GET['id'] == 'video' AND $_GET['cat'] AND $_GET['video_id'])
{
$vidd = mysql_query("SELECT * FROM video WHERE video_id = {$_GET['video_id']}");
$vidd = mysql_fetch_arraY($vidd);
{
echo '
<center>' . $vidd['name'] . '</center>
<BR>
<center>' . $vidd['url'] . '</center>
<BR>
<center>' . $vidd['description'] . '</center>
';
}
}
else
{


if($_GET['id'] == 'video' AND $_GET['cat'])
{
$rez = mysql_query("SELECT * FROM video WHERE cat = {$_GET['cat']}");
while($rr = mysql_fetch_array($rez))
{
echo ' <a href="?id=video&cat='.$_GET['cat'].'&video_id='.$rr['video_id'].'">'.$rr['name'].'</a> | ';
}
}
else
{


$result = mysql_query("select * from video_category ORDER BY id");
while($r=mysql_fetch_array($result))
{
$id=$r["id"];
$name=$r["name"];
$pic=$r["pic"];
echo " <a href='?id=video&cat=$id'>$name</a> | <img src='$pic' alt='$name'>"; }
}
}
?>

Nez, ja vēlies būt tik laipns :)

[anonīms]

ar isset