pomoc oko mysql_fetch

[ smart_boy @ 28.07.2004. 11:43 ] @

pozdravi

imam neku news scriptu. ipload sm je na neti kada idem na neke stranice stalno mi pokazuje ovu gresku :

Citat:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/vedran/public_html/nws1/display.php on line 5

a kod je :

display.php

Code:

<?php
include('db.php');
$q = mysql_query("SELECT * FROM $table ORDER BY id DESC LIMIT $limit"); // That queriesthe database

while($r=mysql_fetch_array($q)){
$title = $r["title"];
$news = $r["news"];
$body = nl2br($news);
$author = $r["author"];
$link = "mailto:".$r["email"];
$date = $r["date"];
echo "<table width=\100%\ cellspacing=\2><tr>
<td><b>$title</b> posted by <a href='$link'>$author</a> on $date</td></tr>
<tr><td>$body</td></tr>
</table>";
}
?>

db.php

Code:
<?php
$user = "user";
$pass = "pass";
$dbase = "db_user";
$table = "news";
$host = "localhost";
$limit = "5";

$cnx = mysql_connect($host, $user, $pass);
mysql_select_db($dbase);
?>

Code:
CREATE TABLE `news` (
`id` INT( 40 ) NOT NULL AUTO_INCREMENT ,
`title` VARCHAR( 200 ) NOT NULL ,
`date` DATE NOT NULL ,
`news` TEXT NOT NULL ,
`author` VARCHAR( 40 ) NOT NULL ,
`email` VARCHAR( 100 ) NOT NULL ,
PRIMARY KEY ( `id` )
);

uvijek mi se prikazuje ista greska KOD SVIH SCRIPTI !!!

molim vas pomozite mi !
thx

[ boccio @ 28.07.2004. 13:06 ] @

cisto na prvu loptu... da li si probao ovako:

$q = "SELECT * FROM {$table} ORDER BY id DESC LIMIT {$limit}";
$sql = mysql_query($q);
while($r=mysql_fetch_array($sql)) {

[ smart_boy @ 28.07.2004. 15:00 ] @

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/vedran/public_html/nws1/display.php on line 16

nista : \ stvarno ne kontam ovo :(

[ _owl_ @ 28.07.2004. 15:32 ] @

Pa ovako odoka cini mi se da su za LIMIT potrebna dva parametra, pocetak i offset. Uglavnom mozes da svom kodu posle mysql_query() dodas:

Code:

echo $sql;
print_r(mysql_error());

[ dr ZiDoo @ 28.07.2004. 16:14 ] @

Citat:

smart_boy: Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/vedran/public_html/nws1/display.php on line 16

nista : \ stvarno ne kontam ovo :(

Funkcija koju si pozvo ne moze da izvrti zadati string zato sto on nije u odgovarajucem formatu, tj. mysql_fetch_array ne moze da izvrti string zato sto on nije niz nego vjerovatno greska u queryu.

[ smart_boy @ 28.07.2004. 17:04 ] @

evo stavio sam to i dobio :

Table "exel_tabela.news' doesn't exist
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/vedran/public_html/nws1/display.php on line 18

napravio sam cijelu tabelu u phpmyadminu osim PRIMARY KEY ( `id` ), kako cu to napravit ?

tebela :

CREATE TABLE news (
id INT( 40 ) NOT NULL AUTO_INCREMENT ,
title VARCHAR( 200 ) NOT NULL ,
date DATE NOT NULL ,
news TEXT NOT NULL ,
author VARCHAR( 40 ) NOT NULL ,
email VARCHAR( 100 ) NOT NULL ,
PRIMARY KEY ( `id` )
);

ili na primjer ovo : ENGINE=MyISAM AUTO_INCREMENT=4

u tabeli

CREATE TABLE `news` (
`id` int(11) NOT NULL auto_increment,
`title` text NOT NULL,
`message` longtext NOT NULL,
`who` text NOT NULL,
`date` text NOT NULL,
`time` text NOT NULL,
PRIMARY KEY (`id`)
) ENGINE=MyISAM AUTO_INCREMENT=4 ;

[ bluesman @ 28.07.2004. 23:15 ] @

exel_tabela.news znaci da tabela news ne postoji u bazi excel_tabela. Da li si siguran da si:
a) selektovao dobru bazu za rad (onu koja ti treba)?
b) ubacio tabelu bas u tu bazu a ne u neku drugu?

Jos ako si iz phpmyadmin - vrlo lako mozes greskom da dodas tabelu u neku sasvim drugu bazu od one u koju si hteo. Proveri.

startuj mysql pa kucaj:
mysql> use excel_tabela
mysql> show tables;

i treba da ti se prikaze spisak svih tabela u bazi. Tvoja tabela treba da je tamo (ako si je dobro kreirao).

[ smart_boy @ 29.07.2004. 00:08 ] @

ehh THX !
sve sam skonto, napravio sam install scriptu sve ql radi, osim ... :D
napravim display.php

Code:

<?php
include ("config.php");

$conn = mysql_connect("$db_host", "$db_user", "$db_pass") or die("Ne mogu se konektovati");
mysql_select_db("$db_name", $conn);

$q = "SELECT name FROM news1";

$sql = mysql_query($q);
echo $sql;
print_r(mysql_error());

while($r=mysql_fetch_array ($sql)) {
$title = $r["news"];

echo "$title"
}
?>

samo mi ispise : Resource id #4
i nista vise ... :\
una li iko sta ovo znaci ??

thx :D

[ bluesman @ 29.07.2004. 00:48 ] @

Pa to je normalno, jer mysq_query vraca resource tip (pogledaj PHP manual) i to ne mozes da printas (mozes ali samo ispise ono sto je tebi ispisao)

osim toga, ti radis SELECT name...
a stampas $r["news"];
kako si to mislio da izvedes? :-)

mysq_errror() vraca string a ne array, pa mozes sa echo (ne moras print_r)

recimo bolji code bi bio ovako:

Code:

<?php
include ("config.php");

$conn = mysql_connect("$db_host", "$db_user", "$db_pass") or die("Ne mogu se konektovati");
mysql_select_db("$db_name", $conn);

$q = "SELECT name FROM news1";
$sql = mysql_query($q) or print("MYSQL ERROR: #".mysql_errno().": ".mysql_error());

if ($sql)
     {
     while($r=mysql_fetch_array ($sql)) {
          echo $r["name"];
     }
else
     echo "some bullshit happened, error in query";
?>

[ smart_boy @ 29.07.2004. 16:17 ] @

thx radi :)
i jos samo jedna stvarcica :D
kako da brisem iz tabele :D

ovako imam ovaj kod:

Code:

<?
if(!isset($cmd))
{

   $result = mysql_query("select * from news order by id");

   while($r=mysql_fetch_array($result))
   {

      $title=$r["title"];//take out the title
      $id=$r["id"];//take out the id

      echo "<a href='delete.php?cmd=delete&id=$id'>$title - Delete</a>";
      echo "<br>";
    }
}

if($_GET["cmd"]=="delete")
{
    $sql = "DELETE FROM news WHERE id=$id";
    $result = mysql_query($sql);
    echo "Row deleted!";
}
?>

(skino sam sa neto ovaj kod)

ubacim sve i namjestim i ispisu mi se rediovi. kad kliknem na delete on otvori novi prozor i kaze "the page cannot..."
a kada napravim delete.php i ubacim kod :

if($_GET["cmd"]=="delete")
{
$sql = "DELETE FROM news WHERE id=$id";
$result = mysql_query($sql);
echo "Row deleted!";
}

kaze mi row deleted al nista ne izbrise

eto jos samo ovo i necu vas vise zamarat:
kako da fino napravim kod da mi brise
thx for any help :)

[ _owl_ @ 29.07.2004. 20:19 ] @

Mozda ovako:

Code:

if($_GET["cmd"]=="delete") {
   $sql = "DELETE FROM news WHERE id='" . $_GET["id"] . "'";
   $result = mysql_query($sql);
   echo "Row deleted!";
}

[ smart_boy @ 29.07.2004. 20:58 ] @

thx, ovo stvarno radi :D
sada cu se zadat i sam skontat kako da editujem :D
one more time thx ;)

pomoc oko mysql_fetch_array():

pomoc oko mysql_fetch_array():