Kļūda

[johanes]

<?php
include $DOCUMENT_ROOT.'/wap/inc/header.php';
include $DOCUMENT_ROOT.'/wap/inc/db/config.php';
#######
if (empty($page)) $page = 0;
if ($page < 0) $page = 0;

$count_query = 'select count(*) from jaunumi where 1;';
$total_mess = mysql_query ($count_query);
$total_count = mysql_fetch_array ($total_mess);
$count = $total_count ['count(*)'];
##########################################
$query = 'select * from jaunumi where 1 ORDER BY `date` DESC  LIMIT '.$page.', '.$max_mess.';';
$result = mysql_query($query);
if (!$result)
{
echo "Error! mysql_query".mysql_error();
exit;
}
echo "Jaunumi: ".$count;
if ($count < 0) # Ja nav zinju neraada cik ierakstu
{
echo 'Nav neviena jaunuma!';
include $DOCUMENT_ROOT.'/wap/inc/end.php';
exit;
}

while ($i < $max_mess) #
{
$row_date = htmlspecialchars(mysql_result($result, $i, "date"));
$row_message = htmlspecialchars(mysql_result($result, $i, "message"));
$row_message = trim($row_message);
if (empty($row_message)) break;
echo "<br/>_________________________<br/>";
echo "".$row_date;
echo "<br /><b>><> </b>".$row_message;
echo "<br/><br/>";
$i++;
}

problēma ir tā ka izmet kļūdu

Warning: mysql_result(): Unable to jump to row 1 on MySQL result index 4 in /index.php on line 32

Edited June 23, 2005 by johanes

[Grey_Wolf]

$count_query = 'select count(*) from jaunumi where 1;';

 

'Paraadiit skaitu (visam) no tabulas jaunumi Ja 1'

Izlasiji ?

saprati?

Mysql arii nesaprata.

:(

peec WHERE janoraada nosacijums !!!

nevis kaads lielums

piem

ja: ieraksts laukss x=y :)

--------

 

Prieciigus sveetkus un lai izdodas....

[bubu]

Grey_Wolf, kverijs ir pareizs. WHERE 1 nozīmē to pašu, ko WHERE True, vai WHERE 1=1. Tb visas rindas (jēga gan nekāda), bet mysql to ļoti labi saprotīs un kļūdu nemetīs.

 

johanes, šī vieta nav pareiza:

$count_query = 'select count(*) from jaunumi where 1;';
$total_count = mysql_fetch_array ($total_mess);
$total_mess = mysql_query ($count_query);
$count = $total_count ['count(*)'];

Vajag apmēram šādi:

$count_query = 'select count(*) as skaits from jaunumi where 1;';
$total_mess = mysql_query ($count_query);
$total_count = mysql_fetch_array ($total_mess);
$count = $total_count ['skaits'];

vai vienkāršāk:

$q = mysql_query('select count(*) from jaunumi';
list($count) = mysql_fetch_row($q);

[johanes]

Liels paldies :)

Prieciigus ligo sveetkus.... :D :D :D

[johanes]

Īstenībā kļūdu rada šeit

$row_date = htmlspecialchars(mysql_result($result, $i, "date"));
$row_message = htmlspecialchars(mysql_result($result, $i, "message"));

[v3rb0]

WHERE 1 nozīmē to pašu, ko WHERE True, vai WHERE 1=1. Tb visas rindas (jēga gan nekāda),

 

laikam pielietojums where true ir tikai tad kad generee visu where dalju un esi par slinku noskaidrot vai where vispar ir, tad raksti $sql .= 'where true '

if (kaut kas) {

$sql .= ' and shis= tas ';

}

 

if (kaut kas cits) {

$sql .= ' and tas=shis ';

}

 

citus gadiijumus kaut kaa nevaru iedomaaties:)

[bubu]

Nez, es labāk šādi tad daru:

$where = array();
if ($kautkas) { $where[] = "abc = xxx"; }
if ($citskautkas) { $where[] = "cits = 4"; }
if ($where) { // pēc Venom metodes :)
 $sql .= 'WHERE ' . implode(' AND ', $where);
}

Citās valodās tas noteikti ir efektīvāk, jo stringus konkatenēt ir lēnāk, nekā masīvam pievienot elementu. Par php nezinu, neesmu testējis, pārsvārā šādu metodi izmanotju JavaScript kodā :)