Php

[SaS^MAN]

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/a3191906/public_html/log.php on line 26

kods : $query = mysql_query("SELECT * FROM Log ORDER BY ID DESC");

kur ir probleema?

[Dj_Stay]

$query = mysql_query("SELECT * FROM log ORDER BY ID DESC");

vai

$query = mysql_query("SELECT id FROM Log ORDER BY ID DESC");

Neesmu php pro masters, bez scripta nevaru palīdzēt.

[evilxdeadly]

paskaties rowus un tablus vai pareizi

$pizda = mysql_query("SELECT * FROM log ORDER BY id desc");
while($row = mysql_fetch_array( $pizda )) {
echo "{$row['id']}<br />\n";
}

[mex]

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/a3191906/public_html/log.php on line 26

kods : $query = mysql_query("SELECT * FROM Log ORDER BY ID DESC");

kur ir probleema?

Ja neesi jau atkodis problēmu, tad problēma ir tāpēc, ka nav ķepas vārdiņam Log.

$query = mysql_query("SELECT * FROM `Log` ORDER BY ID DESC");

Edited July 24, 2011 by mex

[KpC]

Ko tev saka šis?

$query = mysql_query("SELECT * FROM Log ORDER BY id DESC")or die(mysql_error());

[Mazinjais]

Kļūda ir par mysql_fetch_array, kuru tavā kodā neredzu.. Lai nu kā, apsties vai $query nav tukš (isset vai echo).

Edited July 27, 2011 by Mazinjais

[evilxdeadly]

Ja neesi jau atkodis problēmu, tad problēma ir tāpēc, ka nav ķepas vārdiņam Log.

$query = mysql_query("SELECT * FROM `Log` ORDER BY ID DESC");

Ko tev saka šis?

$query = mysql_query("SELECT * FROM Log ORDER BY id DESC")or die(mysql_error());

Kļūda ir par mysql_fetch_array, kuru tavā kodā neredzu.. Lai nu kā, apsties vai $query nav tukš (isset vai echo).

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