Portál AbcLinuxu, 10. května 2025 04:34
$result = mysql_query("SELECT * FROM table"); $num_rows = mysql_num_rows($result); echo "$num_rows Rows\n";Tak podla toho navodu to robim aj ja a funguje to. Neni vsak lepsie pouzit COUNT?
$result = mysql_query("SELECT COUNT(*) FROM table1"); $num_rows = mysql_num_rows($result);ide o to ze potom uz v $num_rows nemam pocet riadkov a ani sa k nemu nemozem nijak dopracovat
Řešení dotazu:
dibi::query('SELECT * FROM `tabulka`')->rowCount();
COUNT(*)
.
<?php $result = mysql_query("SELECT COUNT(*) AS pocet_radku FROM `tabulka`"); $row = mysql_fetch_object($result); echo 'Výsledek dotazu (počet řádků): ' . $row->pocet_radku;
$result2 = MySQL_Query("SELECT DateDiff('2009-01-07', '2009-01-01')"); $datediff_a = MySQL_Fetch_Array($result2); $datediff = $datediff_a[0]; echo "rozdiel datumov je $datediff dni";druhy uz vyhadzuje errory:
$q = mysql_query("SELECT COUNT(*) FROM table;"); $r = mysql_fetch_row($q); echo "Riadkov je ".$r[0];to iste aj ten co mi predtym fungoval:
$result = mysql_query("SELECT * FROM table"); $num_rows = mysql_num_rows($result); echo "$num_rows Rows\n";You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'table' at line 1 resp. bez toho or Die(MySQL_Error()): Warning: mysql_fetch_row(): supplied argument is not a valid MySQL result resource in C:\Program Files\VertrigoServ\www\cotus\test_MYSQL\test.php on line 26
if($q = mysql_query("SELECT COUNT(*) AS cnt FROM `table`")) { if(($r = mysql_fetch_row($q)) !== false) //nebo jiný mysql_fetch_xxxx { echo "Riadkov je ".$r[0]; }else echo "confused error"; }else echo "SQL query error";
Tiskni
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