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kotaku
11-04-03, 04:39 AM
while ((day==$sadd) && (monthyear==$sammyy)) {
$ida = mysql_query("SELECT * FROM $aboutapar",$db);
while ($arr = mysql_fetch_array($ida)) {
$arr[id] = $dayfirst;
}
}

---------

ataupun,cam ner nak cari ID yg mane ia ditetapkan oleh $sadd dan $sammyy?

bustamam
11-04-03, 06:44 AM
Nama table: aboutapar
Field dalam table: day, monthyear, id

Contoh carik id dimana field day=$sadd dan field monthyear=$sammyy:

$ida = mysql_query("SELECT id FROM aboutapar where day='" . $sadd . "' and monthyear='" . $sammyy . "'",$db);
while ($arr = mysql_fetch_array($ida)) {
// komen: $arr[0] mengandungi id
}


Ditukar sebab ralat dapat ditangkap oleh mater tajam mnajem - B.H.

mnajem
11-04-03, 09:19 AM
aboutpar ke aboutapar :p

Nama table: aboutpar

SELECT id FROM aboutapar where day

:D

bustamam
11-04-03, 10:10 AM
Posting asal oleh mnajem
aboutpar ke aboutapar :p

Nama table: aboutpar

SELECT id FROM aboutapar where day

:D Tajam mater ko mnajem :)

Diskuzar
11-04-03, 10:16 AM
tumpang tanye .... id & $dayfirst tu declare kat mane ?? cam tak complete aje ...

$arr[id] = $dayfirst;

- reply from someone with "zero" mySQL knowledge -

bustamam
11-04-03, 10:32 AM
Posting asal oleh Diskuzar
tumpang tanye .... id & $dayfirst tu declare kat mane ?? cam tak complete aje ...

$arr[id] = $dayfirst;

- reply from someone with "zero" mySQL knowledge - Betui ko punye observation disk, setiap variable kene assign dulu sebelum buat checking kat "if".... Pastulak variable tu kene ader "$" kat permulaan. kalu kata if (day ==$sadd) tuh tak betui, sebab kalu "day" tu string kene kata if ("day" == $saddr), kalu "day" tu variable, kene kata if ($day == $saddr). Tapi $day kene ler ader nilai dulu. Dalam SQL SELECT statement lak, select field tu tak payah di initialize atau mulakan dengan "$", contoh SELECT id from aboutapar where ... resultnye plak adalah array, dan indeks array tu kene konstant atau variable. Kalau konstant buat cam ni: $arr[0], kalu variable $arr[$id], (jangan luper assign value kat $id tuh).