msaada wa code za php...
Kumuwezesha mtu kusubmit form mara moja tu kwa siku na ni siku za kazi tu!!! asizidishe zaidi ya mara moja per working day
Are you giving us assignment to do? Where are your efforts?
ah broh ts nat an assigment since u can do it if u can... about putn my effort here its kind hard to me since am using mchina cant code kwa simu bt i can say where i have reached!!!
Are you giving us assignment to do? Where are your efforts?
msaada wa code za php...
Kumuwezesha mtu kusubmit form mara moja tu kwa siku na ni siku za kazi tu!!! asizidishe zaidi ya mara moja per working day
$column_name = "User_Name" ;
$table_name = "Jamii";
$database_name = "Forums";
$con = mysql_connect("localhost", "root", "", true) or die(mysql_error());
mysql_select_db ("$database_name",$con);
$user_found = mysql_query("SELECT $column_name FROM $table_name WHERE $column_name = '$user_name'");
if ( mysql_num_rows ( $user_found ) ) { echo 'You have already logged in today'; exit; }msaada wa code za php...
Kumuwezesha mtu kusubmit form mara moja tu kwa siku na ni siku za kazi tu!!! asizidishe zaidi ya mara moja per working day
Tumia If...else statement.
Are you serius?
usikute hawa ndio wanaojiita mtaani wana IT kumbe wadesaji tu, yaani mtu anatupia swali tu pasipo kuonesha wapi alipofika akakwama.
Swali refu sana, siku za kazi ni simple enough unaweza ukacheki time/date kwenye server, ila mara moja kwa siku itabidi urekodi somewhere kwenye server kama mtu ameshatuma so itabidi utumie database au simple file storage kurekodi kila aliyetuma kisha kucheki kabla ya kuruhusu kutuma tena. Plus kum authenticate user in the first place.
if(todayIsWeekday and hasNotSubmittedToday){
//Ruhusu kusubmit form
}
kama Kang alivyosema hapo juu, kwa kutumia database unaweza call code hii ambayo itaangalia kama kuna rekodi ya 'username'. Unaweza delete hizi rekodi saa sita usiku (kila siku) kwa kutumia cron job command.
PHP:$column_name = "User_Name" ; $table_name = "Jamii"; $database_name = "Forums"; $con = mysql_connect("localhost", "root", "", true) or die(mysql_error()); mysql_select_db ("$database_name",$con); $user_found = mysql_query("SELECT $column_name FROM $table_name WHERE $column_name = '$user_name'"); if ( mysql_num_rows ( $user_found ) ) { echo 'You have already logged in today'; exit; }
hizo function ndo sizijui sasa... hapo todayizweekday ndo unaweka function gani maana kuongea tu nivirahisi!!!
hizo function ndo sizijui sasa... hapo todayizweekday ndo unaweka function gani maana kuongea tu nivirahisi!!!
Hiyo code haijakamilika lakini inakupa idea tuu ya ku query msql kuona kama kuna rekodi ya user_name.kaka hizi code za bila kuweka ata date sa itajuaje kama mara ya mwisho kuingia ilikua ni jana!?? hapo sio itamruhusu kusubmit form once and for all...
if ( mysql_num_rows ( $user_found ) ) {
echo 'You have already logged in today';
exit;
}Unamfundisha checkechea Quantum Physics? Strange!Hiyo code haijakamilika lakini inakupa idea tuu ya ku query msql kuona kama kuna rekodi ya user_name.
Ni hivi, wakati mtu anajaza form unaweza kuchukua moja ya data zake kama 'user_name' na kuiweka katika orodha ya kuhakiki ambayo itakuwa inafutwa kila baada ya masaa 24 au inawekewa time stamp. Mtumiaji huyo huyo kama akirudia kutuma tena katika kipindi cha masaa 24 ile data iliyonakiliwa bado itakuwepo kwenye orodha ya wale waliotuma form kwa siku ile. Kwahiyo mysql_query() ita-return TRUE na kuzuia kutuma form nyingine.
kuna njia nyingi za kufanya hivi unaweza kutumia date(), cookies , $_SESSION au hata microtime(), kuangalia tofauti ya muda siyo lazima utumie cron job. Kumbuka google ni rafiki yako.PHP:if ( mysql_num_rows ( $user_found ) ) { echo 'You have already logged in today'; exit; }