Cracky Posted August 8, 2008 Report Share Posted August 8, 2008 (edited) Sveiki man ir probleema ar webshopu meginaaju izlabot jebkaadigi nu nemaaku!! Kr4 pie lietas. Es aizsutu izzinu atnak kods ievadu to ailee un uzpiezu Apstiprinaatb et man uzmet saadu erroru!! Warning:mysql_result(): Supplied argument is not a valid MySQL result resource in C:\AppServ\www\sms\smscw.phpon line 82 es atveru tur man taa 82 line iskataas saadi! if(mysql_result(mysql_query("SELECT COUNT(*) FROM code WHERE code = '$code'"),0,'COUNT(*)') > 0){ Ludzu kads vareetu paliidzeet! Varetu pateikt ludzu ko darit vajag??? Edited August 8, 2008 by Cracky Link to comment Share on other sites More sharing options...
-=ar4ix8=- Posted August 8, 2008 Report Share Posted August 8, 2008 mysql tabulu esi izveidojis ??? ja nee tad ej taisi savadaak nekas nesanaks Link to comment Share on other sites More sharing options...
Cracky Posted August 8, 2008 Author Report Share Posted August 8, 2008 tu domaaa tipa ieks phpmyadmin? Link to comment Share on other sites More sharing options...
Norek Posted August 8, 2008 Report Share Posted August 8, 2008 Pilnīgs āmurs. Kāda šķirba, caur kurieni taisa ? Jautājums ir, vai Tev ir datubāzes tabula? Ej takš izglītojies pirms ķeries pie tādiem darbiem. Link to comment Share on other sites More sharing options...
Zoliidais- Posted August 8, 2008 Report Share Posted August 8, 2008 Un ja kas NEPAREIZĀ SADALJA AAMUR :@ Link to comment Share on other sites More sharing options...
Cracky Posted August 8, 2008 Author Report Share Posted August 8, 2008 ja nezinu kas par vainu neraksti gaili! MySQL tabulu esmu uztaisijis!! viss ir! bet met araa erroru!! Link to comment Share on other sites More sharing options...
brazer Posted August 8, 2008 Report Share Posted August 8, 2008 Varbūt kautkas nav ar datubāzēm varbūt var nekonektēt,pasties vai kad tu atver to kods.php apsties vai viņs ierakstās Datubāzē! Link to comment Share on other sites More sharing options...
Norek Posted August 8, 2008 Report Share Posted August 8, 2008 Pateikšu priekšā - nav pareizs selekts. Īsti nezinu, vai mysql_result drīkst pielietot tādu count(*) Vispār nesaprotu, kapēc tāds ifs nepieciešams. Kāds ir centies kaut ko nevajadzīgi sarežģīt. Link to comment Share on other sites More sharing options...
martins256 Posted August 9, 2008 Report Share Posted August 9, 2008 var var tā COUNT(*) tur likt iekšā! Link to comment Share on other sites More sharing options...
RaaapuLis Posted August 9, 2008 Report Share Posted August 9, 2008 nu bet tie abi divi COUNT(*) ir pareizās vietās? Link to comment Share on other sites More sharing options...
wap1337 Posted August 9, 2008 Report Share Posted August 9, 2008 Nu ja jau iet, tad kāpēc lai viņi būtu nepareizās vietās? Link to comment Share on other sites More sharing options...
Norek Posted August 9, 2008 Report Share Posted August 9, 2008 kur Tu redzi, ka iet ? ša pamēģināšu tamlīdzīgi kaut ko izselektēt. EDIT : Nav norādīti pareizi mysql pieejas dati vai/un datubāze. Pārbaudi. Pārējais viss ir ok. Link to comment Share on other sites More sharing options...
wap1337 Posted August 9, 2008 Report Share Posted August 9, 2008 EDIT : Nav norādīti pareizi mysql pieejas dati vai/un datubāze. Pārbaudi. Pārējais viss ir ok. Tieši ta , viņš (MySql) Nevar izpildīt kveriju , jo nav norādīti , visi nepieciešamie dati! Link to comment Share on other sites More sharing options...
martins256 Posted August 9, 2008 Report Share Posted August 9, 2008 waplets māk citus atkārtot Link to comment Share on other sites More sharing options...
Norek Posted August 9, 2008 Report Share Posted August 9, 2008 Njā, esmu pamanījis. Diez gan bieži un sen jau viņš to dara. Link to comment Share on other sites More sharing options...
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