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Mysqli query fatal error

Fatal error: Call to a member function prepare( ) on a non- object in C: \ xampp\ htdocs\ syu\ script\ regist. php on line 94 94行目に. ませんか? ・ query( ) では動きませんか?. There is your problem: if ( $ result = $ mysqli- > $ sql( $ sql) ) {. You' re basically calling a method with the same name as your SQL statement, i. SELECT tblConference [. Such a method doesn' t exist, obviously. Rather, looks like you got your variables mixed up. Try the following ( does not prevent injection) : $ link = mysqli_ connect( " localhost", " Username", " Password", " Database" ) ; if ( $ result = $ link- > query( " SELECT * FROM your code you' re mixing both procedural and object- oriented code. Choose either one or the other. Here' s how you would solve the problem the procedural way. $ result = mysqli_ query( $ link, $ sql, MYSQLI_ USE_ RESULT).

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  • Video:Error mysqli fatal

    Query error mysqli

    The culprit is most likely this line: return new mysqli( $ host, $ user, $ pwd, $ db) or die ( ' Cannot open database' ) ;. The do xyz or die( ) construct leads to funny behaviour in conjuction with the return statement ( i. mysqli: : query - - mysqli_ query — データベース上でクエリを実行する. サーバーの max_ allowed_ packet よりも長い ステートメントを mysqli_ query( ) に渡した場合、 返ってくるエラーコードは MySQL Native Driver. " Fatal error: Exception thrown without a stack frame in Unknown on line 0". Your $ mysqli variable is not a member of your class, so it is null inside the scope of your DB: : insert( ) function.