php - Uncaught exception 'mysqli_sql_exception' with message 'No data supplied for parameters in prepared statement' -


i trying write query using prepared statement.

mysqli_report(mysqli_report_all);      $query = $conn->prepare("insert notification(messageid,topicarn,subject,message,timestamp,signatureversion,signature,signingcerturl,unsubscribeurl,topicid)                  values (?, ?, ?, ?, ?,?, ?, ?, ?, ?)");

this throws following exception

fatal error: uncaught exception 'mysqli_sql_exception' message 'no data supplied parameters in prepared statement'

where following query works

$query = $conn->prepare("insert notification(messageid,topicarn,subject,message,timestamp,signatureversion,signature,signingcerturl,unsubscribeurl,topicid)          values (".$message['messageid'].", '".$message['topicarn']."', '".$message['subject']."', '".$message['message']."', '".$message['timestamp']."',".$message['signatureversion'].", '".$message['signature']."', '".$message['signingcerturl']."', '".$message['unsubscribeurl']."', ".$topic.")");

i want use prepared statement bind_param() function. wrong first query? please help.

as suggested - need bind parameters values before can execute sql ~ perhaps:

mysqli_report(mysqli_report_all); $sql='insert notification     (messageid,topicarn,subject,message,timestamp,signatureversion,signature,signingcerturl,unsubscribeurl,topicid)      values     (?, ?, ?, ?, ?,?, ?, ?, ?, ?)';  $stmt = $conn->prepare( sql ); if( $stmt ){     /* assumed strings other `id` in column name */     $stmt->bind_param('issssssssi',         $message['messageid'],         $message['topicarn'],         $message['subject'],         $message['message'],         $message['timestamp'],         $message['signatureversion'],         $message['signature'],         $message['signingcerturl'],         $message['unsubscribeurl'],         $topic     );     $result=$stmt->execute();      /* other code */  } else {     /* investigate why "prepare" method failed */     echo "error:"; }

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