Daten in Server einfügen

Question

Ich habe Probleme beim Einfügen von Daten in meine Datenbank. Die Daten können nicht eingefügt werden und es werden keine Fehler in meinem Logcat angezeigt. Ich arbeite daran, kann aber die Probleme nicht lösen. Das ist mein PHP-Code:

<?php 

    require ("config1.php"); 

    if(!empty($_POST)){ 

     $query = "SELECT * FROM announcement WHERE announceID = :announcementID"; 
     $query_params=array(':announcementID'=> $_POST['announcementID']); 

     try{ 
      $stmt=$db->prepare($query); 
      $stmt->execute($query_params); 


     }catch(PDOException $ex){ 
      $response["success"]=0; 
      $response["message"]="Database Error1. Please try again"; 
      die(json_encode($response)); 
     } 


     $row = $stmt->fetch(); 
     if($row){ 

      $query = "INSERT INTO announcement (title,description,start_date,end_date,time) 
            VALUES (:title,:description,:starDate,:endDate,:time) "; 

        $query_params= array(
         ':title'=>$_POST['title']; 
         ':description'=>$_POST['description']; 
         ':startDate' => $_POST['start_date']; 
         ':endDate' => $_POST['end_date']; 
         ':time' => $_POST['time']; 

        ); 


        try { 
        $stmt = $db->prepare($query); 
        $result = $stmt->execute($query_params); 
       } 

       catch (PDOException $ex) { 
        $response["success"] = 0; 
        $response["message"] = "Database Error2. Please Try Again!"; 
        die(json_encode($response)); 
       } 

       $response["success"] = 1; 
       $response["message"] = "Update successful!"; 
       echo json_encode($response) 

      } 
     } 
    ?> 

Unten ist mein Java-Code:

protected String doInBackground(Void... params) { 
       RequestHandler rh=new RequestHandler(); 
       HashMap<String,String> param= new HashMap<String, String>(); 
       param.put(KEY_TITLE,announcement_title); 
       param.put(KEY_DESCRIPTION,announcement_desc); 
       param.put(KEY_START_DATE,start_date); 
       param.put(KEY_END_DATE,end_date); 
       param.put(KEY_TIME,time); 
       param.put(KEY_IMAGE,announcement_image); 
       String result= rh.sendPostRequest(ANNOUNCEMENT_URL,param); 
       return result; 
      } 

Schätzen jemand die Probleme hinweisen kann.

Answer 1

Sie PHP-Code enthält Syntaxfehler und Mysql preparement Fehler Platzhalter.

Ich habe den Code umgeschrieben, bitte ersetzen Sie ihn mit Ihrem alten Code.

<?php 

    require ("config1.php"); 

    if(!empty($_POST)){ 
//kindly filter the POST value 
     $query = "SELECT * FROM announcement WHERE announcementID = :announcementID"; 
     $query_params=array(':announcementID'=> $_POST['announcementID']); 
      $stmt=$db->prepare($query); 
      $stmt->execute($query_params); 

/* 
     }catch(PDOException $e){ 
      $response["success"]=0; 
      $response["message"]="Database Error1. Please try again"; 
      die(json_encode($response)); 
     } 
*/ 

     $row = $stmt->fetch(); 
     if($row){ 

      $query = "INSERT INTO announcement (title,description,start_date,end_date,time) 
            VALUES (:title,:description,:starDate,:endDate,:time)"; 
//re written by Ajmal PraveeN 
        $query_params= array(
         ':title'=>$_POST['title'], 
         ':description'=>$_POST['description'], 
         ':startDate' => $_POST['start_date'], 
         ':endDate' => $_POST['end_date'], 
         ':time' => $_POST['time'] 
         ); 

/* 
        try { 
*/ 
        $stmt = $db->prepare($query); 
        $result = $stmt->execute($query_params); 
/* 
       } 

       catch (PDOException $ex) { 
        $response["success"] = 0; 
        $response["message"] = "Database Error2. Please Try Again!"; 
        die(json_encode($response)); 
       } 
*/ 
       $response["success"] = 1; 
       $response["message"] = "Update successful!"; 
       echo json_encode($response); 

      } 
     } 
?> 

Wenn Ihre $_POST['announcementID'] eine numerische Nummer i zu bearbeiten die Post mit einem Sanitize und Titel Beitrag Re kann.