試圖獲取MAMP中非對象的屬性

Question

我對此代碼有一個疑問。當我想選擇一個項目（在這種情況下是街道）時，在調試模式下一切都沒有錯誤。 （我使用AJAX來生成消息）

public function placeAvailable() 
    { 
     //open connectie naar database 
     include("Connection.php"); 

     $sSql= "select Street from tblPlaces 
       where Street = '".$this->m_sStreet."'"; 

     $vResult=$mysqli->query($sSql); 
     if($vResult->num_rows>0) 
     { 
      return(false); 
     } 
     else 
     { 
      return(true); 
     } 
     $mysqli->close(); 
    } 

當我想要一個額外的參數添加到我的選擇查詢我得到一個錯誤（注：試圖獲得非對象的屬性/應用程序/ MAMP/htdocs中/ Foodsquare小/資產/班/上線86 places.class.php）

$vResult=$mysqli->query($sSql); 
     if($vResult->num_rows>0) 

額外的參數我想補充的是如下因素。

$sSql= "select Street from tblPlaces 
       where Street = '".$this->m_sStreet."' AND where HouseNumber = '".$this->m_sHouseNumber."'"; 

喬希在這裏你可以看到整個代碼

<?php 

class Places { 

    private $m_sName; 
    private $m_sStreet; 
    private $m_sHouseNumber; 
    private $m_sCity; 
    private $m_sCategory; 

    public function __set($p_sProperty, $p_vValue) { 
     switch($p_sProperty) { 
      case "Name" : 
       $this -> m_sName = $p_vValue; 
       break; 
      case "Street" : 
       $this -> m_sStreet = $p_vValue; 
       break; 
      case "HouseNumber" : 
       $this -> m_sHouseNumber= $p_vValue; 
       break; 
      case "City" : 
       $this -> m_sCity = $p_vValue; 
       break; 
      case "Category" : 
       $this -> m_sCategory = $p_vValue; 
       break; 
     } 
    } 

    public function __get($p_sProperty) { 
     $vResult = null; 
     switch($p_sProperty) { 
      case "Name" : 
       $vResult = $this -> m_sName; 
       break; 
      case "Street" : 
       $vResult = $this -> m_sStreet; 
       break; 
      case "HouseNumber" : 
       $vResult = $this -> m_sHouseNumber; 
       break; 
      case "City" : 
       $vResult = $this -> m_sCity; 
       break; 
      case "Category" : 
       $vResult = $this -> m_sCategory; 
       break; 
     } 
     return $vResult; 
    } 

    public function addPlaces() 
    { 
     include_once("connection.php"); 
      $sSql = "INSERT INTO tblPlaces 
       (Name, 
       Street, 
       HouseNumber, 
       City, 
       Category) 
       VALUES 
       ('" . $mysqli -> real_escape_string($this -> m_sName) . "', 
       '" . $mysqli -> real_escape_string($this -> m_sStreet) . "', 
       '" . $mysqli -> real_escape_string($this -> m_sHouseNumber) . "', 
       '" . $mysqli -> real_escape_string($this -> m_sCity) . "', 
       '" . $mysqli -> real_escape_string($this -> m_sCategory) . "');"; 

     if (!$mysqli -> query($sSql)) 
     { 
      throw new Exception("Er is iets mis gelopen bij het toevoegen van een plaats"); 
     } 

    } 

    public function placeAvailable() 
    { 
     //open connectie naar database 
     include("Connection.php"); 
     global $mysqli; 


     //sql query selecteert straat uit tblplaces waar de straat gelijk is aan het inputveld street uit het formulier 
     $sSql= "select Street from tblPlaces 
       where Street = '".$this->m_sStreet."' AND where HouseNumber = '".$this->m_sHouseNumber."'"; 

     $vResult=$mysqli->query($sSql); 
     if($vResult->num_rows>0) 
     { 
      //Street (place) is al in gebruik 
      return(false); 
     } 
     else 
     { 
      //Hij is nog niet in gebruik 
      return(true); 
     } 
     //database sluiten 
     $mysqli->close(); 
    } 
} 
?> 

Answer 1

你從來沒有定義$mysqli，你需要創建連接。

如果它在你Connection.php，你需要添加一個：

global $mysqli;

後你有，但你的查詢之前。

該錯誤表示$mysqli不是一個對象，並且沒有這樣的屬性，屬性爲query($data)。