處理MySql之後的MySQLi查詢結果到MySqli轉換

Question

我試圖更改以下代碼以使用MySqli而不是MySql。我刪除了一些對我在這裏要解決的問題並不重要的方法。

class db { 
    var $hostname, 
     $database, 
     $username, 
     $password, 
     $connection, 
     $last_query, 
     $last_i, 
     $last_resource, 
     $last_error; 

    function db($hostname=DB_HOSTNAME,$database=DB_DATABASE,$username=DB_USERNAME,$password=DB_PASSWORD) { 
     $this->hostname = $hostname; 
     $this->database = $database; 
     $this->username = $username; 
     $this->password = $password; 

     $this->connection = mysql_connect($this->hostname,$this->username,$this->password) or $this->choke("Can't connect to database"); 
     if($this->database) $this->database($this->database); 
     } 

    function database($database) { 
     $this->database = $database; 
     mysql_select_db($this->database,$this->connection); 
     } 

    function query($query,$flag = DB_DEFAULT_FLAG) { 
     $this->last_query = $query; 

     $resource = mysql_query($query,$this->connection) or $this->choke(); 
     list($command,$other) = preg_split("|\s+|", $query, 2); 

     // Load return according to query type... 
     switch(strtolower($command)) { 
      case("select"): 
      case("describe"): 
      case("desc"): 
      case("show"): 
       $return = array(); 
       while($data = $this->resource_get($resource,$flag)) $return[] = $data; 
       //print_r($return); 
       break; 
      case("replace"): 
      case("insert"): 
       if($return = mysql_insert_id($this->connection)) 
        $this->last_i = $return; 
       break; 
      default: 
       $return = mysql_affected_rows($this->connection); 
      } 

     return $return; 
     } 

    function resource_get($resource = NULL,$flag = DB_DEFAULT_FLAG) { 
     if(!$resource) $resource = $this->last_resource; 
     return mysql_fetch_array($resource,$flag); 
     } 
    } 

這是我到目前爲止有：

class db { 
    var $hostname = DB_HOSTNAME, 
     $database = DB_DATABASE, 
     $username = DB_USERNAME, 
     $password = DB_PASSWORD, 
     $connection, 
     $last_query, 
     $last_i, 
     $last_resource, 
     $last_error; 

    function db($hostname, $database, $username, $password) { 
     $this->hostname = $hostname; 
     $this->database = $database; 
     $this->username = $username; 
     $this->password = $password; 

     $this->connection = new mysqli($this->hostname, $this->username, $this->password, $this->database) or $this->choke("Can't connect to database"); 
if (mysqli_connect_errno()) 
    { 
    echo "Failed to connect to MySQL: " . mysqli_connect_error(); 
    } 


     if($this->database) 
      $this->database($this->database); 
    } 

    function database($database) { 
     $this->database = $database; 
     mysqli_select_db($this->connection, $this->database); 
    } 

    function query($query, $flag = DB_DEFAULT_FLAG) { 
     $this->last_query = $query; 
//print_r($query); 
     $result = mysqli_query($this->connection, $query) or $this->choke("problem connecting to DB"); 
while($row=mysqli_fetch_assoc($result)) { 
$resource[]=$row; 
} 
//print($command); 
//print_r($resource);print("<br>"); 
     list($command, $other) = preg_split("|\s+|", $query, 2); 

     // Load return according to query type... 
     switch(strtolower($command)) { 
      case("select"): 
      case("describe"): 
      case("desc"): 
      case("show"): 
       $return = array(); 
       while($data = $this->resource_get($resource, $flag)) 
        $return[] = $data; 
       //print_r($return); 
       break; 
      case("replace"): 
      case("insert"): 
       if($return = mysqli_insert_id($this->connection)) 
        $this->last_i = $return; 
       break; 
      default: 
       $return = mysqli_affected_rows($this->connection); 
      } 

     return $return; 
     } 

    function resource_get($resource = NULL, $flag = DB_DEFAULT_FLAG) { 
     if(!$resource) 
      $resource = $this->last_resource; 
     return mysqli_fetch_array($resource, $flag); 
    } 

所以這裏的問題：我已經簽了的print_r（結果）和$資源數組正確加載，但使用print_r（）檢查時，$ return的值最終會變成「Array（）」。因此，儘可能接近我自己看着辦，自己是不是被正確的這部分代碼，這就是爲什麼我包括resource_get（）函數調用處理：

  $return = array(); 
      while($data = $this->resource_get($resource, $flag)) 
       $return[] = $data; 
      //print_r($return); 
      break; 

如果我使用mysqli_fetch_row（$資源， $ flag）而不是mysqli_fetch_array（$ resource，$ flag）我仍然得到相同的結果，即print_r（$ return）只產生「Array（）」。

Answer 1

變量$resource在您將其傳遞到$this->resource_get()時不代表mysqli_result資源對象。相反，它已經是二維數組，因爲您之前運行過一個mysqli_fetch_assoc()循環。

爲了與當前的代碼這項工作，您既可以刪除早期取環：

$result = mysqli_query($this->connection, $query) or $this->choke("problem connecting to DB"); 
// Remove this 
//while($row=mysqli_fetch_assoc($result)) { 
// $resource[]=$row; 
//} 

後來，經過$result而不是$resource到您的resource_get()方法，因爲它是$result是資源對象。

或者，您可能完全跳過resource_get()調用，並直接返回$resource，因爲它已包含結果數組。