2014-02-27 29 views
-2

我只在某些浏览器上收到此错误,而我不知道为什么。我希望这是一个简单的修复。听起来应该是这样。这是错误,下面是代码。仅在某些浏览器上使用php mysqli错误

警告:mysqli_fetch_array预计参数1至mysqli_result,在给定的/home/content/yada/html/myapp/main.php上线71

顺便说布尔,这是行71:

while($row = mysqli_fetch_array($result)) 

和下面是完整的代码

$type = $_POST[type]; 
$user="theUser"; 
$password="thePassword"; 
$database="theDatabase"; 
$TABLE = "user"; 

@mysql_connect("mydb.com",$user,$password); 
@mysql_select_db($database) or die("Unable to select database"); 

if($_POST[type]) { 
    $query = "UPDATE $TABLE 
       SET type = $type 
       WHERE fbId = $id"; 

    if(mysql_query($query)) { 
     //echo "Settings saved successfully!"; 
    } else { 
     echo ("MySQL Error: ".mysql_error()); 
    } 
} 

$con=mysqli_connect('localhost',"$user","$password","$database"); 

// Check connection 
if(mysqli_connect_errno($con)) { 
    echo "Failed to connect to MySQL: " . mysqli_connect_error(); 
} 

$result = mysqli_query($con,"SELECT * FROM $TABLE WHERE fbID = $id"); 

while($row = mysqli_fetch_array($result)) { 
    $currentType = $row['type']; 
    //echo $currentType; 
} 

if ($result = mysqli_query($con, "SELECT * FROM $TABLE WHERE fbID = $id", MYSQLI_USE_RESULT)) { 
    //echo "True"; 
    //mysqli_free_result($result); 
} 
+1

这不能是第71行。第71行必须有'mysqli_fetch_array()'。 – Barmar

+0

不应该'$ _POST [type]'是'$ _POST ['type']'? – DarthCaniac

+0

让我仔细检查..谢谢 –

回答

-1

使用 回声mysqli_error($ CON); 以显示在第71行上执行SQL查询时MySQL服务器给出的错误。这将揭示查询出了什么问题。

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