Question

我只在某些浏览器上收到此错误，但我不确定原因。我希望这是一个简单的解决方案。听起来应该是这样。这是错误，下面是代码。

警告：mysqli_fetch_array需要参数1到mysqli_result，第71行/home/content/yada/html/myapp/main.php中给出的布尔值

顺便说一句，这是第71行：

while($row = mysqli_fetch_array($result))

及以下是完整代码

$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);
}

Answer 1

使用 echo mysqli_error（$ con）; 显示MySQL服务器在第71行执行SQL查询时给出的错误。这将揭示查询的错误。

php mysqli只在某些浏览器上出错

1 个答案: