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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);
}
这不能是第71行。第71行必须有'mysqli_fetch_array()'。 – Barmar
不应该'$ _POST [type]'是'$ _POST ['type']'? – DarthCaniac
让我仔细检查..谢谢 –