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給我有一些麻煩檢查如果FB USER_ID在我的數據庫已經存在(如果它不應該再接受用戶作爲一個新的,其他人只是加載畫布應用程序)。我跑了我的託管服務器上,並沒有什麼問題,但在我的本地它給了我下面的錯誤:mysqli_fetch_array()預計參數1被mysqli_result,布爾在
mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given in
這裏是我的代碼:
<?
$fb_id = $user_profile['id'];
$locale = $user_profile['locale'];
if ($locale == "nl_NL") {
//Checking User Data @ WT-Database
$check1_task = "SELECT * FROM `users` WHERE `fb_id` = " . $fb_id . " LIMIT 0, 30 ";
$check1_res = mysqli_query($con, $check1_task);
$checken2 = mysqli_fetch_array($check1_res);
print $checken2;
//If User does not exist @ WT-Database -> insert
if (!($checken2)) {
$add = "INSERT INTO users (fb_id, full_name, first_name, last_name, email) VALUES ('$fb_id', '$full_name', '$first_name', '$last_name', '$email')";
mysqli_query($con, $add);
}
//Double-check, User won't be able to load app on failure inserting to database
if (!($checken2)) {
echo "Excuse us " . $first_name . ". Something went terribly wrong! Please try again later!";
exit;
}
} else {
include ('sorrylocale.html');
exit;
}
我讀過它有話做我的查詢是錯誤的,但它已經在我的託管服務提供商合作,這樣不能吧!
您需要添加錯誤處理您的查詢,這樣你就可以找出到底爲什麼它是失敗的。您還有一個重大的SQL注入漏洞。 – 2013-03-15 18:45:11
請,** **請使用'mysqli'準備好的語句功能[正確逃避你的SQL查詢(http://bobby-tables.com/php)。 – tadman 2013-03-15 18:59:06