这个问题已经在这里有了答案 :
mysqli_fetch_assoc()期望参数/调用成员函数bind_param()错误。如何获取并修复实际的mysql错误? (1个答案)
7年前关闭。
我在检查数据库中是否已存在Facebook User_id时遇到了一些麻烦(如果不存在,则应将用户作为新用户接受,否则仅加载canvas应用程序)。我在托管服务器上运行了它,没有问题,但是在我的本地主机上,它给了我以下错误:
mysqli_fetch_array()期望参数1为mysqli_result,布尔值在
这是我的代码:
<? $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 $checken2; // If the user does not exist @ WT-Database -> insert if (!($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, $add); } // Double-check, the user won't be able to load the app on failure inserting to the database if (!($checken2)) { echo "Excuse us " . $first_name . ". Something went terribly wrong! Please try again later!"; exit; } } else { include ('sorrylocale.html'); exit; }
我读过它与我的查询错误有关,但是它已在我的托管服务提供商上起作用,所以不可能!
该查询失败并返回false。
false
将其放在后面mysqli_query()以查看发生了什么。
mysqli_query()
if (!$check1_res) { printf("Error: %s\n", mysqli_error($con)); exit(); }
欲获得更多信息:
http://www.php.net/manual/zh/mysqli.error.php