数据库查询返回：检查参数是数组还是对象

Question

当尝试使用php和mysql登录此应用程序时，将Authorization failed显示该应用程序中使用的错误。阅读php_error.log时显示：

[05.05.2019 03:31:51 UTC] PHP注意：未定义的索引：结果在第43行的/Applications/MAMP/htdocs/iReporter/api.php中[05-May-2019 03:31:51 UTC] PHP警告：count（）：参数必须是在第43行的/Applications/MAMP/htdocs/iReporter/api.php中实现Countable的数组或对象

登录功能

//login API

function login($user, $pass) {

// try to match a row in the "login" table for the given username and password

$result = query("SELECT IdUser, username FROM login WHERE username='%s' AND pass='%s' limit 1", $user, $pass);

//Line 43

if (count($result['result'])>0) {

    // a row was found in the database for username/pass combination

    // save a simple flag in the user session, so the server remembers that the user is authorized

    $_SESSION['IdUser'] = $result['result'][0]['IdUser'];

    // print out the JSON of the user data to the iPhone app; it looks like this:

    // {IdUser:1, username: "Name"}

    print json_encode($result);

//edit

var_dump($result);

} else {

    // no matching username/password was found in the login table

    errorJson('Authorization failed');

}

}

该应用程序用于让用户无缝登录，但突然停止工作，而无需对代码进行任何更改。即使未更改的备份文件也收到相同的错误。如何更改代码以允许用户成功登录？

更新：index.php。应用程序访问功能的主文件

switch ($_POST['command']) {

case "login":

    login($_POST['username'], $_POST['password']); 

    break;

解决了

更改了iOS应用程序，以将密码而不是固定密码发布到数据库，因为它正在创建错误。

Answer 1

基于该错误，您没有result在$result变量中获取该值。要检查这一点，您可以使用：if (isset($result['result']) && count($result['result']) > 0)首先检查该值是否已设置。您还需要调查为什么无法从数据库中获得预期的结果。为此，您将需要查看$result变量中返回的内容，并查找数据库查询可能返回的任何错误。

Answer 2

function login($user, $pass) {// try to match a row in the "login" table for the given username and password$result = query("SELECT IdUser, username FROM login WHERE username='%s' AND pass='%s' limit 1", $user, $pass);$rows=$result->num_rows;//Line 43if ($rows>0) {    // a row was found in the database for username/pass combination    // save a simple flag in the user session, so the server remembers that the user is authorized    $_SESSION['IdUser'] = $result['result'][0]['IdUser'];    // print out the JSON of the user data to the iPhone app; it looks like this:    // {IdUser:1, username: "Name"}    print json_encode($result);} else {    // no matching username/password was found in the login table    errorJson('Authorization failed');}}