Php 数据库查询返回：检查参数是数组还是对象
当尝试使用php和mysql登录此应用程序时，将显示应用程序中使用的Php 数据库查询返回：检查参数是数组还是对象,php,mysql,sql,Php,Mysql,Sql,当尝试使用php和mysql登录此应用程序时，将显示应用程序中使用的授权失败错误。读取php_error.log时，它显示： [2019年5月5日03:31:51 UTC]PHP通知：未定义索引：导致 /Applications/MAMP/htdocs/iReporter/api.php第43行[2019年5月5日] 03:31:51 UTC]PHP警告：count（）：参数必须是数组或 对象，该对象在 /第43行的Applications/MAMP/htdocs/iReporter/api.p
授权失败php_error.log//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($result);
} else {
// no matching username/password was found in the login table
errorJson('Authorization failed');
}
}
switch ($_POST['command']) {
case "login":
login($_POST['username'], $_POST['password']);
break;
将iOS应用程序更改为向数据库发布密码，而不是附加密码，因为它正在创建错误 根据错误，您没有在
$result结果if (isset($result['result']) && count($result['result']) > 0)
您还需要调查为什么无法从数据库中获得预期结果。为此，您需要查看
$result$resultresultif (isset($result['result']) && count($result['result']) > 0)
您还需要调查为什么无法从数据库中获得预期结果。为此，您需要查看
$resultisset（）is_数组function login($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, $pass);
if (is_array($result) && isset($result) && count($result) > 0) {
if (isset($result['result'])) {
// 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{
errorJson('Sorry! There is a problem with $result["result"]!');
}
} else {
// no matching username/password was found in the login table
errorJson('Authorization failed');
}
}
由于数组不是数组且不可计数，因此返回错误 除了
isset（）is_数组function login($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, $pass);
if (is_array($result) && isset($result) && count($result) > 0) {
if (isset($result['result'])) {
// 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{
errorJson('Sorry! There is a problem with $result["result"]!');
}
} else {
// no matching username/password was found in the login table
errorJson('Authorization failed');
}
}
function login($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 43
if ($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');
}
}
由于数组不是数组且不可计数，因此返回错误 你能检查一下它是否从数据库中提取数据吗。确保它已连接到MySQLdatabases@Sathiraumesh是的。在尝试注册功能时，如果尝试使用现有用户名，则成功显示已使用
用户名用户名[05-May-2019 05:05:26 UTC]PHP解析错误：语法错误，意外的“计数”（T_字符串）。错误日志显示[05-May-2019 05:33:16 UTC]PHP注意：未定义索引：当变量转储（$result）时，第44行的/Applications/MAMP/htdocs/iReporter/api.PHP中出现结果
被添加到登录函数中，就像编辑的问题一样，当试图通过browserYeah访问index.php文件时，在浏览器中看不到任何输出，它以前使用的是$result['result']
，它只是突然停止。当然是从2012年开始的。资料来源：下载在底部Ok。我看到“对不起！$result[“result”]有问题错误。再次感谢。我会看看能做些什么…谢谢你的回复。现在，错误日志在第44行的/Applications/MAMP/htdocs/iReporter/api.PHP中显示[05-May-2019 05:05:26 UTC]PHP解析错误：语法错误，意外的“计数”（T_字符串）。错误日志显示[05-May-2019 05:33:16 UTC]PHP注意：未定义索引：当变量转储（$result）时，第44行的/Applications/MAMP/htdocs/iReporter/api.PHP中出现结果
被添加到登录函数中，就像编辑的问题一样，当试图通过browserYeah访问index.php文件时，在浏览器中看不到任何输出，它以前使用的是$result['result']
，它只是突然停止。当然是从2012年开始的。资料来源：下载在底部Ok。我看到“对不起！$result[“result”]有问题错误。再次感谢。我会看看能做些什么…谢谢你的回复。奇怪的是，它仍然显示授权失败
错误，并且没有任何输出到错误日志谢谢您的回复。奇怪的是，它仍然显示授权失败
错误，并且没有任何输出到错误日志谢谢您的回复。现在，错误日志显示“[05-May-2019 05:18:23 UTC]PHP注意：尝试在第44行的/Applications/MAMP/htdocs/iReporter/api.PHP中获取非对象的属性'num_rows'`您能检查在执行查询后是否填充了$result变量吗；ini设置（“显示错误”，1）已添加到索引文件中。当应用程序运行时，浏览器输出类似于array（0）
。但是现在，当尝试运行带有错误报告语句的应用程序时，将显示（200-299）中的预期状态代码，Get 500
。使用登录功能的位置更新了问题谢谢您的回答。现在错误日志显示“[05-May-2019 05:18
function login($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 43
if ($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');
}
}