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當我輸入了正確的憑據結果是{"success":1,"message":"Login successful!"}
ANDROID&PHP |登錄
,當我故意輸入的密碼不正確,這是結果,我不明白這個登錄腳本
<?php
require("dbConnect.php");
if (!empty($_POST)) {
$query = "SELECT * FROM tb_user WHERE instructor_id = :instructor_id";
$query_params = array(
':instructor_id' => $_POST['instructor_id']
);
try {
$stmt = $db->prepare($query);
$result = $stmt->execute($query_params);
}
catch (PDOException $ex) {
$response["success"] = 0;
$response["message"] = "Database Error1. Please Try Again!";
die(json_encode($response));
}
$row = $stmt->fetch();
if ($row) {
//if we encrypted the password, we would unencrypt it here, but in our case we just
//compare the two passwords
$password = $_POST['password'];
if (md5($password) == $row['password']) {
$response["success"] = 1;
$response["message"] = "Login successful!";
die(json_encode($response));
}else {
$response["success"] = 0;
$response["message"] = "Invalid Credentials!";
die(json_encode($response));
}
}
} else {
?>
<h1>Login</h1>
<form action="login.php" method="post">
Username:<br />
<input type="text" name="instructor_id" placeholder="instructor_id" />
<br /><br />
Password:<br />
<input type="password" name="password" placeholder="password" value="" />
<br /><br />
<input type="submit" value="Login" />
</form>
<?php
}
?>
{"success":0,"message":"Invalid Credentials!"}
但問題是當我輸入兩個錯誤的用戶名和傳遞沒有JSON顯示結果,我期待的結果仍然應該這樣{"success":0,"message":"Invalid Credentials!"}
,因爲我輸入了錯誤的登錄憑據
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