[英]Having trouble retrieving values from MySQL database
I'm very new to using MySql and am having trouble retrieving values from my database. 我对使用MySql非常陌生,无法从数据库中检索值。 I was under the impression that i was going about it the correct way but my echo statements don't print anything.
我的印象是我正在以正确的方式进行操作,但是我的echo语句不打印任何内容。
I'd appreciate some help. 我将不胜感激。 My code is below.
我的代码如下。 I know i'll have to add security later on like sanitizing user input.
我知道以后必须像清理用户输入一样增加安全性。
<?php
$email = $_POST['email'];
$password = $_POST['password'];
$hashedPass = sha1($password);
if ((!isset($email)) || (!isset($password))) {
//Visitor needs to enter a name and password
echo "Data not provided";
} else {
echo "Received details $email and $password <br/>";
// connect to mysql
$mysql = mysqli_connect("localhost", "root", "root");
if(!$mysql) {
echo "Cannot connect to PHPMyAdmin.";
exit;
} else {
echo "Connected to phpmyadmin <br/>";
}
}
// select the appropriate database
$selected = mysqli_select_db($mysql, "languageapp");
if(!$selected) {
echo "Cannot select database.";
exit;
} else {
echo "DB Selected";
}
// query the database to see if there is a record which matches
$query = "select count(*) from user where email = '".$email."' and password = '".$hashedPass."'";
$result = mysqli_query($mysql, $query);
if(!$result) {
echo "Cannot run query.";
exit;
}
$row = mysqli_fetch_row($result);
$count = $row[0];
$userdata = mysqli_fetch_array($result, MYSQLI_BOTH);
echo $userdata[3];
echo $userdata['firstName'];
if ($count > 0) {
echo "<h1>Login successful!</h1>";
echo "<p>Welcome.</p>";
echo "<p>This page is only visible when the correct details are provided.</p>";
} else {
// visitor's name and password combination are not correct
echo "<h1>Login unsuccessful!</h1>";
echo "<p>You are not authorized to access this system.</p>";
}
?>
I believe the problem is that you call twice the * fetch * family function which will cause the $userdata to be empty. 我相信问题是您调用了两次* fetch *系列函数,这将导致$ userdata为空。
From the documentation mysql_fetch_row will fetch the next row and move the internal data pointer ahead. 从文档mysql_fetch_row将获取下一行并将内部数据指针向前移动。 So when you call mysqli_fetch_array($result, MYSQLI_BOTH) and I suppose the user/password is unique there is nothing to retrieve.
因此,当您调用mysqli_fetch_array($ result,MYSQLI_BOTH)时,我想用户名/密码是唯一的,没有任何可检索的内容。 Also another mistake you did is that your SELECT doesn't retrieve the actual user data, but just the count number for the user/password combination.
您犯的另一个错误是,SELECT不会检索实际的用户数据,而只会检索用户/密码组合的计数。 So your userdata will be always incorrect, even if you fetch the data right.
因此,即使您正确获取数据,您的用户数据也总是不正确的。
So change your query to something like that: 因此,将查询更改为类似的内容:
$query = "select * from user where email = '".$email."' and password = '".$hashedPass."' LIMIT 1";
Then use mysql_fetch_array to check if the entry exist and then retrieve the user data. 然后使用mysql_fetch_array检查条目是否存在,然后检索用户数据。
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