从Mysql / PHP获取名称或UserID的用户信息

问题描述：

Can I know what's mistake in here? I have two tables in my database. it is this

table

table2

I have written the code as search player thing. I'll put the name or userid in the form and it'll process the information of user.

Here is my code

<html lang="en">
 <head>
 <meta charset="utf-8" />
   <title></title>
 </head>
 <body>
 <form action="" method="post">
 Search: <input type="text" name="term" />
 <input type="submit" value="Submit" />
 </form>
 <?php
 include('config.php');
 if (!empty($_REQUEST['term']))
    {
    $term = mysql_real_escape_string($_REQUEST['term']);
    $sql = " select u.* from users u inner join ranks r ON (u.UserID = r.UserID) where u.UserID = '%" . $term . "%'";
    $r_query = mysql_query($sql);
    while ($row = mysql_fetch_array($r_query))
        {
        echo 'Name: ' . $row['Name'];
        echo '<br /> Cash: ' . $row['Cash'];
        echo '<br /> Score: ' . $row['Score'];
        echo '<br /> Race: ' . $row['Race'];
        echo '<br /> Horseshoe: ' . $row['Horseshoe'];
        }
    }
?>
</body>
</html>

</div>

答

First of all you should update your config.php to mysqli_ functions. and the mysqli_real_escape_string() and mysqli_query() functions need 2 Parameters.
First $conn, second: the variable

finally your code should look like this:

<html lang="en">
<head>
    <meta charset="utf-8"/>
    <title></title>
</head>
<body>
<form method="POST">
    Search: <input title="searchfield" required type="text" name="term"/>
    <input type="submit" name="submit" value="Submit"/>
</form>
<?php

include('config.php');

if (isset($_POST["submit"]) && !empty($_POST["submit"])) {
    $term = mysqli_real_escape_string($conn, $_REQUEST["term"]); //make sure the $conn isset
    $sql = "SELECT u.* FROM users u INNER JOIN ranks r ON (u.UserID = r.UserID) WHERE u.UserID LIKE '%" . $term . "%'"; // change = to LIKE
    $r_query = mysqli_query($conn, $sql);

    while ($row = mysqli_fetch_array($r_query)) {
        echo 'Name: ' . $row['Name'];
        echo '<br /> Cash: ' . $row['Cash'];
        echo '<br /> Score: ' . $row['Score'];
        echo '<br /> Race: ' . $row['Race'];
        echo '<br /> Horseshoe: ' . $row['Horseshoe'];
    }
}
?>

</body>
</html>

your config.php should look like this:

<?php
$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "idkw0t";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);

// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}
echo "Connected successfully";
?>

答

you should use like in query as follows

$sql = " select u.* from users u inner join ranks r ON (u.UserID = r.UserID) where u.UserID like '%" . $term . "%'";

从Mysql / PHP获取名称或UserID的用户信息

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