Question

我有一个搜索栏，可以返回我的成员表中的用户名。然而，不仅仅是显示他们的用户名，我希望结果能够链接到他们各自的个人资料。

以下是搜索栏的PHP代码：

<?php

$servername = "localhost";
$username = "root";
$password = "root";
$dbname = "coursework_db";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
} 

$partialSearch = "%". $_POST['partialSearch'] ."%";
$stmt = $conn->prepare("SELECT username FROM members WHERE username LIKE ? 
                    LIMIT 3");

$stmt->bind_param('s',$partialSearch);
$stmt->execute();
$stmt->bind_result($username);

while ($stmt->fetch()) {

echo $username;

echo "<div>".$searchResults."</div>";
}
?>

这是实际的搜索栏，其中结果div是ajax的目标div：

<li><input type="text" name="partialSearch"onkeyup="search(this.value)" placeholder="Search for other users"/></li>

<section id="divider">
<div id="results"></div>

AJAX：

function search(partialSearch){
$.ajax({url:"searchbar.php",type:"POST",data:{partialSearch:partialSearch},success:function(result){
$("#results").html(result);
}});
};

以下是我用于生成用户个人资料的代码：

<?php

    function user_profile(){

            $servername = "localhost";
            $username = "root";
            $password = "root";
            $dbname = "coursework_db";


            // Create connection
            $conn = new mysqli($servername, $username, $password, $dbname);
            // Check connection
            if ($conn->connect_error) {
            die("Connection failed: " . $conn->connect_error);
            } 

            $sql = 'SELECT memberID, username FROM members;';
            $result = $conn->query($sql);

            $users = array();

             while($row = $result->fetch_assoc()) {
                $user[]= $row;
            }

            return $user;

            foreach (user_profile() as $user){
        ?>  
            <p>
            <a href="ViewProfile.php?uid=<?php echo $user['memberID'];?>"><?php echo $user['username'];?></a>
            </p>
        <?php


    }
    }

?>

提前感谢您的帮助。希望得到意见和建议

Answer 1

你非常接近完成申请。我重构了您的代码，并添加了详细的注释以指导您完成更改。

搜索页面

    <form action="#">
        <input type="text" name="partialSearch" onkeyup="search(this.value)" placeholder="Search for other users"/>
    </form>

    <section id="divider">
        <div id="results"></div>
    </section>

    <!-- always place scripts just before the ending <body> tag for better performance -->
    <script src="http://code.jquery.com/jquery-latest.min.js"></script>
    <script type="text/javascript">
        function search(partialSearch) {
            $.ajax({
                url: "search_results_ajax.php",
                type: "POST",
                data: {partialSearch: partialSearch},
                success: function (result) {
                    $("#results").html(result);
                }
            });
        }
    </script>

Ajax搜索

<?php
/**
 * Your function names should match their purpose. user_profile() does not accurately define the operations
 * this function will be performing.
 *
 * @param array $usernameToSearch - The username to compare with the database.
 *
 * @return array - A list of users.
 */
function get_users_by_username($usernameToSearch)
{
    $servername = "localhost";
    $username   = "root";
    $password   = "root";
    $dbname     = "coursework_db";

    // Create connection
    $conn = new mysqli($servername, $username, $password, $dbname);
    // Check connection
    if($conn->connect_error) {
        die("Connection failed: " . $conn->connect_error);
    }

    /**
     * The query below will fetch all of your users from the database.
     *
     *      SELECT memberID, username FROM members
     *
     * You can narrow the results by using the WHERE clause with the LIKE operator:
     *
     *      $sql = 'SELECT memberID, username FROM members WHERE username LIKE "%' . $username . '%"'
     *
     * Learn more about the LIKE clause at: http://www.w3schools.com/sql/sql_like.asp
     */
    $sql    = 'SELECT memberID, username FROM members WHERE username LIKE "%' . $usernameToSearch . '%"';
    $result = $conn->query($sql);

    $users = array();
    while($row = $result->fetch_assoc()) {
        /** Renamed from $user to $users to match variable definition. */
        $users[] = $row;
    }

    /** Renamed from $user to $users to match variable definition. */
    return $users;
}


/**
 * Your AJAX request sends a parameter called "partialSearch", which contains the value of the textbox
 * on the search page. You can access that parameter through the PHP global, $_POST.
 *
 * (or $_GET, if the parameter is stored in the url. Such as: www.mywebsite.com?partialSearch=my+search+here)
 *
 * Before you execute get_users_by_username(), you must check if the "partialSearch" parameter exists. This
 * can be done through the empty() function, which essentially checks if the value if $_POST['partialSearch'] is:
 *  - false
 *  - an empty string
 *  - null
 *
 * The NOT operator "!" in front of empty() will flip the value returned by empty(). So if empty() returns true, the
 * NOT operator will flip it to false. This important because we only what the code within the "if" statement to
 * execute if $_POST['partialSearch'] is NOT empty.
 */
if(!empty($_POST['partialSearch'])) {
    /** Store the results of get_users_by_username(), so we can count the number of results without re-running the function. */
    $users = get_users_by_username($_POST['partialSearch']);

    /**
     * Foreach loops with through warnings if user_profile() returns an empty value. Make sure count($users) is > 0
     * before running the loop.
     */
    if(count($users) > 0) {
        foreach($users as $user) {
            print '<p>
                  <a href="ViewProfile.php?uid=' . $user['memberID'] . '">
                      ' . $user['username'] . '
                  </a>
               </p>';
        }
    } else {
        print "<p>No Results</p>";
    }
}

/**
 * In files with only PHP code, you do not need the closing PHP tag.
 *
 * "?>"
 */

用户个人资料

<?php
/**
 * Fetch user from database by their memberID.
 *
 * @param $id - The user's ID
 *
 * @return array - The user database record.
 */
function get_single_user_by_id($id)
{
    $servername = "localhost";
    $username   = "root";
    $password   = "root";
    $dbname     = "coursework_db";

    // Create connection
    $conn = new mysqli($servername, $username, $password, $dbname);
    // Check connection
    if($conn->connect_error) {
        die("Connection failed: " . $conn->connect_error);
    }

    $sql    = 'SELECT memberID, username FROM members WHERE memberID = ' . $id;
    $result = $conn->query($sql);

    //Return single user
    return $result->fetch_assoc();
}
?>

<!DOCTYPE html>
<html>
<head>
    <title>Search Demo</title>
</head>
<body>

<?php
    if(!empty($_GET['uid'])) {
        $user = get_single_user_by_id($_GET['uid']);

        if(!empty($user)) {
            print "<h1>Welcome " . $user['username'] . "!";
        } else {
            print "User with ID " . $_GET['uid'] . " doesn't exist.";
        }
    } else {
        //No UID was provided
        print "Please provide a UID to a valid user.";
    }
?>

</body>
</html>

提示1 不要在应用程序中使用mysql或mysqli数据库连接。这两个连接都已被弃用，并且非常不安全。查看this和this以了解SQL注入攻击。要防止此类攻击，请使用PHP Data Objects (PDO)。

提示2 将您的功能定义在另一个文件中，与HTML分开。在HTML 旁边放置功能定义会很快变得混乱。您可以使用PHP的include和require函数来执行此操作。（learn about it here）

提示3 使用重复代码通常是不好的做法。您使用相同的代码在两个页面（搜索ajax和配置文件页面）上连接到您的数据库。我建议构建一个数据库类来处理所有应用程序的数据库事务，这样你就不会有重复的代码。 This will help you with that.

更新

您的搜索栏PHP代码存在一些问题：

<强>第一

您的代码中的以下行只会获取完全匹配：

$stmt = $conn->prepare("SELECT username FROM members WHERE username LIKE ? 
                LIMIT 3");

为了实现所需的自动完成搜索，您需要为其添加通配符：

$stmt = $conn->prepare("SELECT username FROM members WHERE username LIKE '?%' LIMIT 3");

＆＃39;％＆＃39;还将包含以搜索值开头的用户名。

<强>第二

在同一文件中，未定义$searchResults变量。您的结果字符串将始终为空。

您需要更改循环迭代结果并为每个结果打印链接。

while ($stmt->fetch()) {

    echo $username;

    echo "<div>".$searchResults."</div>";
}

为：

$results = $stmt->fetchAll();

if($results > 0) {
    foreach($results as $user) {
        print '<p>
            <a href="ViewProfile.php?uid=' . $user['memberID'] . '">
                ' . $user['username'] . '
            </a>
        </p>';
    }
} else {
    print "<p>No Results</p>";
}

搜索栏中的结果是指向各个配置文件的链接

1 个答案:

更新