Question

我能够使用PHP创建一个MySQL数据库，用我的数据填充它，然后在我的网页上弹出它。但是，我希望用户能够从Options DropDown中点击Football中的位置，比如QuarterBack，然后网页将只显示团队中四分卫的球员。

我非常确定我需要的是与我的SELECT * FROM

行相关的事情

编辑 - 此外，我在网页上的代码显示了Unidentified Index - Position，即使我定义了它

相关代码信息在此处：

<form action="FinalProject.php" method="get">
    <select name="position" id="position" value="position">
        <option value="">Select a position:</option>
        <option value="1">Quarterback</option>
        <option value="2">Wide Receiver</option>
        <option value="3">Linebacker</option>
        <option value="4">Tight End</option>
    </select>
</form>

<?php
//                     --- CREATE THE DATABASE
$db_conn = mysqli_connect("localhost", "root", "");
if (!$db_conn)
    die("Unable to connect: " . mysqli_connect_error());

mysqli_query($db_conn, "CREATE DATABASE IF NOT EXISTS sportDatabase;");

// --- CREATE THE TABLE
mysqli_select_db($db_conn, "sportDatabase");
$cmd = "CREATE TABLE bearsRoster (
          number int(2) NOT NULL PRIMARY KEY,
          playerName varchar(20),
          position char(2),
          height varchar(5),
          weight int(3),
          age int(2),
          experience int(2),
          collegeName varchar(50)
          );";

mysqli_query($db_conn, $cmd);
$cmd = "LOAD DATA LOCAL INFILE 'bearsRoster.csv' INTO TABLE bearsRoster FIELDS TERMINATED BY ',';";
mysqli_query($db_conn, $cmd);
$selectedOption = $_GET["position"];
$cmd = "SELECT * FROM bearsRoster ";
//                    ORDER BY position WHERE".$selectedOption.";";
$records = mysqli_query($db_conn, $cmd);

echo("<table border = 'black' align:'center'> 
        <tr> 
        <th>Number</th> 
        <th>Player Name</th> 
        <th>Position</th> 
        <th>Height</th> 
        <th>Weight</th> 
        <th>Age</th>
        <th>Experience</th>
        <th>College Name</th>
        </tr>" . PHP_EOL);

while ($row = mysqli_fetch_array($records)) {
    echo("<tr> 
            <td id = 'yellow'>" . $row['number'] . "</td> 
            <td id = 'red'>" . $row['playerName'] . "</td> 
            <td id = 'red'>" . $row['position'] . "</td> 
            <td id = 'blue'>" . $row['height'] . "</td>
            <td id = 'blue'>" . $row['weight'] . "</td>
            <td id = 'pink'>" . $row['age'] . "</td>
            <td id = 'pink'>" . $row['experience'] . "</td>
            <td id = 'pink'>" . $row['collegeName'] . "</td>
          </tr>" . PHP_EOL);
}
echo("</table>");
mysqli_close($db_conn);
?>

Answer 1

我编辑了你的代码，我应该这样做。

<form action="FinalProject.php" method="POST">
    <select name="position" id="position" value="position">
        <option value="">Select a position:</option>
        <option value="QB">Quarterback</option>
        <option value="WR">Wide Receiver</option>
        <option value="LB">Linebacker</option>
        <option value="TE">Tight End</option>
    </select>

    <input type="submit" name="submit" value="Click me!" />
</form>

<?php
$db_conn = mysqli_connect("localhost", "root", "");
if (!$db_conn)
    die("Unable to connect: " . mysqli_connect_error());

mysqli_query($db_conn, "CREATE DATABASE IF NOT EXISTS sportDatabase;");
mysqli_select_db($db_conn, "sportDatabase");

$cmd = "CREATE TABLE bearsRoster (number int(2) NOT NULL PRIMARY KEY, playerName varchar(20), position char(2), height varchar(5), weight int(3), age int(2), experience int(2), collegeName varchar(50));";
mysqli_query($db_conn, $cmd);

$cmd = "LOAD DATA LOCAL INFILE 'bearsRoster.csv' INTO TABLE bearsRoster FIELDS TERMINATED BY ',';"; 
mysqli_query($db_conn, $cmd);

$cmd = "SELECT * FROM bearsRoster";
if(isset($_POST['submit'])) {
    if($_POST['position'] == 'QB' || $_POST['position'] == 'WR' || $_POST['position'] == 'LB' || $_POST['position'] == 'TE')
        $selectedOption = $_POST['position'];
    else
        $selectedOption = null;

    if($selectedOption !== null)
        $cmd = "SELECT * FROM bearsRoster WHERE position = ".$selectedOption.";";
}

$records = mysqli_query($db_conn, $cmd);

echo("<table border = 'black' align:'center'><tr><th>Number</th><th>Player Name</th><th>Position</th><th>Height</th><th>Weight</th><th>Age</th><th>Experience</th><th>College Name</th></tr>" . PHP_EOL);

while($row = mysqli_fetch_array($records)){
    echo("<tr><td id = 'yellow'>" . $row['number'] . "</td><td id = 'red'>" . $row['playerName'] . "</td><td id = 'red'>" . $row['position'] . "</td><td id = 'blue'>" . $row['height'] . "</td><td id = 'blue'>". $row['weight'] . "</td><td id = 'pink'>". $row['age'] . "</td><td id = 'pink'>". $row['experience'] . "</td><td id = 'pink'>". $row['collegeName'] . "</td></tr>". PHP_EOL);
}

echo("</table>");
mysqli_close($db_conn);

Answer 2

在表创建中使用IF NOT EXISTS，因为每次运行脚本时都会删除并创建表。您可以使用AJAX从服务器获取数据，而无需重新加载页面或重定向。

AJAX是开发人员的梦想，因为你可以：


更新网页而不重新加载页面

在页面加载后从服务器请求数据

在页面加载后从服务器接收数据

将数据发送到服务器 - 在后台

在select添加onchange属性中，该属性将调用Javascript函数，该函数将从数据库中获取数据并在页面上显示而不重新加载。

<强> HTML

<!DOCTYPE html>
<html>
<head>

</head>
<body>
<select name="position" id="position" onchange="getPosition(this)" value="position">
    <option value="">Select a position:</option>
    <option value="QB">Quarterback</option>
    <option value="WR">Wide Receiver</option>
    <option value="LB">Linebacker</option>
    <option value="TE">Tight End</option>
</select>

<div id="results"></div>
<script>
    function getPosition(position) {
        var xhttp = new XMLHttpRequest();
        xhttp.onreadystatechange = function () {
            if (this.readyState == 4 && this.status == 200) {
                document.getElementById("results").innerHTML = this.responseText;
            }
        };
        xhttp.open("POST", "FinalProject.php", true);
        xhttp.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
        xhttp.send('position=' + position.value);
    }
</script>
</body>
</html>

<强> FinalProject.php

<?php
//                     --- CREATE THE DATABASE
$db_conn = mysqli_connect("localhost", "root", "");
if (!$db_conn)
    die("Unable to connect: " . mysqli_connect_error());

mysqli_query($db_conn, "CREATE DATABASE IF NOT EXISTS sportDatabase;");

// --- CREATE THE TABLE
mysqli_select_db($db_conn, "sportDatabase");
$cmd = "CREATE TABLE IF NOT EXISTS bearsRoster (
          number int(2) NOT NULL PRIMARY KEY,
          playerName varchar(20),
          position char(2),
          height varchar(5),
          weight int(3),
          age int(2),
          experience int(2),
          collegeName varchar(50)
          );";

mysqli_query($db_conn, $cmd);
$cmd = "LOAD DATA LOCAL INFILE 'bearsRoster.csv' INTO TABLE bearsRoster FIELDS TERMINATED BY ',';";
mysqli_query($db_conn, $cmd);
$selectedOption = $_POST["position"];
$cmd = "SELECT * FROM `bearsRoster` WHERE position = '$selectedOption' ORDER BY position";
$records = mysqli_query($db_conn, $cmd);

if ($records && mysqli_num_rows($records)) : ?>
    <table border='black' align="center">
        <tr>
            <th>Number</th>
            <th>Player Name</th>
            <th>Position</th>
            <th>Height</th>
            <th>Weight</th>
            <th>Age</th>
            <th>Experience</th>
            <th>College Name</th>
        </tr>

        <?php while ($row = mysqli_fetch_array($records))  : ?>
            <tr>
                <td id='yellow'><?php echo $row['number']; ?></td>
                <td id='red'><?php echo $row['playerName']; ?></td>
                <td id='red'><?php echo $row['position']; ?></td>
                <td id='blue'><?php echo $row['height']; ?></td>
                <td id='blue'><?php echo $row['weight']; ?></td>
                <td id='pink'><?php echo $row['age']; ?></td>
                <td id='pink'><?php echo $row['experience']; ?></td>
                <td id='pink'><?php echo $row['collegeName']; ?></td>
            </tr>
        <?php endwhile; ?>
    </table>
<?php else: ?>
<h1>No Data found</h1>
<?php endif;
mysqli_close($db_conn);
?>

如何让用户选择一个玩家位置并让mySQL向我的数据库中的每个玩家显示该位置？

2 个答案: