Question

我对这个研究做了很多研究但没有成功。所以我现在要做的是：

根据数据库中的表填充下拉列表（已完成）
从数据库生成的下拉列表中选择一个值后，我希望允许用户通过自动填充当前的内容来编辑记录，并允许他们更改存储在数据库中的内容，或者只需单击按钮即可删除记录。（未完成）

数据库对象是一个具有以下属性的相册

+-------------+-------------+------+-----+---------+----------------+
| Field       | Type        | Null | Key | Default | Extra          |
+-------------+-------------+------+-----+---------+----------------+
| AlbumID     | int(11)     | NO   | PRI | NULL    | auto_increment |
| AlbumName   | varchar(75) | YES  |     | NULL    |                |
| Label       | varchar(75) | YES  |     | NULL    |                |
| Genre       | varchar(30) | YES  |     | NULL    |                |
| ReleaseDate | date        | YES  |     | NULL    |                |
| Artist      | varchar(75) | YES  |     | NULL    |                |
+-------------+-------------+------+-----+---------+----------------+

我目前正在尝试使用的HTML是：

<!DOCTYPE HTML>
<html>
<head>
    <title>MyMuzik Library</title>
    <meta charset="utf-8" />
    <link rel="stylesheet" type="text/css" href="content/site.css">
    <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.3/jquery.min.js"></script>
    <script type="text/javascript" src="scripts/mask.js"></script>
    <script type="text/javascript">
        jQuery(function($){
            $("#txtReleaseDate").mask("9999/99/99",{placeholder:"yyyy/mm/dd"});
        });
    </script>
</head>
<body>
<form action="insertAlbum.php" method="post">
    <div class="nav">
        <ul>
            <li><a href="fphome.php">Home</a></li>
            <li><a href="fpsongs.php">Songs</a></li>
            <li><a href="#">Albums</a></li>
        </ul>
    </div>
    <div class="content-container">
    <form action="insertAlbum.php" method="post" autocomplete="off">
        <h1>Albums</h1>
        <h3>This is where you add new albums</h3>
        <div class="form-group">
        <label for="ddlAlbumSelect" id="lblAlbumSelect">Select Album to Edit:</label>
        <div class="form-control">
            <select name="ddlAlbumSelect" id="ddlAlbumSelect">
                <option value="0"> Select Album...</option>
                <?php
                    mysql_connect('localhost', 'root', '1978afc247a');
                    mysql_select_db('finalproject');
                    $sql = mysql_query("SELECT * FROM albums ORDER BY AlbumName");
                    while ($row = mysql_fetch_array($sql)){
                    echo "<option value='" . $row['AlbumID'] . "'>" . $row['AlbumName'] . "</option>";
                    }
                ?>
            </select>
        </div>
        </div>
        <div class="form-group">
        <label for="txtAlbumName" id="lblAlbumName">Album Name:</label>
        <div class="form-control">
            <input type="text" <?php if(ddlAlbumSelect.Value != 0) echo "<p>Selected a value!</p>" ?>  name="txtAlbumName" id="txtAlbumName" />
        </div>
        </div>
        <div class="form-group">
        <label for="txtArtist" id="lblArtist">Artist:</label>
        <div class="form-control">
            <input type="text" name="txtArtist" id="txtArtist" />
        </div>
        </div>
        <div class="form-group">
        <label for="txtLabel" id="lblLabel">Label:</label>
        <div class="form-control">
            <input type="text" name="txtLabel" id="txtLabel" />
        </div>
        </div>
        <div class="form-group">
        <label for="txtGenre" id="lblGenre">Genre:</label>
        <div class="form-control">
            <input type="text" name="txtGenre" id="txtGenre" />
        </div>
        </div>
        <div class="form-group">
        <label for="txtReleaseDate" id="lblReleaseDate">Release Date:</label>
        <div class="form-control">
            <input type="text" name="txtReleaseDate" id="txtReleaseDate" />
        </div>
        </div>
        <input style="margin-left:30px" type="submit" />
</form>
<div class="clearfix"></div>
<br />
    </div>
    <footer>
        <p>MyMuzik Library&copy; <?php echo date("Y"); ?>
    </footer>
</body>
</html>

引用的表单操作文件insertAlbum.php如下所示：

<html>
<body>


<?php
$con = mysql_connect("localhost","root","1978afc247a");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("finalproject", $con);

$sql="INSERT INTO albums (AlbumName, Label, Genre, ReleaseDate, Artist)
VALUES
('$_POST[txtAlbumName]','$_POST[txtLabel]', '$_POST[txtGenre]','$_POST[txtReleaseDate]','$_POST[txtArtist]')";

if (!mysql_query($sql,$con))
  {
  die('Error: ' . mysql_error());
  }
echo "1 record added <br/>";

mysql_close($con)
?>
<a href="<?php echo $_SERVER['HTTP_REFERER']; ?>"><< Back</a>
</body>
</html>

我尝试过使用AJAX，但一直没有成功，只是朝着正确的方向寻找机会！感谢。

Answer 1

您至少应该提供Ajax次努力。无论如何，下面是我的答案。我已将我的解释放在评论表中。

基于SELECTED ALBUM的广告文字字段：

$(document).on("change", "#ddlAlbumSelect", function(){ /* WHEN AN ALBUM HAS BEEN SELECTED */

  var albumid = $(this).val(); /* GET THE ID OF THE SELECTED ALBUM */

  $.ajax({ /* START OF AJAX */
    type: "POST", /* METHOD TO USE TO PASS THE DATA */
    url: "getalbum.php", /* THE FILE WHERE WE WILL PROCESS THE PASSED DATA */
    data: {"albumid": albumid}, /* DATA TO BE PASSED */
    dataType: "json", /* TYPE OF DATA THAT WILL BE RETURNED FROM getalbum.php */
    success: function(result){ /* GET RETURNED DATA FROM getalbum.php */
      /* BELOW WILL POPULATE THE CORRESPONDING TEXT FIELDS IN YOUR FORM */
      $("#txtAlbumName").val(result.albumName);
      $("#txtArtist").val(result.artist);
      $("#txtLabel").val(result.label);
      $("#txtGenre").val(result.genre);
      $("#txtReleaseDate").val(result.releaseDate);
    } /* END OF GETTING THE RETURNED DATA FROM getalbum.php */
  }); /* END OF AJAX */

});

然后是getalbum.php：

/* ESTABLISH CONNECTION */
$con = new mysqli("localhost", "root", "1978afc247a", "finalproject");

/* CHECK CONNECTION */
if (mysqli_connect_errno()) {
    printf("Connect failed: %s\n", mysqli_connect_error());
    exit();
}

$stmt = $con->prepare("SELECT txtAlbumName, txtLabel, txtGenre, txtReleaseDate, txtArtist FROM albums WHERE AlbumID = ?");
$stmt->bind_param("i", $_POST["albumid"]);
$stmt->execute();
$stmt->bind_result($albumname, $label, $genre, $releasedate, $artist);
$stmt->fetch();
$stmt->close();

echo json_encode(array("albumName" => $albumname, "label" => $label, "genre" => $genre, "releaseDate" => $releasedate, "artist" => $artist"));

不要使用deprecated mysql_* API并使用mysqli_*或PDO。

更新专辑：

我以为你正在尝试更新所选专辑？为什么您的表单会转到insertAlbum.php？

根据下拉列表php中的选择填写表单

1 个答案:

基于SELECTED ALBUM的广告文字字段：

更新专辑：