截至目前，我可以在一列中显示我的图像，包括图像，标题和简短的描述。所有这些都来自同一个数据库。我不太擅长编码，需要一些指导，您将如何添加到此现有代码中 1) 允许图片显示在多列中......2)允许单击缩略图，这将加载一个单独的页面，然后我可以在该页面上设置样式并列出完整的食谱。
我一直在弄乱代码，我对我创建的内容感到困惑。我不知道如何进行。
<h2>index.php:</h2>
<section class="gallery-links">
<div class="wrapper">
<div class="gallery-container">
<?php
include_once 'includes/dbh.inc.php';
$sql = "SELECT * FROM gallery ORDER BY orderGallery DESC"
$stmt = mysqli_stmt_init($conn);
if (!mysqli_stmt_prepare($stmt, $sql)) {
echo "SQL statment failed!";
} else {
mysqli_stmt_execute($stmt);
$result = mysqli_stmt_get_result($stmt);
while ($row = mysqli_fetch_assoc($result)) {
echo '<a href="#">
<div style="background-image: url(img/gallery/'.$row["imgFullNameGallery"].');"></div>
<h3>'.$row["titleGallery"].'</h3>
<p>'.$row["descGallery"].'</p>
</a>';
}
}
?>
</div>
<?php
echo '<div class="gallery-upload">
<form action="includes/gallery-upload.inc.php" method="post" enctype="multipart/form-data">
<input type="text" name="filename" placeholder="File name...">
<input type="text" name="filetitle" placeholder="Image title...">
<input type="text" name="filedesc" placeholder="Image description...">
<input type="file" name="file">
<button type="submit" name="submit">Upload</button>
</form>
</div>'
?>
</div>
</section>
<h2>gallery-upload.inc.php:</h2>
<?php
if (isset($_POST['submit'])) {
$newFileName = $_POST['filename'];
if (empty($newFileName)) {
$newFileName = "gallery";
} else {
$newFileName = strtolower(str_replace(" ", "-", $newFileName));
}
$imageTitle = $_POST['filetitle'];
$imageDesc = $_POST['filedesc'];
$file = $_FILES["file"];
$fileName = $file["name"];
$fileType = $file["type"];
$fileTempName = $file["tmp_name"];
$fileError = $file["error"];
$fileSize = $file["size"];
$fileExt = explode(".", $fileName);
$fileActualExt = strtolower(end($fileExt));
$allowed = array("jpg", "jpeg", "png");
if (in_array($fileActualExt, $allowed)) {
if ($fileError === 0) {
if($fileSize < 2000000) {
$imageFullName = $newFileName . "." . uniqid("", true) . "." . $fileActualExt;
$fileDestination = "../images/gallery/" . $imageFullName;
include_once "dbh.inc.php";
if (empty($imageTitle) || empty($imageDesc)) {
header("Location: ../index.php?upload=empty");
exit();
} else {
$sql = "SELECT * FROM gallery;";
$stmt = mysqli_stmt_init($conn);
if (!mysqli_stmt_prepare($stmt, $sql)) {
echo "SQL statement failed!";
} else {
mysqli_stmt_execute($stmt);
$result = mysqli_stmt_get_result($stmt);
$rowCount = mysqli_num_rows($result);
$setImageOrder = $rowCount + 1;
$sql = "INSERT INTO gallery (titleGallery, descGallery, imgFullNameGallery, orderGallery) VALUES (?, ?, ?, ?);";
if (!mysqli_stmt_prepare($stmt, $sql)) {
echo "SQL statement failed!";
} else {
mysqli_stmt_bind_param($stmt, "ssss", $imageTitle, $imageDesc, $imageFullName, $setImageOrder);
mysqli_stmt_execute($stmt);
move_uploaded_file($fileTempName, $fileDestination);
header("Location: ../index.php?upload=success");
}
}
}
} else {
echo "File size is too big!";
exit();
}
} else {
echo "You had an error!";
exit();
}
} else {
echo "You need to upload a proper file type!";
exit();
}
<h2>dbh.inc.php:</h2>
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "gallery";
$conn = mysqli_connect($servername, $username, $password, $dbname);
基本上，您需要在anchor标签上创建一个链接，该标签将具有图像详细页面url，image id如下所示：
while ($row = mysqli_fetch_assoc($result)) {
// assuming that imgId is your primary key
echo '<a href="detail.php?imageId="'.$row["imgId"].' target="_blank">
<div style="background-image: url(img/gallery/'.$row["imgFullNameGallery"].');"></div>
<h3>'.$row["titleGallery"].'</h3>
<p>'.$row["descGallery"].'</p>
</a>';
}
之后，您需要创建一个新文件detail.php，您可以在其中通过$_GET['imgId']获取image id，然后可以对其进行查询，并能够获得完整的图像详细信息。您还需要创建一个用于查看的HTML，并且可以显示详细信息。
希望对您有所帮助！！