[英]Retrieving variable from one foreach statement and displaying in another foreach statement
我試圖在另一個數組的foreach語句中使用一個數組中的變量。
我有一個網格視圖,其中顯示用戶的個人資料照片,他們的顯示名稱,當前位置和可用性,以及使用他們的用戶名來完成應用於該框的鏈接。
用戶表保存; 用戶名(和用戶ID)將來是否應該使用。
配置文件表保存; 顯示名稱。
profilephotos表保存; 個人資料照片。
位置表保存; 當前位置。
在表中,所有這些都通過與users表匹配的user_id和username列進行鏈接。
我為用戶框提供的代碼是;
<?php foreach($rows as $row): ?>
<div class="box">
<div class="boxInner">
<a href="profile.php?username=<?php echo htmlentities($row['username'], ENT_QUOTES, 'UTF-8'); ?>&uid=<?php echo htmlentities($row['id'], ENT_QUOTES, 'UTF-8'); ?>">
<img src="uploads/profile-photos/<?php echo htmlentities($pphoto['profilephoto_file'], ENT_QUOTES, 'UTF-8'); ?>" />
<div class="titleBox" style="text-align:left; line-height:20px;">
<span style="font-size:18px; font-weight:bold;"><i class="fa fa-user"></i> <?php echo htmlentities($row['profile_displayname'], ENT_QUOTES, 'UTF-8'); ?>,
<?php echo htmlentities($row['profile_displayage'], ENT_QUOTES, 'UTF-8'); ?></span>
<br />
<span style="font-size:14px;"><i class="fa fa-map-marker"></i> City Name | <i class="fa fa-clock-o"></i> Now</span>
</div></a>
</div>
</div>
<?php endforeach; ?>
我的SQL查詢和數組代碼是;
$query = "
SELECT
users.id,
users.username,
users.email,
profiles.profile_displayname,
profiles.profile_displayage,
profiles.profile_photo
FROM users, profiles
WHERE users.id = profiles.user_id;
";
try
{
// These two statements run the query against your database table.
$stmt = $db->prepare($query);
$stmt->execute();
}
catch(PDOException $ex)
{
// Note: On a production website, you should not output $ex->getMessage().
// It may provide an attacker with helpful information about your code.
die("Failed to run query: " . $ex->getMessage());
}
// Finally, we can retrieve all of the found rows into an array using fetchAll
$rows = $stmt->fetchAll();
$query = "
SELECT
users.id,
users.username,
profilephotos.user_id,
profilephotos.profilephoto_file
FROM users, profilephotos
WHERE users.id = profilephotos.user_id;
";
try
{
// These two statements run the query against your database table.
$stmt = $db->prepare($query);
$stmt->execute();
}
catch(PDOException $ex)
{
// Note: On a production website, you should not output $ex->getMessage().
// It may provide an attacker with helpful information about your code.
die("Failed to run query: " . $ex->getMessage());
}
// Finally, we can retrieve all of the found rows into an array using fetchAll
$profilephotos = $stmt->fetchAll(PDO::FETCH_ASSOC);
無論我嘗試什么,我都無法獲得正確的印象。 我設法插入圖像,但是無論我指示查詢$profilephotos
查找的ID如何,foreach語句都將同一圖像應用於每個用戶。
我究竟做錯了什么? 我什至會以正確的方式這樣做嗎?
任何幫助將不勝感激-請注意,盡管我是PHP新手。
您需要一個查詢而不是兩個查詢。 該查詢可能看起來像這樣:
SELECT
users.id AS id,
users.username AS username,
users.email as email,
profilephotos.profilephoto_file AS file_photo,
profiles.profile_displayname AS file_displayname,
profiles.profile_displayage AS displaypage,
profiles.profile_photo AS photo
FROM users
JOIN profilephotos ON users.id = profilephotos.user_id
JOIN profiles ON users.id = profiles.user_id;
您需要兩次使用聯接-更好的實踐。 並注意關鍵字“ AS”-有助於消除不同表中相同的列名稱的歧義。
如果要查看查詢變體之一(帶有最新的個人資料圖片),就是這樣:
SELECT
users.id AS id,
users.username AS username,
users.email as email,
profilephotos.profilephoto_file AS file_photo,
profiles.profile_displayname AS file_displayname,
profiles.profile_displayage AS displaypage,
profiles.profile_photo AS photo
FROM users
LEFT JOIN profiles ON users.id = profiles.user_id
LEFT JOIN profilephotos ON users.id = profilephotos.user_id
WHERE profilephotos.id in (select max(id) from profilephotos group by user_id)
ORDER BY id
但是我認為它太復雜了,可能根本是錯誤的,因為我不知道您的表架構。
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