我有两个看起来像这样的表:
发布表格:
ID | photo | ident
-------------------------------------
80 | img/photo1 | ACH3882
81 | img/photo2 | SHD8837
82 | img/photo3 | SFF4837
83 | img/photo4 | DLL3266
荣誉表:
ID | photo_id | ident_id
-------------------------------------
1 | 80 | SHD8837
2 | 83 | ACH3882
3 | 82 | SHD8837
工作原理:我正在尝试向我的网站添加荣誉系统。我已经设置了上面显示的表格。每个用户都有自己的ident
。当用户按下荣誉按钮时,用户的ident
和photo
的ID将存储在荣誉表中,如下所示:
$id = $_GET['id']; // get id through query string
$ident = $_SESSION["ident"];
$sql = "INSERT INTO kudos (photo_id, ident_id) VALUES ('$id', '$ident')";
if(mysqli_query($link, $sql)){
mysqli_close($link); // Close connection
header("location:index.php"); // redirects to all records page
exit;
} else{
echo "ERROR: Could not able to execute $sql. " . mysqli_error($link);
}
现在,我想显示每张照片的荣誉数量。目前,我只查询一个表,但是我还需要查询第二个表,以从该表中获取值。这就是我查询第一个表的方式:
<?php
$query = "SELECT * FROM post;
if ($result = $mysqli->query($query)) {
$num_rows = 0;
while ($row = $result->fetch_assoc()) {
$num_rows++;
/* <!-- Feed start --> */
echo "{$row['photo']}.";
echo '<a class="btn" href="kudos.php?id=';
echo "{$row['id']}";
echo '">';
echo '★ Give kudos</a>';
echo ' 0'; <- This is where I want the number count of kudos gives to show up.
... and so on
有人可以帮我吗?
答案 0 :(得分:0)
从帖子p中选择p.id,p.photo,p.indent 左加入荣誉k on p.id = k.photo_id
答案 1 :(得分:-1)
也许您可以尝试联接表?
SELECT Post.id as post, count(Kudos.ident_id) as kudos
FROM Post
LEFT JOIN Kudos ON Post.id=Kudos.photo_id
GROUP BY Post.id;
这应该使您像:
post_id ... kudosCount
在此处阅读有关联接的更多信息: SQL joins