您好,我想加入3个表,并从他们那里得到的所有信息表名:最爱,图片,帖子MySQL查询错误3表连接
最爱表存储:用户ID,帖子ID(此表是最喜欢的广告)
图像表店:帖子ID,路径(1-5)
职位表用于存储后的所有信息:ID,用户ID,职称等
所以我现在做的SQL查询巫应选择一切fav和joing帖子和图片在$_session["userid"] = to userid
这怎么它看起来像:
<?php
$userid = $_SESSION["userid"];
$sql = "SELECT * FROM fav f JOIN posts p ON f.postid = p.id JOIN images i ON p.id = i.postid WHERE userid='$userid' ";
$res = mysqli_query($connect,$sql);
while ($row = mysqli_fetch_assoc($res)) {
$postid = $row["postid"];
?>
<div id="ads">
<div id="titlepic">
<a href="review.php?id=<?php echo $postid; ?>"><?php echo $row["title"]; ?></a><br>
<a href="review.php?id=<?php echo $postid; ?>"><img src="<?php if(!empty($row["path1"])) { echo $row["path1"]; } else echo "image/noimage.png"; ?>" height="100px" width="150px"></a>
</div>
<div id="dates">
<b>Date Added:</b> <?php echo date('m/d/Y H:i', $row["dateadded"]); ?><br>
<b>Renew Date:</b> <?php if($row["renewdate"] > 0){ echo date('m/d/Y H:i', $row["renewdate"]); } ?><br>
<b>Location:</b> <?php echo $row["location"]; ?><br>
<b>Price:</b> <?php echo $row["price"]."£"; ?><br>
</div>
</div>
<hr width="100%">
<?php
}
?>
,但我得到的错误后,我把这样"Warning: mysqli_fetch_assoc() expects parameter 1 to be mysqli_result, boolean given in C:\wamp\www\project\fav.php on line 129"
哪里出了问题我不明白,因为如果我不这样做WHERE部分WHERE userid='$userid'
错误的声音一切正常,没有警告和错误
您的查询失败,然后您需要找出是什么。请参阅以下链接http://php.net/manual/en/mysqli.error.php和http://php.net/manual/en/function.error-reporting.php 并将其应用于您的代码。 –