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我一直在努力解決我的問題，因爲早上:)MYSQL INNER JOIN語法錯誤
我有兩個表。 table1 =發佈table2 =用戶。現在我需要從兩個表中提取數據。搜索如何使用INNER JOIN，找到一些代碼。我跑裏面phpMyAdmin的查詢，並在這裏工作得很好是我的SQL查詢
SELECT post.topic_id, post.category_id, post.topic_id, post.post_creator, post.post_date, post.post_content, users.username, users.gender, users.id FROM post INNER JOIN users ON post.post_creator=users.id WHERE post.post_creator=users.id and post.topic_id=19 ORDER BY post.post_date DESC
，但是當我用我的PHP這裏面的SQL查詢它給了我錯誤
下面你有一個錯誤的SQL語法;檢查對應於你的MySQL服務器版本使用附近的1號線
「INNER JOIN用戶ON post.post_creator = users.id ORDER BY post.post_date ASC」正確的語法手冊是我的PHP代碼
<?php
include_once './forum_Scripts/connect_to_MySql.php';
$cid = $_GET['cid'];
if(isset($_SESSION['uid'])){
$logged = " | <a href ='create_topic.php?cid=".$cid."'>Click here to create a new topic</a> ";
}else{
$logged = " | Please log in to create topics in this forum.";
}
$tid = $_GET['tid'];
$sql = "SELECT * FROM topics WHERE category_id='".$cid."' AND id='".$tid."' LIMIT 1";
$res = mysql_query($sql) or die(mysql_error());
if(mysql_num_rows($res) == 1){
echo "<table width='100%'> ";
if(isset($_SESSION['uid'])){
echo "<tr><td colspan='2'><input type='submit' value='Add Reply' onClick=\"window.location = 'post_reply.php?cid=".$cid."&tid=".$tid."'\" /> | <a href = 'http://amaforum.net63.net/'>Return to Forum Index</a><hr />";
}else{
echo "<tr><td colspan='2'><p>Please log in to add your reply</p><hr /></td>";}
while ($row = mysql_fetch_assoc($res)){
$sql2 = "SELECT post.topic_id, post.category_id, post.topic_id, post.post_creator, post.post_date, post.post_content, users.username, users.gender, users.id"
. "FROM post INNER JOIN users ON post.post_creator=users.id ORDER BY post.post_date ASC" ;
$res2 = mysql_query($sql2) or die(mysql_error());
while ($row2 = mysql_fetch_assoc($res2))
{
echo "<tr><td valign='top' style='border:2px solid #000000'><div style='min-height: 125px;'>".$row['topic_title']."<br />
by ".$row2['post_creator']." - ".$row2['post_date']."<hr />".$row2['post_content']."</div></td>"
. "<td width='200' valign='top' style='border: 1px solid #000000;'>"
. "<input id='".d_text."' type='".text."' name='".the_creator."' value='".$row2['post_creator']."' >"
. "echo '$user_info' "
. "</td></tr>"
. "<tr><td colspan='2'><hr /></td></tr> ";
}
$old_views = $row['topic_views'];
$new_views = $old_views + 1;
$sql3 = "UPDATE topics SET topic_views='".$new_views."' WHERE category_id='".$cid."' AND id='".$tid."' LIMIT 1 ";
$res3 = mysql_query($sql3) or die (mysql_error());
}
echo "</table>";
}else{
echo "<p>This topic does not exist</p>";
}
mysql_close();
?>
我可以讓它提前工作，我真的需要從你們的幫助..
感謝
這裏需要一個空格：'users.id''爲'users.id''特別是：'$ sql2 =「選擇post.topic_id，post.category_id，post.topic_id，post.post_creator，post.post_date， post.post_content，users.username，users.gender，users.id「 。 「FROM post INNER JOIN users on post.post_creator = users.id ORDER BY post.post_date ASC」;' – xQbert 2015-04-01 14:57:53
您在查詢中缺少空格：'users.idFROM' – 2015-04-01 14:57:59
''users.id'後面提供一些空格' – 2015-04-01 14:58:01