php - Ajax 聊天 - 登录问题

Question

在我的Ajax 聊天脚本中，我收到了以下警告：

Warning: mysql_fetch_array() expects parameter 1 to be resource,
boolean given in /home/mychat/public_html/c/listChats.php on line 7

第 7 行：

while($row = mysql_fetch_array($sql)){

listChats.php：

<?php 
$chats = "";

include_once("scripts/connect_db.php");

$sql = mysql_query("SELECT * FROM (
    SELECT * FROM chatBox order by id DESC LIMIT 20
) TMP ORDER BY tmp.id ASC");

while($row = mysql_fetch_array($sql)){
    $chat = $row['chatBody'];
    $username = $row['username'];
    $chats = '<p><span id="un">' . $username . ':</span> ' . $chat . '</p>';
    echo "$chats";
}
?>

Answer 1

您的查询语法错误

尝试这个

$sql = mysql_query("SELECT *,(

    SELECT * FROM chatBox order by id DESC LIMIT 20

) as data FROM TMP ORDER BY tmp.id ASC");

更新的查询

select *,(select username from chatBox where id=".$_SESSION['userid'].") from chatBody where user_id=".$_SESSION['userid']

Answer 2

我认为您使用了错误的表名。

尝试这个

$sql = mysql_query("SELECT *,(

    SELECT * FROM chatBox order by id DESC LIMIT 20

) as data FROM TMP ORDER BY TMP.id ASC");

如果你的桌子是TMP那么使用TMP.id如果它是tmp那么使用tmp.id