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mysql_fetch_array()/mysql_fetch_assoc()/mysql_fetch_row()/mysql_num_rows etc... expects parameter 1 to be resource (31 answers)

Closed 11 years ago.

我无法让我的无名小卒 Mysql 声明工作, 它快把我逼疯了:

$ancestors = mysql_query( 
    SELECT * FROM comments e
        WHERE 
            ancestors = "  . $comment["id"] .  " AND 
            user_id != "  . $user->user_object["id"] .  " AND
                NOT EXISTS
                    (
                        SELECT  null 
                        FROM    notifications d
                        WHERE   d.target_id = e.id
                    )
 , $database->connection_handle);

有什么想法吗?

错误:

Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in /<>/<>/functions.php on line 785
Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in /<>/<>/functions.php on line 785
Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in /<>/<>/functions.php on line 785
Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in /<>/<>/functions.php on line 785
Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in /<>/<>/functions.php on line 785
Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in /<>/<>/functions.php on line 785
Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in /<>/<>/functions.php on line 785
Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in /<>/<>/functions.php on line 785
Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in /<>/<>/functions.php on line 785

第785行:

    while($reply = mysql_fetch_array($ancestors, MYSQL_ASSOC)){

如果我这样做:

$ancestors = mysql_query( SELECT * FROM   . $database->db_prefix .  comments 
    WHERE    
        ancestors = "  . $comment["id"] .  " AND 
        user_id != "  . $user->user_object["id"] .  " , 
        $database->connection_handle
);

它返回我预期的结果。

通知表是否包含一个条目

i>==================================================================================================================

string(46) "Table  whatever_co.comments  doesn t exist"

//SOLVED::: . $database->db_prefix . was missing from my table selectors.

Answer 1

我错过了我的表格前缀 : $database- & gt; db_ prefix. from comments and notice. 我错过了表格前缀 : $database- & gt; db_ prefix. from compublications and not understand. 。之所以出现这个问题,是因为我第一次没有使用..., 并且假设那部分是不正确的。感谢您一直以来的感谢。这是庆祝的原因。

Answer 2

您的 < code> NOT exists 是好的。我认为这与您的选择可能有问题。您在该表格中是否有名为“ null” 的字段? 如果不是, 您的问题所在。

我想你正在寻找的东西类似于

Select * 
FROM notifications d
WHERE d.target_id = e.id
AND e.id IS NULL

尝试使用类似的东西。

Answer 3

Where Exists 和 Where> NOTISTS 不看下层返回的行的内容,他们只检查是否有行返回:http://dev.mysql.com/doc/refman/5.0/en/exists-and-not-exists-subqueries.html>http://dev.mysql.com/doc/refman/5.0/en/existences-and-not-exist-subqueries.html/a>。

所以,尽管你选择了一个硬码的无效, 通常会“不存在”, MySQL并没有看那个无效, 它只看那列内有无效的行, 并且把它评为“嘿, 存在” 。

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