mysql_fetch_array MYSQL_NUM

Question

我收到以下警告：

warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given. 

当我尝试此代码：

$wyn_sport = mysql_query("SELECT a.title, a.content, a.id, FROM article a 
LEFT JOIN section s 
ON a.id`enter code here`_section=s.id 
WHERE s.name='Sport' 
ORDER BY date_art desc"); 
$sport = mysql_fetch_array($wyn_sport, MYSQL_NUM); 

$wyn_film = mysql_query("SELECT a.title, a.content, a.id FROM article a 
LEFT join section s on a.id_section=s.id 
WHERE s.name='Film' 
ORDER BY date_art desc"); 
$film = mysql_fetch_array($wyn_film, MYSQL_NUM); 

$wyn_nauka = mysql_query("SELECT a.title, a.content, a.id FROM article a 
LEFT join section s on a.id_section=s.id 
WHERE s.name='Nauka' 
ORDER BY date_art desc"); 
$nauka = mysql_fetch_array($wyn_nauka, MYSQL_NUM); 

Answer 1

这可能意味着你的查询失败，返回false。然后，您尝试对false的值执行mysql_fetch_array()。

Answer 2

@winterblood已经指出了这一点，但是这显然是问题...您的查询的一些随机enter_code_here它不属于

$wyn_sport = mysql_query("SELECT a.title, a.content, a.id, FROM article a 
    LEFT JOIN section s 
    ON a.id =s.id 
    WHERE s.name='Sport' 
    ORDER BY date_art desc"); 
    if(!$sport = mysql_fetch_array($wyn_sport, MYSQL_NUM)){ 
    echo mysql_error(); 

} 

上面应该工作