Question

PHP数据库调用出错怎么办？

function Rand_news() {
$mysqli = new mysqli ( "localhost", "root", "", "nimei", 3306 );
$query_news = "select id,title,content from content where type=0 order by id desc LIMIT 1";
$result_news = $mysqli->query ( $query_news );
list ( $id,$title,$content ) = $result_news->fetch_row ();
$news_content[0][0]=trim($title);
$news_content[0][1] = trim(str_replace("
","
",$content));
$query_news = "update nimei.content set type=1 where id='$id'";
$result_news = $mysqli->query ( $query_news );
return $news_content;
}

调用nimei数据库中表content的id，title,content

Answer 1

遇见这种情况：“Fatal error: Call to a member function fetch_row() on a non-object”

然后显示出：“严重错误：调用fetch_row()的不是一个对象”。

就要立即开始检查 ：　　　

1. 首先，来执行代码$result_news = $mysqli->query ( $query_news );

2. 第二步开始检查结果是否正确，也就是说，看　$result_news　是否为同一个对象，写下代码： print_r($result_news)来查看;

3. 正确结果为： mysqli_result Object ( [current_field] => 0 [field_count] => 8 [lengths] => [num_rows] => 1 [type] => 0 )