SQL日期选取

我的数据库里有个表是日期的
news_last_modified
2013-04-12 13:04:18

是这个样式的。

如何只选择日期而不选择时间呢？
前台的代码是这样的

require_once(DIR_WS_CLASSES . 'news.php');

$news_list_by_author_sql='select n.article_id,n.news_last_modified, nt.news_article_name from ' . TABLE_NEWS_ARTICLES . ' n left join ' . TABLE_NEWS_ARTICLES_TEXT . ' nt on (n.article_id = nt.article_id and nt.language_id = \'' . (int)$_SESSION['languages_id'] . '\') and n.news_status = \'1\'' . ' order by n.article_id DESC limit 6 ';
$news = $db->Execute($news_list_by_author_sql);

if ($news->RecordCount() > 0) {
$i = 0;
$article_array = array();
while (!$news->EOF) {
$article_array[$i] = array(
'articleId' => $news->fields['article_id'],
'articledate' => $news->fields['news_last_modified'],
'articleName' => stripslashes($news->fields['news_article_name']),
);
$i++;
$news->MoveNext();
} //end while
} else {
$articlesNotFound = true;
}

?>
-------------------------------------
'articledate' => $news->fields['news_last_modified']

麻烦大家帮个忙。我是个新手。正在学习中

回复讨论(解决方案)

date()

	$str="2013-04-12 13:04:18";	$ar=explode(" ",$str);	print_r($ar);//Array ( [0] => 2013-04-12 [1] => 13:04:18 )		$t=explode(" ",$news->fields['news_last_modified']);	'articledate' => $t[0],

$news_list_by_author_sql='select n.article_id, date_format(n.news_last_modified,"%Y-%m-%d") as news_last_modified , nt.news_article_name from ' . TABLE_NEWS_ARTICLES . ' n left join ' . TABLE_NEWS_ARTICLES_TEXT . ' nt on (n.article_id = nt.article_id and nt.language_id = \'' . (int)$_SESSION['languages_id'] . '\') and n.news_status = \'1\'' . ' order by n.article_id DESC limit 6 ';

或者在php里面转也可以

'articledate' => date("Y-m-d",strtotime($news->fields['news_last_modified'])),

$news_list_by_author_sql='select n.article_id,date_format(n.news_last_modified,"%Y-%m-%d") as news_last_modified , nt.news_article_name from ' . TABLE_NEWS_ARTICLES . ' n left join ' . T……


感谢.现在明白了大概要怎么弄了.调试了一下也出来相应的效果.满分