我已设置此查询,我从我的数据库中筛选出一些结果。 它实际上显示了应该的方式 - 但我仍然得到这个错误:
“警告:mysql_numrows()期望参数1是资源,字符串在......”
中给出我的问题是什么? ; - )
$dblink = mysqli_connect("localhost", "*", "*", "*") or die ("Fejl: Kan ikke etablere forbindelse til databasen..");
$query= "SELECT * FROM udgivelse";
$count= mysql_numrows($query);
$query2= "SELECT * FROM (SELECT release_id, release_pic, release_artist, release_title, release_info, release_tracks, release_catalognr, beatportlink, ituneslink, traxsourcelink, alternativelink FROM udgivelse ORDER BY release_id ASC LIMIT "+$count-3+";) ORDER BY release_id DESC;";
$oldrelease = mysqli_query($dblink, $query) or die( "Forespørgslen kunne ikke udføres: " . mysqli_error($dblink));
提前致谢!
答案 0 :(得分:1)
$query = "SELECT * FROM udgivelse";
$count= mysql_numrows(mysql_query($query));
问题出在QUERY 2
上,子查询应该有ALIAS
$query2= " SELECT *
FROM
(SELECT release_id,
release_pic,
release_artist,
release_title,
release_info,
release_tracks,
release_catalognr,
beatportlink,
ituneslink,
traxsourcelink,
alternativelink
FROM udgivelse
ORDER BY release_id ASC
LIMIT " . $count-3 .") xx
ORDER BY release_id DESC; ";
答案 1 :(得分:1)
$query= "SELECT * FROM udgivelse"; need to execute using mysql_query.
//use this
$query= mysql_query("SELECT * FROM udgivelse");
$numrows = mysql_num_rows($query);
答案 2 :(得分:0)
与警告消息一样,mysql_num_rows不希望查询字符串作为参数,而是资源句柄(mysql_query调用的返回值)。
请参阅下面的代码(来自php.net):
$link = mysql_connect("localhost", "mysql_user", "mysql_password");
mysql_select_db("database", $link);
$result = mysql_query("SELECT * FROM table1", $link);
$num_rows = mysql_num_rows($result);
echo "$num_rows Rows\n";