有feeds和likes表,首先我从feeds表中查询数据,然后尝试为抓取的feeds获取likes表的likes。TEXT NOT NULL,点赞表格:like_useridlike_postid// IN((like_postid,like_type),(like_postid,like_type),(like_posti
此查询应给出已取消(post_appt_status_id = 3)且尚未重新预订的约会。appt_date appointments appts2 WHEREappointments.post_appt_status_id = '3'
AND contacts.id NOT IN (SELECT contact_id FROM appointments WHERE
在下面的代码中,第四行创建了一个错误"AND 'user‘IN(SELECT 'user_s’FROM‘SELECT’WHERE sex='$iaminterestedin')“。NOT IN query可以很好地工作,但是有像WHERE ...IN query这样的东西吗?$result = queryMysql("SELECT `user` FROM `members`
WHERE `user` NOT IN(SELECT `ilike`FROM
我有一个sql查询的问题。该查询一直有效,直到我添加了第三行:AND procurements.activity1 IN("'.$arr.'")'。我猜'和"没有正确设置。organisations.city, FROM countries, procurements, organisations
WHERE
我正在使用MySQL/XAMPP供参考。("SELECT * FROM user_info WHERE FNAME = $FNAME" ); print "<BR>There is no such user with the user number of $FNAME($result, MYSQL_ASSOC);
?mysql_num_rows(
可能重复:
mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in /home($_POST['edit2']) . "' WHERE id='" . mysql_real_escape_string($_GET['id']) . "'");
echo 'Met su
我有一些问题与分页的数据,我从一个MySQL数据库在PHP。我真的卡住了。这是我第一次尝试分页。$data = mysql_query("SELECT * FROM `articles` WHERE `content` = '' AND `requestedby` != '$id'") or die(mysql_error());
$rows =