$sql="SELECT count(actid) AS tr
FROM useractions ua
WHERE qid=-1
OR qid IN (
SELECT qid
FROM questions q
WHERE q.visible=".VISIBLE."
)
AND ua.actid =".$actid;
上面的查询给出了这样的错误:SQL查询错误,不能弄明白
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 1
哪些错误的发言?
我做了转储和得到这个:
string "SELECT count(actid) AS tr FROM useractions ua WHERE qid=-1 OR qid IN ( SELECT qid FROM questions q WHERE q.visible=1 ) AND ua.actid =" (length=270)
$actid
是另一个查询,如下图所示的结果。然后将它传递给上面查询显示的函数。
foreach ($_POST['q'] as $qid) {
list($actid) = mysql_fetch_row(mysql_query("SELECT actid FROM useractions WHERE qid='$qid'"));
upd_facts_status($actid);
}
做一个var_dump($ sql);并张贴它。 –
我该如何解决它? – salmanhijazi
如果我是你,我会这样做:$ array = mysql_fetch_row(...);的var_dump($阵列); $ actid = $ array [0]; pd_facts_status($ ACTID); –