select * from member where data!='' order by time desc Limit 0,10上面得出前十行倒序排列的数据,如何才能进一步比对这十行数据包含关健词的条数?即:data like '%$keyword%' 得出上面10条中有几条包含该关健词?
select * from member where data!='' order by time desc Limit 0,10上面得出前十行倒序排列的数据,如何才能进一步比对这十行数据包含关健词的条数?即:data like '%$keyword%' 得出上面10条中有几条包含该关健词?
from (
select * from member where data!='' order by time desc Limit 0,10
)T
where data like '%$keyword%'
(select * from member where data!=''order by time desc limit 0,10
)a
where data like'%$keyword%'
测试了一下,出错:Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in麻烦再帮看看
T换成a也同样出错:Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in麻烦再帮看看!谢
一开始我也是这样想的,但这样只能得到data=$keyword的数值,比如有一千条数据,不管如果它都会搜索出包含$keyword的总条数的前10条。 而我要先得到倒序10条,再比较这10条中有几条包含$keyword
$query="select * from
(select * from member where data!=''order by time desc limit 0,10
)a
where data like'%$keyword%'
";
$result=$obj_db->simplequery($query);
while ($arr=$obj_db->fetchrow($result,DB_FETCHMODE_ASSOC))
{
$list.="$arr[data]<br>";
}MYSQL本地安装调试的,有点老:
MySQL 4.0.24-nt
select * from member where data!='' order by time desc Limit 0,10输出10条了,再
foreach ($list as $v){
比对出与$keyword相同的数据 得出 $count
}if ($count>0){
echo "<script type='text/javascript'> alert('最近发布的前10条已包含 $count 条与你提交的 $keyword 相同的内容!');history.go(-1);</script>";
exit();
}只有思路,哪位高手能把代码写出?或者有更简便的方式?
与这个无关,整个都改成 `mydata` `mytime`一样也出错:
select * from
(select * from member where mydata!=''order by mytime desc limit 0,10
)a
where mydata like'%$keyword%'
$count=0;
foreach ($list as $v){
if($v==$keyword) $count++;
}if ($count>0){
echo "<script type='text/javascript'> alert('最近发布的前10条已包含 $count 条与你提交的 $keyword 相同的内容!');history.go(-1);</script>";
exit();
}
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in
我要比对:%$keyword%
比如:一二三四五六,二三四五六七,五六七八九,关键词$keyword是“四五”,那前面就有二条包含关键词if($v==$keyword)能实现吗?
$result=$obj_db->simplequery($query) or die (mysql_error()); //这样提示什么错误foreach ($list as $v){
if(strpos($v,$keyword) !== false) $count++;
}这样查找。
晕,又出这个错了,帮看看:
Warning: Invalid argument supplied for foreach() in e:\appserv\www\index.php on line 133代码:
$query="select * from member where data!='' order by time desc Limit 0,10";
$result=$obj_db->simplequery($query);
while ($arr=$obj_db->fetchrow($result,DB_FETCHMODE_ASSOC))
{
$list.="$arr[data]";
}$count=0;
foreach ($list as $v){
if(strpos($v,$keyword) !== false) $count++;
}if ($count>0){
echo "<script type='text/javascript'> alert('最近发布的前10条已包含 $count 条与你提交的 $keyword 相同的内容!');history.go(-1);</script>";
exit();
}
$list[]=$arr[data];