Php 选择标识符并显示其中的行/列
可能重复:Php 选择标识符并显示其中的行/列,php,iphone,mysql,sql,Php,Iphone,Mysql,Sql,可能重复: 我的问题是: $sql = "SELECT * FROM $table WHERE "; switch($type) { case "global": $sql .= "1 "; break; case "device": $sql .= "udid = '$udid' "; break; case "name":
我的问题是:
$sql = "SELECT * FROM $table WHERE ";
switch($type) {
case "global":
$sql .= "1 ";
break;
case "device":
$sql .= "udid = '$udid' ";
break;
case "name":
$sql .= "name = '$name' ";
break;
}
$sql .= "ORDER BY $sort ";
$sql .= "LIMIT $offset,$count ";
$result = mysql_query($sql,$conn);
udid是唯一的标识符。循环:
while ($row = mysql_fetch_object($result)) {
echo '<tr>
<td>
'.$rank.'
</td>
<td>
'.$row->name.'
</td>
<td>
'.$row->score.'
</td>
<td>
'.$row->udid.'
</td>
</tr>';
$rank++;
}
这很好,但是如果我想显示基于udid的分数和排名,那么如果我想要udid:119d62e398bf6f1cea89b235b5c8ecf95255442e,我想要排名2,名字john和分数2.41940
我该怎么做
Joe获取使用udid查询的行的分数,并使用以下查询获得排名
$sql = "SELECT Count(*) FROM $table WHERE score < '$score'";
$sql=“从得分<'$score''的$table中选择Count(*);
排名将是从上述查询返回的值+1如果运行相同的查询,但将
限制$offset,$count
更改为限制“($offset+$rank)。”,1
?或实际上($offset+$rank-1)
$sql = "SELECT Count(*) FROM $table WHERE score < '$score'";