调用多个查询作为一个查询

Question

我对数据库中的一个表有两个查询，唯一的区别是限制。我喜欢输出，但我知道一定有更实际的方法来做这件事。有没有人能提个建议？

$sql1 = mysql_query("SELECT * FROM products ORDER BY id DESC LIMIT 20");
$count = mysql_num_rows($sql1);
if ($count > 0) {
    while($row = mysql_fetch_array($sql1)){ 
}
}

$sql2 = mysql_query("SELECT * FROM products ORDER BY id DESC LIMIT 1");
$count = mysql_num_rows($sql2);
if ($count > 0) {
    while($row = mysql_fetch_array($sql2)){ 
}
}

Answer 1

创建一个函数并将限制作为参数传递：

function doQuery($limit = 1) {
    $sql = mysql_query("SELECT * FROM products ORDER BY id DESC LIMIT $limit");
    $count = mysql_num_rows($sql);
    if ($count > 0) {
        while($row = mysql_fetch_array($sql)){ 

        }
    }
};

doQuery(1);
doQuery(20);

Answer 2

$sql1 = mysql_query("SELECT * FROM products ORDER BY id DESC LIMIT 20");
$count = mysql_num_rows($sql1);
if ($count > 0) {
    while($row = mysql_fetch_array($sql1)){
        //loop over all records...
    }
}

$count = mysql_num_rows($sql1);
if ($count > 0) {
    if($row = mysql_fetch_array($sql1)){ 
        //retrieve one record
    }
}