返回并比较最高计数值php mysql

Question

是的，我会在这里尽可能具体-首先概述我需要的东西，然后再完成我的事情-我通常可以can足这些，因此请记住，我只是在问我的大脑被炸了！

1. 我的数据库中有一个表格-我们只需要知道它具有以下字段（性别-男性或女性，布局-0或1）

2. 我想找到男性最常用的布局（0或1）-通过以下方式做到这一点：

   $result = mysqli_query($conn, "SELECT layout,COUNT(*) as num FROM style where gender = 'male' group by layout order by num DESC LIMIT 1" );

3. 我想检查返回的结果（最频繁）是否为0或1，所以我可以在IF语句中使用它（到目前为止，我只是使用echo进行测试）

很抱歉，如果这很琐碎，或者我错过了任何内容，如果您需要任何其他信息，请告诉我。

Answer 1

$resultResource = mysqli_query($conn, 'SELECT layout,COUNT(*) as num FROM style where gender = "male"');

while ($row = mysqli_fetch_assoc($resultResource)){ 
  $maleCount = $row['num'];
}

$resultResource = mysqli_query($conn, 'SELECT layout,COUNT(*) as num FROM style where gender = "female"');

while ($row = mysqli_fetch_assoc($resultResource)){ 
  $femaleCount = $row['num'];
}

if ($femaleCount > $maleCount){
    //more females in database
}elseif($femaleCount < $maleCount){
    //more males in database
}else{
    //same amount of both;
}

Answer 2

PHP：

$DTB->new mysqli($Mysql_Server,$Mysql_User,$Mysql_Password,$Mysql_Database);
$Layout0=($DTB->query("SELECT layout FROM style WHERE gender ='male' AND layout=0 "))->num_rows;
$Layout1=($DTB->query("SELECT layout FROM style WHERE gender ='male' AND layout=1 "))->num_rows;
IF ($Layout0>$Layout1){
  ...
  }

Answer 3

您的代码可以很好地确定男性是喜欢布局0还是1，您只需要查看输出值即可：

$result = mysqli_query($conn, "SELECT layout, COUNT(*) as num 
                               FROM style
                               WHERE gender = 'male' 
                               GROUP BY layout 
                               ORDER BY num DESC 
                               LIMIT 1" );
$row = mysqli_fetch_assoc($result);
if ($row['layout'] == 0)
    echo "males prefer layout 0";
else
    echo "males prefer layout 1";