返回并比较最高计数php mysql的值

Right, I'll be specific as possible here - I'll first outline what I need, then what I've done - I can usually plod on with these so bear in mind I'm only asking as my brain is fried!

1. I have a table in my database - we only need to know it has these fields (gender - male or female, layout - 0 or 1)

2. I want to find the most used layout (0 or 1) for males - I've done this by:

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

3. I want to check if the returned result (so most frequent) is 0 or 1, so i can use that in an IF statement (so far i'm just using an echo to test)

I'm sorry if this if very trivial, or if i've missed anything out - if you need any extra info let me know.

Your code is fine for determining whether males prefer layout 0 or 1, you just need to look at the output value:

$result = mysqli_query($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";

$resultResource = mysqli_query($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;
}

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){
  ...
  }