この文に関して何か問題はありますか？何が問題なのか分かりません

この文に何か間違いはありますか？何が問題なのかわかりません

PHP コード

<!--

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--><?php
$con = mysql_connect("localhost","root","");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("test", $con);
$sql1=$result = SELECT `a1` FROM `tab` WHERE 1;
$sql2=$result = SELECT `a3` FROM `tab` WHERE 1;
$result 1=mysql_query($sql1," $con");
$result 2=mysql_query($sql2," $con");
$con1=mysql_fetch_array($result 1);
$con2=mysql_fetch_array($result 2);
echo $con1["con"]/$con2["con"];
//你要用百分比的可心像下面一样,把注释去掉就行了
//$m= $con1["con"]/$con2["con"]*100;
//echo $m."%" ;
?>

-----解決策-------------- ------

PHP コード

/*
$sql1=$result = SELECT `a1` FROM `tab` WHERE 1;
$sql2=$result = SELECT `a3` FROM `tab` WHERE 1;
$result1=mysql_query($sql1," $con");
$result2=mysql_query($sql2," $con");
$con1=mysql_fetch_array($結果 1);
$con2=mysql_fetch_array($result 2);
echo $con1["con"]/$con2["con"];
*/
//本当に尊敬します。上記のコードのすべての行にエラーがあります。 ！

$sql1 = 'SELECT `a1` FROM `tab` WHERE 1';
$sql2 = 'SELECT `a3` FROM `tab` WHERE 1';
$result1=mysql_query($sql1,$con);
$result2=mysql_query($sql2,$con);
$con1=mysql_fetch_array($result1);
$con2=mysql_fetch_array($result2);
echo $con1['a1'] / $con2['a3'];