这句 有什么问题么？不知道错在哪

PHP code

<?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."%" ;?>

PHP code

/*$sql1=$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($result 1);$con2=mysql_fetch_array($result 2);echo $con1["con"]/$con2["con"];*///太佩服你了，上面这段代码每一行都有错误！！$sql1 = 'SELECT `a1` FROM `tab` WHERE 1';$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'];