設置PHP與MySQL

Question

多變量，這樣我有我想要使用

mysql_select_db("website", $con); 
    $result=mysql_query("SELECT * FROM characters where online='1'"); 
    while ($row= 
       mysql_fetch_array($result)) 
       { 
       echo $row['name']; 
       } 
       if ($row['race'] == "1"); 
       { 
       echo '<img src="img/8-0.gif" />'; 
       } 
       if ($row['class'] == "3"); 
       { 
       echo '<img src="img/3.gif" ?>'; 
       } 
      mysql_close($con); 
     ?> 

現在我只希望這兩個圖像顯示，如果網上的場爲一個代碼，但他們表現出不管是什麼。有誰知道我該如何解決這個問題？謝謝。

Answer 1

  if ($row['race'] == "1" AND $row['online'] == 1); 
      { 
      echo '<img src="img/8-0.gif" />'; 
      } 
      if ($row['class'] == "3" AND $row['online'] == 1); 
      { 
      echo '<img src="img/3.gif" ?>'; 
      } 

Answer 2

對於參數的緣故：

mysql_select_db("website", $con); 
$result=mysql_query("SELECT * FROM characters where online='1'"); 
while ($row= mysql_fetch_array($result)) 
{ 
    echo $row['name']; 
//} This bracket would immediately close your query processing and only display the last images. Or is that the desired behaviour? 
    if($row['isOnline'] == '1') { //Makes sure, that 'isOnline' is set before displaying. 
      if ($row['race'] == "1"); 
      { 
      echo '<img src="img/8-0.gif" />'; 
      } 
      if ($row['class'] == "3"); 
      { 
      echo '<img src="img/3.gif" ?>'; 
      } 
    } 
} //This bracket closes the actual query result handling 
mysql_close($con); 
?>