如果没有结果，PHP echo代码不返回'0'

Question

以下是查询：

$query = "SELECT name,deleted,COUNT(message) as message FROM guestbook_message WHERE name='".$name."' AND deleted=1 GROUP BY name";  
$result = mysql_query($query) or die(mysql_error());
$deletedtotal_row = mysql_fetch_array($result);

我在这里使用它：

echo "You have had ".($deletedtotal_row['deleted']) ? $deletedtotal_row['deleted'] : '0'." messages deleted";

此错误，不显示语法错误，但不显示任何结果。

但是当我使用时：

echo ".$totaldeleted_row['deleted'].";

工作正常。但是如果没有为该用户删除消息，则它不返回任何内容，但我希望它显示为“0”。

我出错的任何想法？

Answer 1

echo "You have had ".intval($deletedtotal_row['deleted'])." messages deleted";

Answer 2

改变这个：

echo "You have had ".($deletedtotal_row['deleted'] ? $deletedtotal_row['deleted'] : '0')." messages deleted";

此外，echo ".$totaldeleted_row['deleted'].";非常糟糕。这样做：

echo $totaldeleted_row['deleted'];

Answer 3

if(mysql_num_rows($result)==0) {
   echo 'no rows';
} else {
   echo "You have had " . $deletedtotal_row['deleted'] ." messages deleted";
}

Answer 4

谢谢大家。我得到了它：

echo "You have had ".($deletedtotal_row['deleted'] ? $deletedtotal_row['deleted'] : '0')." messages deleted";

Answer 5

使用：

sprintf("You have had %d messages deleted", $deletedtotal_row['deleted');