mysql查询返回所有其他列的第一列中的值

Question

您好，我对下面的一个表进行了查询，当我尝试获取每一列中的值时，它从第一列中为所有其他列返回相同的值。要详细说明

在我的数据库表中，我有以下内容：

owner_id = 21
pet_id = 1
name = fluffy
color = green
type = dog
sub_type = boxer
location = LA

但是，每当我尝试访问一列时，它返回21，这是对应于该pet_id的owner_id列中的值。我不确定为什么会发生这种情况。

$query = sprintf("SELECT * FROM `petAttributes` where pet_id ='%d'",$p_id);
 $result = performQuery($query);

 $owner_id = stripslashes(mysql_result($result,"owner_id"));
 $pet_id = stripslashes(mysql_result($result,"pet_id"));
 $name = stripslashes(mysql_result($result,"name"));
 $color = stripslashes(mysql_result($result,"color"));
 $type = stripslashes(mysql_result($result,"type"));
 $sub_type = stripslashes(mysql_result($result,"sub_type"));
 $loc = stripslashes(mysql_result($result,"location"));

我的环境信息

PHP版本5.2.14

MYSQL版本5.0.67

Answer 1

我相信，如果使用mysql_result，在指定列之前，还必须指定行索引号(在本例中是第0行？)。

$name = stripslashes(mysql_result($result, 0, "name"));

Answer 2

参考http://php.net/manual/en/function.mysql-result.php mysql_result，它的参数如下：mysql_result($result,$rownumber,$fieldname or $fieldnumber)

这应该是可行的：

$query = sprintf("SELECT * FROM petAttributes where pet_id ='%d'",$p_id);
$result = performQuery($query);

$owner_id = stripslashes(mysql_result($result,0,"owner_id"));
$pet_id = stripslashes(mysql_result($result,0,"pet_id"));
$name = stripslashes(mysql_result($result,0,"name"));
$color = stripslashes(mysql_result($result,0,"color"));
$type = stripslashes(mysql_result($result,0,"type"));
$sub_type = stripslashes(mysql_result($result,0,"sub_type"));
$loc = stripslashes(mysql_result($result,0,"location"));

顺便说一句，如果你取多个行，mysql_result的效率会变得非常低。然后，您应该使用mysql_fetch_row、mysql_fetch_array或mysql_fetch_assoc

Answer 3

你也可以；

仅适用于第一行

$query = sprintf("SELECT * FROM petAttributes where pet_id ='%d'",$p_id);
$result = performQuery($query);
$row = mysql_fetch_array($result);

extract($row);

或所有返回的行；

$query = sprintf("SELECT * FROM petAttributes where pet_id ='%d'",$p_id);
$result = performQuery($query);
while($row = mysql_fetch_array($result))
{
  foreach ($row as $value)  echo $value."<br>";
}