mysql查询返回第一列中所有其他列的值

Question

嗨，我对下表进行了查询，当我尝试获取每列中的值时，对于所有其他列，它从第一列返回相同的值。 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（例如name列）时，它都会返回21，这是owner_id列中与该pet_id对应的值。 我不确定为什么会这样。

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

有关我的环境的信息
PHP版本5.2.14
MYSQL版本5.0.67

Answer 1

我相信，如果您使用mysql_result，则在指定列之前，还必须指定行索引号（在您的情况下为0行？）。

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

Answer 2

引用http://php.net/manual/en/function.mysql-result.php mysql_result具有如下参数： mysql_result($result,$rownumber,$fieldname or $fieldnumber)

这应该工作：

$query = sprintf("SELECT * FROM petAttributes where pet_id ='%d'",$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，如果您占用多行，mysql_result的效率将非常低。 然后，您应该使用mysql_fetch_row，mysql_fetch_array或mysql_fetch_assoc

Answer 3

您也可以；

仅第一行

$query = sprintf("SELECT * FROM petAttributes where pet_id ='%d'",$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>";
}