Question

考虑这两个数据库表：

  Users
UID Name
1   Shawn
2   Sean

  Photos
UID Description
1   Beauty
2   Ugliness

我想要检索一个这样的数组：

Array
(
    [0] => Array
        (
            [UID] => 1
            [Description] => Ugliness
        )
)

请注意，UID为1，但Description对应于Photos表的UID 2。事实上，我想要的是检索表用户的UID 1的信息，但是对于表照片的UID 2。所以从理论上讲，我会像这样编写我的SQL：

SELECT * FROM Users, Photos WHERE Users.UID = '1' AND Photos.UID = '2'

以下是我尝试获得我希望的内容：

$myArray_both = array();
$sql = "SELECT * FROM Users, Photos WHERE Users.UID = '1' AND Photos.UID = '2'";
$results = mysql_query($sql);
while($result = mysql_fetch_array($results, MYSQL_BOTH))    // MYSQL_BOTH is the default behaviour but I added it in for clarity
{
    array_push($myArray_both, $result);
}
echo("<h2>Using MYSQL_BOTH</h2>");
echo("<pre>");
print_r($myArray_both);
echo("</pre>");
/* This has the following output:
Array
(
    [0] => Array
        (
            [0] => 1
            [UID] => 2
            [1] => Ugliness
            [Description] => Ugliness
        )
)
**which is not good for me because I want to be able to access the UID with $result['UID'], but I want the UID returned to be 1, not 2.**
*/

$myArray_assoc = array();
$sql = "SELECT * FROM Users, Photos WHERE Users.UID = '1' AND Photos.UID = '2'";
$results = mysql_query($sql);
while($result = mysql_fetch_array($results, MYSQL_ASSOC))
{
    array_push($myArray_assoc, $result);
}
echo("<h2>Using MYSQL_ASSOC</h2>");
echo("<pre>");
print_r($myArray_assoc);
echo("</pre>");
/* This has the following output:
Array
(
    [0] => Array
        (
            [UID] => 2
            [Description] => Ugliness
        )
)
**This also doesn't work because I want to obtain UID 1, not 2**
*/

$myArray_num = array();
$sql = "SELECT * FROM Users, Photos WHERE Users.UID = '1' AND Photos.UID = '2'";
$results = mysql_query($sql);
while($result = mysql_fetch_array($results, MYSQL_NUM))
{
    array_push($myArray_num, $result);
}

echo("<pre>");
print_r($myArray_num);
echo("</pre>");
/* This has the following output:
Array
(
    [0] => Array
        (
            [0] => 1
            [1] => Ugliness
        )
)
**This gets the correct values, but I can't access them by name..**
*/
mysql_close($liendb);   // Closes the database connection
?>

原来如此！我该怎么办？

Answer 1

SELECT Users.UID AS userID, Users.Name, Photos.UID AS photoID, Photos.Description WHERE [blah blah blah]

HTH

如何获取table1.UID = 1和table2.UID = 2的值并按名称访问它们？

1 个答案: