php - 使用来自两个 mysql 查询的数据填充选择下拉列表

Question

我确信对于大多数了解 PHP 和 mysql 的人来说这很容易弄清楚，但我无法弄清楚。我会尽力以最好的方式解释

如果我错过了什么，请告诉我

相关表结构

Table | Fields
category | id,group_n (more fields exist but are not required now)
p_group | id,group_n

what I want to achieve:

Query to find currently selected group_n field in category table if available based on the current _isset id taken from the _get URL in relation to the category.

Example of work flow :

sample rows for category

ID | category             | short     | group_n
1  | Skin Whitening Pills | pills     | Skin Whitening
2  | Skin Whitening Cream | cream     | Skin Whitening
3  | Weight Gain / Loss   | wgainl    | Weight Gain / Loss

As you can see multiple categories can have a single group_n. I use the group_n to show an expert for that group.

1. select option is populated with the field group_n from category
2. select options are populated with all options from p_group (except selected so there is no duplicate"

category with id 2 has group_n as Skin Whitening edit_category.php?id=2 would display all input options for for id 2 that are currently in the table row in order to edit them and update the query. Now when the select options for the group_n are populated the selected should be Skin Whitening but all rows in p_group should be available to select.

code

<?php
$qry=mysql_query("SELECT group_n from category where id=$id", $con);
if(!$qry)
{
die("Query Failed: ". mysql_error());
}

?>

<select name="group" id="group">
    <?php
    // First, define the desired default value
    $default_value = $row['group_n'];
    while($row=mysql_fetch_array($qry))
    {
        // Then you can mark that one as selected in your "while" loop
        $selected = ($row['group_n'] == $default_value) ? ' selected' : '';

                $qry2=mysql_query("SELECT * from p_group", $con);

    while ($row2 =  mysql_fetch_array($qry2))
        {
                    echo "<option value='" . $row2['group_n'] . "'" . $selected . ">" . $row2['group_n'] . "</option>";
        }


    }
    ?>
</select>

我还尝试使用单个查询左连接两个表，但以下查询将仅列出一行，该行不会自动填充，但会给出正确的值。

$qry=mysql_query("
SELECT a.group_n,
b.group_n from p_group a 
left join category b
on a.group_n=b.group_n
group by b.group_n where b.id=$id", $con);

我不确定我的查询或 php 是否需要更改，但我一次只能让其中一个查询与选择一起使用。我需要知道如何使两个查询都正常工作，因此选择的选项是类别表中的一个和 p_group 表中的下拉列表。

理想情况下，我希望类别表与 p_group 中的 id 相关联，但现在我愿意将其保留为文本。

Answer 1

在等待本周异常安静的用户社区的输入后，我四处搜索并摆弄代码，直到找到 UNION sql 连接。这基本上需要两组数据并呈现为一组，从而达到了目的。不确定这是否是最好的方法，也许有一个纯 PHP 的方法，但这确实成功了。

SQL 片段

   $qry=mysql_query("select group_n as group_a from category where id=$id union select group_n from p_group group by group_n", $con);

基本上，这会选择左侧的查询并将其与右侧的查询结合起来并显示所有记录。

限制 1.选择的字段数必须匹配 2. UNION不会选择重复项 3. UNION ALL 将选择重复项

完整的相关代码:

   $qry=mysql_query("select group_n as group_a from category where id=$id union select group_n from p_group group by group_n", $con);
if(!$qry)
{
die("Query Failed: ". mysql_error());
}
    // First, define the desired default value
    $default_value = $row['group_a'];
    while($row=mysql_fetch_array($qry))
    {

$selected = ($row['group_a'] == $default_value) ? ' selected' : '';

        // Then you can mark that one as selected in your "while" loop
     echo "<option value='" . $row['group_a'] . "'" . $selected . ">" . $row['group_a'] . "</option>";


 }   
?>
</select>