Question

我试图找出如何通过检查其值来选择下拉选项，但该值来自另一个查询。

我从名为$FK_TopicID的查询中获取$quickedit。下拉列表由名为$topresult的其他查询生成。我有一个IF / ELSE语句，当SELECTED等于<option value="the Topic ID" SELECTED>时，该语句应该在$FK_TopicID之类的选项内打印$row['TopicID']。

我只是不确定如何检查$FK_TopicID while循环中的$topresult。有什么想法吗？

    <?php

       $NewsID = $_GET["n"];
       $quickedit = mysql_query("SELECT * FROM News LEFT JOIN Topics on Topics.TopicID = News.FK_TopicID WHERE NewsID = $NewsID ORDER BY TopicName ASC, NewsTitle");
       $row = mysql_fetch_array($quickedit);

    echo "<p>" . $FK_TopicID . "</p>";

    /* additional php... */

    $topresult = mysql_query("SELECT * FROM Topics WHERE FK_UserID=$_SESSION[user_id] ORDER BY TopicSort, TopicName");

    while($row = mysql_fetch_array($topresult)) {
                if ( $row['TopicID'] == $FK_TopicID){ /* $FK_TopicID not printing value here */
                   $selected = " SELECTED";
                } else {
                   $selected = "";
                }
                echo '<option value=\"' . $row['TopicID'] . '" ' . $selected . '>' . $row['TopicName'] . '</option>';
        }

    ?>

Answer 1

我对数据库的结构一无所知，但我会试一试

这会输出你想要的东西吗？

<?php
$NewsID = $_GET['n'];
$quickedit = mysql_query("SELECT * FROM News LEFT JOIN Topics on Topics.TopicID = News.FK_TopicID WHERE NewsID = $NewsID ORDER BY TopicName ASC, NewsTitle");
$topresult = mysql_query("SELECT * FROM Topics WHERE FK_UserID=$_SESSION[user_id] ORDER BY TopicSort, TopicName");

echo "<p>" . $FK_TopicID . "</p>";

while($row = mysql_fetch_array($quickedit)) {
    while($row2 = mysql_fetch_array($topresult)) {
        if ( $row2['TopicID'] == $row['FK_TopicID']){
           $selected = " SELECTED";
        } else {
           $selected = "";
        }
        echo '<option value=\"' . $row2['TopicID'] . '" ' . $selected . '>' . $row['TopicName'] . '</option>';
    }
}

?>

此代码循环遍历两个查询，如果TopicID中的$topresult等于FK_TopicID中的$quickedit，则会选中该代码。

根据两个查询将选项元素设置为已选中

1 个答案: