我试图弄清楚如何通过检查它的值来标记下拉选项，但该值来自另一个查询。
我从名为$quickedit的查询中获得$FK_TopicID。下拉列表是由另一个名为$topresult的查询生成的。我有一个IF/ELSE语句，当$FK_TopicID等于$row['TopicID']时，应该在<option value="the Topic ID" SELECTED>选项中打印SELECTED。
我只是不知道如何检查while循环中的$FK_TopicID为$topresult。什么好主意吗?
<?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>';
}
?>
我对数据库的结构一无所知，但我将尝试一下
这个输出是你想要的吗?
<?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 . '>' . $row['TopicName'] . '</option>';
}
}
?>
这段代码循环遍历两个查询，如果$topresult中的TopicID等于$quickedit中的FK_TopicID，它将被选中。