Question

我正在尝试使用下拉菜单在用户点击提交时保留其选定的值，但由于表单上的错误而失败。

我有一个while循环从数据库返回值来构建下拉列表的选项，但是如何在右侧选项上回显“selected”？

我已经尝试了if($district == $row["name"]) { echo "selected";}，如下所示，但它不起作用。

<?php

$result = mysql_query("SELECT dist.name FROM districts AS dist JOIN int_bd AS ibd ON dist.id = ibd.districts_id WHERE banners_id = 6 GROUP BY dist.id ORDER BY dist.id ASC", $connection);
if (!result) {
    die("Database query failed: " . mysql_error());
}

while ($row = mysql_fetch_array($result)) {
    echo '<option value="{$row["name"]}"'; if($district == $row["name"]) { echo "selected";} ; echo '>' . $row["name"] . "</option>";
}
?>

抱歉延误。所有建议的答案都不适合我。还有其他想法吗？

Answer 1

你能试试吗，

<?php

$result = mysql_query("SELECT dist.name FROM districts AS dist JOIN int_bd AS ibd ON dist.id = ibd.districts_id WHERE banners_id = 6 GROUP BY dist.id ORDER BY dist.id ASC", $connection);
if (!result) {
    die("Database query failed: " . mysql_error());
}

$district = $_REQUEST['name']; // You need pass the value you have been submitted

while ($row = mysql_fetch_array($result)) {
      $selected ="";
     if(trim($district) == trim($row["name"])) {  $selected = "selected";} 

    echo '<option value="{$row["name"]}" '.$selected.' >' . $row["name"] . "</option>";
}
?>

Answer 2

试试这个..

if($district == $row["name"])
{
echo "<option value='$district' selected>$district</option>";
}

Answer 3

我刚刚找到答案。这就是我所做的：

while ($row = mysql_fetch_array($result)) {
    echo '<option value="' . $row["name"] . '"';
    if($row["name"] == $district) { echo 'selected';} ;
    echo '>' . $row["name"] . '</option>';
}

似乎是这一行

echo '<option value="{$row["name"]}"';

导致了这个问题。

在while循环中查找以前选择的选项

3 个答案: