如何獲取此代碼以顯示哪個選項已被選中?如何顯示PHP中已經選擇了哪些表單元素
這基本上是一個編輯頁面,它是拉動從數據庫信息和填充相關領域
我有一個下拉菜單,多選框,並在頁面上的單選按鈕的一些元素一起。信息正在顯示在元素中,但我無法弄清楚如何讓s和單選按鈕顯示選中的內容,如果它們與數據庫中的信息匹配的話。
代碼:向剛選擇選定= 「選擇」
<select name="client">
<option value="empty">Change Client...</option>
<?php
$result2 = mysql_query("SELECT name FROM clients") or die("Database query failed: " . mysql_error());
while($row = mysql_fetch_assoc($result2)) {
$clientlist = $row['name'];
$clientname = htmlspecialchars($row['name']);
if ($_POST['client'] == $clientlist)
{
echo '<option value="' . $clientlist . '" selected="selected" >' . $clientname . '</option>' . '\n';
}
else{
echo '<option value="' . $clientlist . '" >' . $clientname . '</option>' . '\n';
}
}
?>
</select>
</p>
<p class="subheadsmall">Core Classification</p>
<?php
switch ($niche) {
case "brand":
echo '<input type="radio" name="niche" value="Brand" checked="checked" />Brand';
echo '<input type="radio" name="niche" value="Marketing" />Marketing';
echo '<input type="radio" name="niche" value="Communication" />Communication';
break;
case "marketing":
echo '<input type="radio" name="niche" value="Brand" />Brand';
echo '<input type="radio" name="niche" value="Marketing" checked="checked" />Marketing';
echo '<input type="radio" name="niche" value="Communication" />Communication';
break;
case "communication":
echo '<input type="radio" name="niche" value="Brand" />Brand';
echo '<input type="radio" name="niche" value="Marketing" />Marketing';
echo '<input type="radio" name="niche" value="Communication" checked="checked" />Communication';
break;
default;
echo '<input type="radio" name="niche" value="Brand" />Brand';
echo '<input type="radio" name="niche" value="Marketing" />Marketing';
echo '<input type="radio" name="niche" value="Communication" />Communication';
break;
}
?>
<p class="subheadsmall">Strategies</p>
<p class="sidebargrey">
<?php
$result = mysql_query("SELECT strategies FROM studies WHERE id = '$id';
if (!$result) {
die("Database query failed: " . mysql_error());
}
while($row = mysql_fetch_array($result)) {
$strategyname = $row['strategies'];
echo $strategyname.'<br />';
}
?>
<p class="subheadsmall">Add a strategy... (hold down command key to select more than one)</p>
<select name="strategies[]" multiple="multiple">
<?php
$result = mysql_query("SELECT * FROM strategies");
if (!$result) {
die("Database query failed: " . mysql_error());
}
while($row = mysql_fetch_array($result)) {
$strategylist = $row['name'];
$strategyname = htmlspecialchars($row['name']);
$pagelink = str_replace(" ","_",$strategylist);
echo '<option value="<a href="strategies.php?strategy=' . $pagelink . '">'.$strategyname.'</a>" >' . $strategyname . '</option>' . '\n';
}
?>
</p>
萬一OP不知道,這個JS需要jQuery的。 – 2009-07-01 15:07:07