我有一個頁面,其中我的數據庫中的表中的值正在被拉出並顯示在下拉列表中。一旦選擇了一個值並提交表單,除了下拉列表之外的每個數據都會提交給我的mysql數據庫。代碼如下:下拉列表值不被髮送到數據庫
<?
$sql="SELECT user_id, firstname FROM Users WHERE role = 'chairperson'";
$result=mysql_query($sql);
$options="";
while ($row=mysql_fetch_array($result)) {
$id=$row["user_id"];
$thing=$row["firstname"];
$options.="<OPTION VALUE=\"$id\">".$thing;
}
?>
<form action="meetingsinserted.php" method="post">
...
<tr>
<td> <label for="chairperson">Chairperson:</label>
</td>
<td><span id="spryselect1">
<select name="thing" id="chairperson">
<OPTION VALUE=0>
<?=$options?>
</select>
<span class="selectRequiredMsg">You Must Choose A Chairperson For This Meeting</span></span></td>
</tr>
...
meetingsinserted.php頁面如下:
<?php
$title = $_REQUEST['title'];
$chairperson = $_REQUEST['chairperson'];
$secretary = $_REQUEST['secretary'];
$tof = $_REQUEST['tof'];
$occurances = $_REQUEST['occurances'];
$con = mysql_connect("*********","***","****");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db('mdb_hj942', $con);
$sql="INSERT INTO Meetings (title, chairperson, secretary, tof, occurances) VALUES ('$title','$chairperson', '$secretary','$tof','$occurances')";
if (!mysql_query($sql,$con))
{
echo '<h1>Meeting Has Been Sent To Chairperson For Approval</h1>';
die('Error: ' . mysql_error());
}
?>
任何想法的傢伙?感謝..
嗨感謝您的回覆。它現在有效,只是最後一個問題....一旦我的表單被提交,它會提出「錯誤」一詞,而不是確認它提交給數據庫?它不應該這樣做,因爲數據正在成功提交。我所說的代碼是..... if(!mysql_query($ sql,$ con)) { echo'
會議已發送給主席批准
'; die('Error:'。mysql_error()); } – user1114080 2012-01-01 22:16:37$ _REQUEST ['chairperson'];從來沒有設置,你試圖訪問它,所以你看到了這個錯誤。現在,您已經聲明瞭變量$ chairperson,並將其設置爲null,它將在數據庫中創建條目。此外,列主席未設置爲NOT NULL,因此它已成功添加條目。 – Mo3z 2012-01-01 22:36:31
即時通訊對不起,你剛剛把我弄糊塗了!我該怎麼改變...... – user1114080 2012-01-01 22:40:30