我有一个页面,其中我的数据库中的表中的值正在被拉出并显示在下拉列表中。一旦选择了一个值并提交表单,除了下拉列表之外的每个数据都会提交给我的mysql数据库。代码如下:下拉列表值不被发送到数据库
<?
$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页面如下:
<?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());
}
?>
任何想法的家伙?感谢..
嗨感谢您的回复。它现在有效,只是最后一个问题....一旦我的表单被提交,它会提出“错误”一词,而不是确认它提交给数据库?它不应该这样做,因为数据正在成功提交。我所说的代码是..... if(!mysql_query($ sql,$ con)) { echo'
会议已发送给主席批准
'; die('Error:'。mysql_error()); } – user1114080 2012-01-01 22:16:37$ _REQUEST ['chairperson'];从来没有设置,你试图访问它,所以你看到了这个错误。现在,您已经声明了变量$ chairperson,并将其设置为null,它将在数据库中创建条目。此外,列主席未设置为NOT NULL,因此它已成功添加条目。 – Mo3z 2012-01-01 22:36:31
即时通讯对不起,你刚刚把我弄糊涂了!我该怎么改变...... – user1114080 2012-01-01 22:40:30