【发布时间】：2015-04-06 12:38:40
【问题描述】：
我的网站中有一个表格，我的网站中有一个下拉列表，根据从下拉列表中选择的值，它将更改表格的输出。 我进行了搜索，找到了this on stackoverflow，我尝试做类似的事情，但它不起作用。这是我想要显示表格的名为公告的 php 文件。
<?php
include_once 'header.php';
$connection = mysql_connect("localhost", "root", "smogi")?>
<html>
<head>
<script type="text/javascript" src="javascript.js"></script>
</head>
<body>
<?php
$result = queryMysql("SELECT * FROM doctor WHERE username='$username'");
if (mysql_num_rows($result)):
?>
<!-- Koumpi pou se metaferei sti selida gia tin dimiourgia neas anakoinwseis -->
<form action="new_announcement.php">
<input type="submit" value="Create New Announcement">
</form>
<br />
Select Category :<select id="SelectDisease" name="category">
<option value="*">All</option>
<!--emfanizei tis epiloges gia ta specialties me basi auta p exoume sti basi mas -->
<?php
$sql = mysql_query("SELECT name FROM disease");
while ($row = mysql_fetch_array($sql)) {
echo "<option value='" . $row['name'] . "'>" . $row['name'] . "</option>";
}
?>
</select><br><br />
<table border="1" style="width:100%">
<tr>
<td><b>Author</b></td>
<td><b>Category</b></td>
<td><b>Subject</b></td>
<td><b>Content</b></td>
</tr>
<?php
if(isset($_GET["selected"])){
$type = $_GET["selected"];
$query = "SELECT author,category,subject,content FROM announcements WHERE category='" . $type . "'";
$announcements = mysql_query($query, $connection);
$counter = 0;
$z = 0;
if ($announcements == FALSE) {
die(mysql_error()); // To get better errors report
}
while ($row = mysql_fetch_assoc($announcements)) {
while ($row = mysql_fetch_assoc($announcements)) {
$counter++;
?>
<tr>
<td><?php echo $row['author'];?></td>
<td><?php echo $row['category']; ?></td>
<td><?php echo $row['subject']; ?></td>
<td><?php echo $row['content']; ?></td>
</tr>
<?php }
}
}
?>
<?php
else:
?>
Select Category :<select id="SelectDisease" name="category">
<option value="*">All</option>
<!--emfanizei tis epiloges gia ta specialties me basi auta p exoume sti basi mas -->
<?php
$sql = mysql_query("SELECT name FROM disease");
while ($row = mysql_fetch_array($sql)) {
echo "<option value='" . $row['name'] . "'>" . $row['name'] . "</option>";
}
?>
</select><br><br />
<table border="1" style="width:100%">
<tr>
<td><b>Author</b></td>
<td><b>Category</b></td>
<td><b>Subject</b></td>
<td><b>Content</b></td>
</tr>
<?php
if(isset($_GET["selected"])){
$type = $_GET["selected"];
$query = "SELECT author,category,subject,content FROM announcements WHERE category='" . $type . "'";
$announcements = mysql_query($query, $connection);
$counter = 0;
$z = 0;
while ($row = mysql_fetch_assoc($announcements)) {
$counter++;
?>
<tr>
<td><?php echo $row['author'];?></td>
<td><?php echo $row['category']; ?></td>
<td><?php echo $row['subject']; ?></td>
<td><?php echo $row['content']; ?></td>
</tr>
<?php } } endif;?>
</table>
</body>
</html>
这是我的 javascript.js 文件
$(document).ready(function() {
$('#SelectDisease').change(function() {
var selected=$(this).val();
$.get("announcements.php?selected="+selected, function(data){
$('.result').html(data);
});
});
});
感谢您的宝贵时间:)
PS：INCLUDE_ONCE 代码创建与数据库的连接
【问题讨论】：
-
打开错误报告，并在执行查询后检查 MySQL 错误。 $announcements 很可能（很可能）是错误的，因为您的查询失败。
-
mysql_select_db();你忘了
-
@satishrajak 我从
include_once 'header.php';中选择我的数据库 -
那么首先你写 $connection = mysql_connect("localhost", "root", "smogi");包含include_once'header.php';
-
如果我按照你说的做，我会得到这个
Parse error: syntax error, unexpected 'include_once' (T_INCLUDE_ONCE)
标签： javascript php jquery mysql