我需要选择一个选项(图书类别),然后在数据库中搜索该类别的所有图书,并将数据库显示为一个表。
我可以使用HTML, JavaScript, PHP和MySQL在这个代码。
这是HTML代码:<p>
<select name="genre" size="1">
<option value='0' id='AA'>Art & Architecture</option>
<option value='1' id='BG'>Biography</option>
<option value='2' id='CH'>Children</option>
<option value='3' id='DR'>Drama</option>
<option value='4' id='ER'>Erotica</option>
<option value='5' id='HS'>History</option>
<option value='6' id='ML'>Military</option>
<option value='7' id='MU'>Music</option>
<option value='8' id='NE'>Non-English</option>
<option value='9' id='NV'>Novels</option>
<option value='10' id='OC'>Occult</option>
<option value='11' id='PS'>Philosophy</option>
<option value='12' id='PG'>Photography</option>
<option value='13' id='PT'>Poetry</option>
<option value='14' id='PE'>Politics & Economics</option>
<option value='15' id='RG'>Religion</option>
<option value='16' id='SE'>Science & Engineering</option>
<option value='17' id='SP'>Sport</option>
<option value='18' id='TE'>Travel & Exploration</option>
</select>
<input type="submit" value="Search" onClick="">
</p>
这是PHP和MySQL的代码:
<?php
$mysqli = new mysqli('localhost','root','','bookstore');
if(mysqli_connect_errno())
{
$problem = mysqli_connect_error();
echo "Error opening database";
die ($problem);
}
$genre = getElementByName('genre');
$query = "SELECT category(*) FROM book_list WHERE category=$genre";
echo "<h2>These are the books available</h2>";
// Execute Query
$result = $mysqli->query($query);
// echo "<table border="1">";
while($row = $result->fetch_array(mysql_query))
{
$Title = $row["Title"];
$Author = $row["Author"];
$Price = $row["Price ($)"];
echo "<tr><td>$Title</td><td>$Author</td><td>$Price</td></tr>";
echo "</table>";
}
$result->close();
?>
你知道哪里出了问题吗?
注:PHP新手,请使用简单的语言和解释。谢谢你!
你好像把php和javascript (jquery)混在一起了。不过,对于你的问题,我得出了以下结论…
创建test.php文件
<script src="https://code.jquery.com/jquery-3.1.0.min.js" integrity="sha256-cCueBR6CsyA4/9szpPfrX3s49M9vUU5BgtiJj06wt/s=" crossorigin="anonymous"></script>
<script>
$(document).ready(function(){
$("#genre").on('change', function postinput(){
var genre = $(this).val(); // this.value
$.ajax({
url: 'test2.php',
data: { genre: genre },
type: 'post'
}).done(function(responseData) {
console.log('Done: ', responseData);
$("#your_html_response").html(responseData);
}).fail(function() {
console.log('Failed');
});
});
});
</script>
<p>
<select name="genre" id="genre" size="1">
<option value='0' id='AA'>Art & Architecture</option>
<option value='1' id='BG'>Biography</option>
<option value='2' id='CH'>Children</option>
<option value='3' id='DR'>Drama</option>
<option value='4' id='ER'>Erotica</option>
<option value='5' id='HS'>History</option>
<option value='6' id='ML'>Military</option>
<option value='7' id='MU'>Music</option>
<option value='8' id='NE'>Non-English</option>
<option value='9' id='NV'>Novels</option>
<option value='10' id='OC'>Occult</option>
<option value='11' id='PS'>Philosophy</option>
<option value='12' id='PG'>Photography</option>
<option value='13' id='PT'>Poetry</option>
<option value='14' id='PE'>Politics & Economics</option>
<option value='15' id='RG'>Religion</option>
<option value='16' id='SE'>Science & Engineering</option>
<option value='17' id='SP'>Sport</option>
<option value='18' id='TE'>Travel & Exploration</option>
</select>
</p>
<div id="your_html_response">
</div>
在上面,我已经包含了一个jquery库,并使用下拉菜单的onchange函数发送一个值到test2.php文件。
你还可以看到空白的<div id="your_html_response"></div>
,它将添加来自test2.php文件
也删除了<input type="submit" value="Search" onClick="">
,因为它不是必需的,因为你没有在你的代码中创建一个表单。
创建test2.php文件
<?php
$mysqli = new mysqli('localhost','root','','bookstore');
if(mysqli_connect_errno())
{
$problem = mysqli_connect_error();
echo "Error opening database";
die ($problem);
}
$genre = $_POST['genre'];
$query = "SELECT category(*) FROM book_list WHERE category=$genre";
$html_response = "<h2>These are the books available</h2>";
// Execute Query
$result = $mysqli->query($query);
$html_response .= "<table border='1'>";
while($row = $result->fetch_array(mysql_query))
{
$Title = $row["Title"];
$Author = $row["Author"];
$Price = $row["Price ($)"];
$html_response.= "<tr><td>$Title</td><td>$Author</td><td>$Price</td></tr>";
}
$html_response .= "</table>";
$result->close();
echo $html_response;
?>
在上面,我已经从下拉菜单中获得了张贴的值,并连接了一个数据库,并使用mysqli查询返回记录(如果有的话)。我在这里返回了HTML响应…
在test.php中可以找到以下代码done(function(responseData) {
console.log('Done: ', responseData);
$("#your_html_response").html(responseData);
}
其中我使用了$("#your_html_response").html(responseData);
,这是从我们从test2.php文件中获得的响应中添加数据