Javascript 基于所选选项和onclick的PHP MySQL显示数据库
我需要选择一个optionbook类别,然后搜索数据库以查找具有该类别的所有书籍,并将数据库显示为表 我可以在这段代码中使用HTML、JavaScript、PHP和MySQL 这是HTML代码:Javascript 基于所选选项和onclick的PHP MySQL显示数据库,javascript,php,html,mysql,forms,Javascript,Php,Html,Mysql,Forms,我需要选择一个optionbook类别,然后搜索数据库以查找具有该类别的所有书籍,并将数据库显示为表 我可以在这段代码中使用HTML、JavaScript、PHP和MySQL 这是HTML代码: <p> <select name="genre" size="1"> <option value='0' id='AA'>Art & Architecture</option>
<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>";
}
$result->close();
?>
你知道哪里出了问题吗
附言:PHP非常新,所以请使用简单的语言和解释。谢谢大家! 这里似乎把php和javascript jquery混为一谈了。。 不过,对于你的提问,我得出了以下结论 创建了一个test.php文件 其中我使用了$your_html_response.htmlresponseData;这是从test2.php文件得到的响应中添加的数据
希望这能解决您的困惑。显示您的getElementByName getElementByName您在php文件中创建了此函数吗?仅供参考,有一个内置的javascript函数,名为getElementsByName,因此请确保您不会对此感到困惑。..getElementByName是通过javascript创建的,这不是一个函数,您不能从PHPthank调用客户端javascript函数!但是有更简单的方法吗?我不允许使用jquery。。。
<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>
<?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;
?>
done(function(responseData) {
console.log('Done: ', responseData);
$("#your_html_response").html(responseData);
}