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[英]How do I use PHP to construct an SQL query to sort and select from multiple rows in my database
[英]How do I use PHP to query my SQL database using SELECT tag in html
基本上我想使用用户从不同的选择标签中选择的选项,并且实质上将它们存储为变量,然后我可以使用它来在SQL中查询我的数据库。
我的HTML代码在这里:
<div id ="search_elements">
<img src="UniSelect.jpeg">
<select>
<option selected disabled>Select a university</option>
<option value="ucl">UCL</option>
<option value="kings">Kings College</option>
<option value="imperial">Imperial College</option>
<option value="lse">London School of Economics</option>
</select>
<img src="PriceSelect.jpeg">
<select>
<option selected disabled>Select a weekly rent price</option>
<option value="50">0-£50</option>
<option value="100"> £100-£150</option>
<option value="150">£150-200</option>
<option value="200"> £200+</option>
</select>
</div>
我想要使用的php的类型是:
//$con=mysqli_connect("localhost","adam","YjM3ZTYwOTQ5OWRmYWZh","adam_...");
//if (mysqli_connect_errno())
// {
// echo "Failed to connect to MySQL: " . mysqli_connect_error();
// }
// Perform queries
//$sql=mysqli_query($con,"SELECT CONTENT FROM Blog WHERE ID = 01");
//$result=mysqli_fetch_assoc($sql);
//echo $result['CONTENT'];
//mysqli_close($con);
为了再次清楚,我想使用用户选择的不同值,在单击搜索按钮时,将这些查询结果显示在表格中。
这个解决方案与你的解决方案略有不同,因为你没有提供你的表单,提交按钮等等。但是一般来说,经过一些改动之后,它也应该有效:
<form method="post" action="">
<img src="UniSelect.jpeg">
<select name="university">
<option selected disabled>Select a university</option>
<option value="ucl">UCL</option>
<option value="kings">Kings College</option>
<option value="imperial">Imperial College</option>
<option value="lse">London School of Economics</option>
</select>
<img src="PriceSelect.jpeg">
<select name="rent_price">
<option selected disabled>Select a weekly rent price</option>
<option value="50">0-£50</option>
<option value="100"> £100-£150</option>
<option value="150">£150-200</option>
<option value="200"> £200+</option>
</select>
<input type="submit" value="Submit form">
</form>
现在,要获取这些值(类似这样的东西,我建议将它放在您的表单上方):
if (!empty($_POST)) {
// Checking connection in here
$university_id = mysqli_real_escape_string($con, $_POST['university']);
$rent_price = mysqli_real_escape_string($con, $_POST['rent_price']);
$sql = mysqli_query($con, "SELECT * FROM university WHERE name = '".$university_id."'");
$result = mysqli_fetch_assoc($sql);
// Same thing goes for rent price
}
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