php - 需要我的 HTML 搜索表单来搜索在搜索表单上选择的选项，然后显示我的结果？

Question

我有一个有 4 个页面的网站:主页、销售页面、租赁页面和联系我们页面。

主页上有一个搜索表单，需要搜索他们选择的选项。这是一个房地产网站，因此如果他们点击搜索按钮，他们需要进入他们选择的销售页面或租赁页面，例如，如果他们选择待售、独立屋、3 间卧室、2 间浴室，然后选择任何金额最低价格/最高价格，然后需要转到销售页面并根据他们选择的值(value)检索房屋。

这是搜索表单的图像:

谁能帮我解决这个问题吗？

这是 HTML 搜索表单编码:

<form id="advanced_search" action="search.php" class="clearfix" name="advanced_search" method="post">

<div class="col-lg-6 col-md-6 col-sm-6 col-xs-6">
<label for="type"><b>Type</b></label>
<select>
  <option value="House">House</option>
  <option value="Apartment">Apartment</option>
  <option value="Flat">Flat</option>
  <option value="Townhouse">Townhouse</option>
  <option value="Plot">Plot</option>
  <option value="Farm">Farm</option>
</select>

</div>

<div class="col-lg-6 col-md-6 col-sm-6 col-xs-6">
<label for="status"><b>Status</b></label>
<select>
  <option value="For Sale">For Sale</option>
  <option value="For Rent">For Rent</option>
</select>

</div>


<div class="col-lg-6 col-md-6 col-sm-6 col-xs-6">
<label for="bedrooms"><b>Bedrooms</b></label>
<select>
  <option value="1">1</option>
  <option value="2">2</option>
  <option value="3">3</option>
  <option value="4">4</option>
  <option value="5">5</option>
  <option value="6">6</option>
  <option value="7">7</option>
  <option value="8">8</option>
  <option value="9">9</option>
  <option value="10">10</option>
</select>

</div>
<div class="col-lg-6 col-md-6 col-sm-6 col-xs-6">
<label for="baths"><b>Bathrooms</b></label>
<select>
  <option value="1">1</option>
  <option value="2">2</option>
  <option value="3">3</option>
  <option value="4">4</option>
  <option value="5">5</option>
  <option value="6">6</option>
  <option value="7">7</option>
  <option value="8">8</option>
  <option value="9">9</option>
  <option value="10">10</option>
</select>


</div>
<div class="col-lg-6 col-md-6 col-sm-6 col-xs-6">
<label for="min_price"><b>Min Price</b></label>
<select>
  <option value="1">1</option>
  <option value="2">2</option>
  <option value="3">3</option>
  <option value="4">4</option>
  <option value="5">5</option>
  <option value="6">6</option>
  <option value="7">7</option>
  <option value="8">8</option>
  <option value="9">9</option>
  <option value="10">10</option>
</select>

</div>
<div class="col-lg-6 col-md-6 col-sm-6 col-xs-6">
<label for="max_price"><b>Max Price</b></label>
<select>
  <option value="1">1</option>
  <option value="2">2</option>
  <option value="3">3</option>
  <option value="4">4</option>
  <option value="5">5</option>
  <option value="6">6</option>
  <option value="7">7</option>
  <option value="8">8</option>
  <option value="9">9</option>
  <option value="10">10</option>
</select>

</div>

<div class="clearfix"></div>
<div class="col-lg-12 col-md-12 col-sm-12 col-xs-12">
<button name="search" type="submit" class="btn btn-success">SEARCH PROPERTY</button>
</div>
</form>

这是我正在努力解决的 PHP 编码:

<?php
if(isset($_POST['search']))
{
$key=$_POST["search"];  //key=pattern to be searched
$con=mysqli_connect("localhost", "root", "");

// Check connection
if (mysqli_connect_errno()) {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result=mysqli_query($con,"SELECT * FROM properties WHERE");




while($row=mysqli_fetch_assoc($result))
{




}
}

?>

Answer 1

试试这个:

<?php

$type = $_POST['type'];
$status = $_POST['status'];
$minPrice = $_POST['min_price'];
$maxPrice = $_POST['max_price'];
$bedrooms = $_POST['bedrooms'];
$baths = $_POST['baths'];

$con=mysqli_connect("localhost","root","","house");
// Check connection
if (mysqli_connect_errno()) {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

$result = mysqli_query($con,"SELECT * FROM properties  WHERE tipo='$type' and baths = '$baths' and bedrooms = '$bedrooms' and status = '$status' and price between '$minPrice' and '$maxPrice'");

while($row = mysqli_fetch_array($result)) {
  echo $row['type'] . " " . $row['status'];
  echo "<br>";
}

mysqli_close($con);

我在这里创建了一个表并且它正在工作。 不要忘记更改 html code "标签并在其上添加属性 "name"