来自HTML下拉菜单的Mysql SELECT / query [英] Mysql SELECT / query from HTML dropdown menu

问题描述

我已创建用于MySQL数据库查询的HTML表单，查询结果将显示在我的Google Map API上。



SELECT方法在我使用普通文本时起作用，但在使用HTML下拉菜单时没有用



SUCCEED QUERY：

Hi,

I have create HTML Form which will used for Query on MySQL database, the query result will displaying on my Google Map API's.

The SELECT method is work when i have used normal text, but didnt work when using dropdown menu from HTML

SUCCEED QUERY :

$query = "SELECT * FROM table_p WHERE nama_provinsi='JAWA TIMUR' ";

UNCUCCEED QUERY ：

UNSUCCEED QUERY :

$query = "SELECT * FROM table_p WHERE nama_provinsi='$nama_provinsi' ";

我尝试过：

What I have tried:

<?php
require("koneksi.php");

$nama_provinsi = $_POST['nmprovinsi'];
echo $nama_provinsi;

function parseToXML($htmlStr)
{
$xmlStr=str_replace('<','<',$htmlStr);
$xmlStr=str_replace('>','>',$xmlStr);
$xmlStr=str_replace('"','"',$xmlStr);
$xmlStr=str_replace("'",''',$xmlStr);
$xmlStr=str_replace("&",'&',$xmlStr);
return $xmlStr;
}

$connection=mysqli_connect ($db_host, $db_user, $db_pass);
if (!$connection) {
  die('Not connected : ' . mysqli_error());
}

$db_selected = mysqli_select_db($connection,$db_name);
if (!$db_selected) {
  die ('Can\'t use db : ' . mysqli_error());
}

$query = "SELECT * FROM table_p WHERE nama_provinsi='$nama_provinsi' ";

$result = mysqli_query($connection,$query);
if (!$result) {
  die('Invalid query: ' . mysqli_error());
}

header("Content-type: text/xml");

// Start XML file, echo parent node
echo "<?xml version='1.0' ?>";
echo '<table_p>';
$ind=0;
// Iterate through the rows, printing XML nodes for each
while ($row = @mysqli_fetch_assoc($result)){
  // Add to XML document node
  echo '<marker ';
  echo 'id="' . $row['id'] . '" ';
  echo 'Properti="' . parseToXML($row['Properti']) . '" ';
  echo 'Alamat="' . parseToXML($row['Alamat']) . '" ';
  echo 'Luas="' . parseToXML($row['Luas']) . '" ';
  echo 'Harga="' . parseToXML($row['Harga']) . '" ';
  echo 'status="' . parseToXML($row['status']) . '" ';
  echo 'Fasilitas="' . parseToXML($row['Fasilitas']) . '" ';
  echo 'Nama="' . parseToXML($row['Nama']) . '" ';
  echo 'Kontak="' . parseToXML($row['Kontak']) . '" ';
  echo 'lat="' . $row['lat'] . '" ';
  echo 'lng="' . $row['lng'] . '" ';
  echo '/>';
  $ind = $ind + 1;
}

// End XML file
echo '</table_p>';

?>

以下是我选择的HTML表单用于此查询

Below is the HTML Select form i have used for this query

<select name="nmprovinsi" id="idprovinsi" class="dropdown_item_select">
    <option value="1" >JAWA TIMUR</option>
          <?php
          // Load file koneksi.php
          include "koneksi.php";

          $sql = mysqli_query($konekan, "SELECT DISTINCT(nama_provinsi) AS nama_provinsi FROM table_p ORDER BY nama_provinsi");

          while($data = mysqli_fetch_array($sql)){
            echo "<option value='".$data['nama_provinsi']."'>".$data['nama_provinsi']."</option>";
          }

          ?>
</select>

推荐答案

query =SELECT * FROM table_p WHERE nama_provinsi ='JAWA TIMUR';

query = "SELECT * FROM table_p WHERE nama_provinsi='JAWA TIMUR' ";

UNCUCCEED QUERY：

UNSUCCEED QUERY :

query =SELECT * FROM table_p WHERE nama_provinsi ='

query = "SELECT * FROM table_p WHERE nama_provinsi='

nama_provinsi';

nama_provinsi' ";

我的尝试：

What I have tried:

<?php
require("koneksi.php");

这篇关于来自HTML下拉菜单的Mysql SELECT / query的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！