下拉菜单 PHP

Question

I have this html code for a menu with sub-menu:

<li class="has-submenu">
                                <a href="#">Services</a>
                                <div class="mainmenu-submenu">
                                    <div class="mainmenu-submenu-inner"> 
                                        <div>
                                            <h4>Home</h4>
                                            <ul>
                                                <li><a href="#">Plumbers</a></li>
                                                <li><a href="#">Painters</a></li>
                                            </ul>
                                            <h4>People</h4>
                                            <ul>
                                                <li><a href="#">Babysitter</a></li>
                                                <li><a href="#">Trainer</a></li>

                                            </ul>
                                            <h4>School</h4>
                                            <ul>
                                                <li><a href="#">Teacher</a></li>

                                            </ul>
                                        </div>
                                        <div>
                                            <h4>Machine</h4>
                                            <ul>
                                                <li><a href="#">Mecanic</a></li>
                                            </ul>
                                            </div>
                                </div>
                            </div>
                        </li>

Then I have this table called "services":

id  |    name   | service   | description    

  1       Mario    Plumber         aaa        
  2       Luigi    Plumber         aaa      
  3       Link     Painter         aaa      
  4       Zelda    Babysitter      aaa      
  5       Sonic    Trainer         aaa      
  6       Marth    Teacher         aaa      
  7        Ike     Trainer         aaa      
  8     Little Mac Mecanic         aaa 

I want to create a code that shows me the results only associated with that sub-menu. For example, if I press Plumbers in the sub-menu, I want it to only show me the plumbers from my table. The PHP code I use is:

$query = "SELECT * FROM services WHERE service LIKE service";
                    $result = mysql_query($query) or die(mysql_error());
                    $casaArray = array();

But I can only get it to show me all the services. I'm new to PHP.

Thanks for your help.

Answer 1

You can simply create a link in your menu:

<a href="information.php?action=Plumber">Plumbers</a>

Then, for your information.php. (I've used mysql_real_escape_string() function to prevent SQL injections )

if(!empty($_GET["action"])){

  $action = mysql_real_escape_string($_GET["action"]);

  $query = "SELECT * FROM services WHERE service='$action'";
  $result = mysql_query($query);
  while($row = mysql_fetch_array($result)){
    echo $row["name"]." - ".$row["description"]."<br>";
  } /* END OF WHILE LOOP */

} /* END OF IF NOT EMPTY ACTION */

Note:

• You must have a structured database, a schema that you have, etc. in order to achieve what you want. You can update your post in order for people here in SO to help you.
• As far as I can see with your given code, user will click a link that they prefer, and they will be redirected to a page that contains information, in a form of list or some sort.
• If you don't have yet a code that would connect to your database, I assume that you want to use PHP and mysql because of the tag you used in your post. Unfortunately, mysql is already deprecated and we recommend that you use mysqli prepared statement .

To start, you have this menu, with links that would redirect them to another page. Unfortunately, again, you only gave us one table that contains the information you want to show on the next page.

First, let us create a table, which we will call main_services and it will have two column: service_id (auto-increment and your primary key) and service .

service_id  |  service
    1       |  Plumber
    2       |  Painter
    3       |  Babysitter
    4       |  Trainer
    5       |  Teacher
    6       |  Mechanic
and so on...

Then, let us alter the table structure of your services table. Your services table will still have four columns: id , service_id , name and description . We will index the main_services.service_id to your services.serviceid

id  |   service_id |    name    |  description    

  1         1        Mario           aaa        
  2         1        Luigi           aaa      
  3         2        Link            aaa      
  4         3        Zelda           aaa      
  5         4        Sonic           aaa      
  6         5        Marth           aaa      
  7         4        Ike             aaa      
  8         6        Little Mac      aaa 

After we have established your tables, we will create the links on your menu.

<div>
  <h4>Options</h4>
  <ul>
    <?php

      $con = new mysqli("YourHost", "Username", "Password", "Database"); /* REPLACE NECESSARY DATA */

      if (mysqli_connect_errno()) {
        printf("Connect failed: %s\n", mysqli_connect_error());
        exit();
      }

      if($stmt = $con->prepare("SELECT service_id, service FROM main_service")){
        $stmt->execute();
        $stmt->bind_result($serviceid,$service);
        while($stmt->fetch()){
          ?>
            <li>
              <a href="information.php?id=<?php echo $serviceid; ?>"><?php echo $service; ?></a>
            </li>
          <?php
        }
        $stmt->close();

      }
    ?>
  </ul>
</div>

Then let us create your information.php file:

<?php

  /* LET US FIRST ESTABLISH YOUR CONNECTION TO YOUR DATABASE */
  $con = new mysqli("YourHost", "Username", "Password", "Database"); /* REPLACE NECESSARY DATA */

  if (mysqli_connect_errno()) {
    printf("Connect failed: %s\n", mysqli_connect_error());
    exit();
  }

  if(!empty($_GET["id"])){

    if($stmt = $con->prepare("SELECT name, description FROM services WHERE service_id = ?")){
      ?>
        <table>
          <tr>
            <th>Name</th>
            <th>Description</th>
          </tr>
      <?php

      $stmt->bind_param("i",$_GET["id"]);
      $stmt->execute();
      $stmt->bind_result($name,$description);
      while($stmt->fetch()){
        ?>
          <tr>
            <td><?php echo $name; ?></td>
            <td><?php echo $description; ?></td>
          </tr>
        <?php
      } /* END OF WHILE LOOP */

      ?>
        </table>
      <?php
      $stmt->close();
    } /* END OF PREPARED STATEMENT */

  } /* END OF IF NOT EMPTY ID */

?>

Answer 2

Fist add jquery files (learn link)

      $con = new mysqli("YourHost", "Username", "Password", "Database"); /* REPLACE NECESSARY DATA */

      if (mysqli_connect_errno()) {
        printf("Connect failed: %s\n", mysqli_connect_error());
        exit();
      }
      if($stmt = $con->prepare("SELECT service_id, service FROM main_service")){
        $stmt->execute();
        $stmt->bind_result($serviceid,$service);
        while($stmt->fetch()){
      ?>

  <li class="ui-state-disabled">
      <a href="information.php?id=<?php echo $serviceid; ?>"><?php echo $service; ?></a>
      <---------------here set sub menu sql and take the result if submenu not empty---->
      <ul>
      <li class="ui-state-disabled">Ada</li>
      <li>Saarland</li>
      <li>Salzburg an der schönen Donau</li>
    </ul>
    <---------------here close the submenu loop---->
  </li>
  <---------------here close the menu loop---->
</ul>

I think you can solve this (jQuery menu).. go through this url and learn https://jqueryui.com/menu/