一个显示所选选项的按钮

Question

According to my code you could choose a game from the games table, so I am trying to have a button which is called Show to show the chosen Game. Something like:

<input type ="submit" value= "Show" name="Show">

I tried this

    `<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.12.0/jquery.min.js"></script>
    <script>
    $(document).ready(function(){
        $("#hide").click(function(){
            $("p").hide();
        });
        $("#show").click(function(){
            $("p").show();
        });
    });
    </script>`

But Somehow something was missing or it did not work. Here is my code:

    <div id="content">
<article>
<h2> Games</h2>

<form action='Games.php' method='post'>
<p>Show</p>
<select size= "1"  name='gameid' >
<option value ="Show"> Choose Game </option>

Here is my code

    <?php
    include('dbaccess.php');
        $mysqli = new mysqli($dbhost,$dbusername,$dbpassword,$dbname);
        $mysqli->set_charset('utf8');
        if ($mysqli->connect_error) {print("<p>Fail: ".$mysqli->connect_error."</p>");} 
        $result = $mysqli->query("SELECT * FROM Game;"); 

    if ($mysqli->error) {print("<p>Fail: ".$mysqli->error."</p>");}
    if ($result->num_rows == 0) 
        {
        print("<p>Ingen data.</p>");
        }
    else
        {
        while($row = $result->fetch_array())
          {

       print "<option value=$row[0]>$row[1]:$row[2]:$row[3]:</option>";



        }
        }
    $mysqli->close();
    ?>



    </form>

Answer 1

Your function expects a tag with an id of show :

$("#show").click(function(){
    $("p").show();
});

So you need to add the id to the button:

<input type ="submit" value= "Show" name="Show" id="show">

Likewise with the hide id.