通过选择下拉列表过滤mysql结果

Question

I'm a new member of StackOverflow, and although I've been using the website for a long time, it's my first time posting a question, in a hope that someone will be able to help me. I'll start by saying that my knowledge of PHP and MySQL is basic, but what I'm trying to do isn't too complex in my opinion, so hopefully I won't be asking for much. I've done a lot of prior research, but I just couldn't find the right answer.

In short, this is what I'm trying to do:

I've got an html form, which upon submission writes data to a database, and then publishes a table on a separate html page. With each successful submission a new table gets generated and published, while the old one gets pushed underneath. This all works fine, and I've also implemented pagination so that only 5 tables are visible per page.

What I'd like to be able to do is allow people to ONLY view/display results (tables) based on a specific criteria, in this case "rating", by selecting a rating from a drop-down on the page where tables are published. Rating is one of the fields in my form which gets submitted to a database and then published in one of the rows in a table.

Below is the code which publishes tables. Thanks in advance for your help!

<?php
include('dbconnect.php'); 

mysql_select_db("vtracker", $con);
$result  = mysql_query("SELECT * FROM userdata");
$age = "Age:";
$rating = "Rating:";
$country = "From:";
$name = "Name:";

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

                        echo "<table id='mft_table' cellspacing='0'>";
                        echo "<tbody>";
                        echo "<tr>";
                        echo "<td class='row1'>" .$name . " " . $row['personsname'] . "</td>";
                        echo "<td rowspan='4'>";
                        echo "<div class='mft_column'>" . $row['mft'] . "</div>";
                        echo "</td>";
                        echo "</tr>";
                        echo "<tr>";
                        echo "<td class='row2'>" . $country . " " . $row['nationality'] . "</td>";
                        echo "</tr>";
                        echo "<tr>";
                        echo "<td class='row3'>" . $age . " " . $row['personsage'] . "</td>";
                        echo "</tr>";
                        echo "<tr>";
                        echo "<td class='row4'>" . $rating . " " . $row['rating'] . "</td>";
                        echo "</tr>";
                        echo "</tbody>";
                        echo "<br>";
                        echo "</table>"; 

        }
    ?>

Answer 1

for both true and false use can add thid in your code:

if($_POST['rating_dropdown']!='')
{
    $temp_rating = $_POST['rating_dropdown'];
    $query=mysql_query("SELECT * FROM userdata WHERE rating = '$temp_rating'");
}
else
{
    $query=mysql_query("SELECT * FROM userdata");
}

Answer 2

Dunno if this works, it's just a hinch. haha. It will see if the rating is true(not null), if it's true it will echo the results.

while($row = mysql_fetch_array($result))
{
if ($rating)
                echo "<table id='mft_table' cellspacing='0'>";
                echo "<tbody>";
                echo "<tr>";
                echo "<td class='row1'>" .$name . " " . $row['personsname'] . "</td>";
                echo "<td rowspan='4'>";
                echo "<div class='mft_column'>" . $row['mft'] . "</div>";
                echo "</td>";
                echo "</tr>";
                echo "<tr>";
                echo "<td class='row2'>" . $country . " " . $row['nationality'] . "</td>";
                echo "</tr>";
                echo "<tr>";
                echo "<td class='row3'>" . $age . " " . $row['personsage'] . "</td>";
                echo "</tr>";
                echo "<tr>";
                echo "<td class='row4'>" . $rating . " " . $row['rating'] . "</td>";
                echo "</tr>";
                echo "</tbody>";
                echo "<br>";
                echo "</table>"; 
}

}

Answer 3

Once the dropdown gets selected and posted to your display page, use this code:

$temp_rating = $_POST['rating_dropdown'];
mysql_query("SELECT * FROM userdata WHERE rating = '$temp_rating'");

Keep in mind, however, that you should be using PDO or mysqli extension, not the mysql extension. According to PHP's website:

This extension is deprecated as of PHP 5.5.0, and will be removed in the future. Instead, the MySQLi or PDO_MySQL extension should be used. See also MySQL: choosing an API guide and related FAQ for more information.