Displaying database contents from drop down selection

Question

In my website I have a page to display testimonials. I wrote this code to display my all testimonials from database.

This is my code so far :

while ( $row = mysqli_fetch_array( $r, MYSQLI_ASSOC)) {
    $testimonial = $row['testimonial'];
    //echo $testimonial;
    $mytestimonial = nl2br($testimonial);
    $city               = $row['city_name'];
    $name               = $row['name'];
    $url                = $row['web_address'];
    $imageName      = $row['image_name'];
    $type               = $row['membership_type'];
}

With this code I can get all my testimonials to the page. Its pretty working for me. My problem is now I need to filter my testimonials according to its type. I have 3 different kind of testimonials in my database. (tutor, institute, student)

I am going to use a select box to filter the data. When selecting an option from select box I need to display testimonials according to that selected type.

<div class="filter-box">    
    <div id="select_box">
        <form method="post" action="">          
            <div class="variation2">
                <label>Filter By</label>
                <select class="select">
                    <option>Tutor</option>
                    <option>Institute</option>
                    <option>Student</option>
                </select>
            </div>
        </form> 
    </div>  
</div>

Can anyone get me going in a direction here?

Thank You

Insert a `where` clause in your query. Also I would suggest use Ajax. — Ravi, Mar 05 '13 at 13:41
1. Give a name to your select box 2. accept this value as type 3. Use this type variable to filter in the query -- "SELECT testimonial, city_name, name, web_address, image_name, membership_type FROM testimonials INNER JOIN city ON city.city_id = testimonials.city_id Where type = $type ORDER BY date_added DESC LIMIT $start, $display";" — Jaspal Singh, Mar 05 '13 at 13:41

score 0 · Answer 1 · answered Mar 05 '13 at 13:42

0

Why not reload the select box dynamically (using AJAX) once the user has selected one of the 3 options and display your required testimonial.I guess that will solve your problem.

answered Mar 05 '13 at 13:42

Atanu

61
2
12

UnholyRanger · Answer 2 · 2013-03-05T13:54:30.690

0

So you want to then select the testimonials based on their type

$q = "SELECT testimonial, city_name, name, web_address, image_name, membership_type
        FROM testimonials 
        INNER JOIN city ON city.city_id = testimonials.city_id
        WHERE type = '$type'
        ORDER BY date_added DESC LIMIT $start, $display";

Now you also want to get the type from the user

<select name="type" class="select">

With the POST

$type = $_POST['type'];

Auto submit the form on change example

Javascript:

<script type="text/javascript">
    function submitform(){
        document.frmType.submit();
    }
</script>

Form:

<form name="frmType" method="post" action="">
    <select name="type" class="select" onchange="submitform()">

Further examples can be found here

edited Mar 05 '13 at 13:54

answered Mar 05 '13 at 13:43

UnholyRanger

1,977
14
24

In this case, I can't use a submit button to my form. Without a button is it possible? – TNK Mar 05 '13 at 13:46
why can't you use a submit button? ``. Yes it is possible, by javascript or Ajax – UnholyRanger Mar 05 '13 at 13:47
With my page I can't display a submit button. That's why I am looking for a different solution. – TNK Mar 05 '13 at 13:50
Thanks for reply. May I know can I use jqery for this? – TNK Mar 05 '13 at 13:59
I don't know much about JQuery but you can check out [this](http://stackoverflow.com/questions/7261305/javascript-select-onchange-this-form-submit) – UnholyRanger Mar 05 '13 at 14:09

SeanWM · Answer 3 · 2013-03-06T16:08:08.070

0

You have a couple options. The first is with ajax. Second is submitting the form onchange or onsubmit:

<select class="select" name="type" onchange="document.forms[0].submit();">
    <option value="Tutor">Tutor</option>
    <option value="Institute">Institute</option>
    <option value="Student">Student</option>
</select>

And your query:

$type = '';
if(!empty($_POST) && isset($_POST[type])){
    $type = " WHERE testimonials.type = '".$_POST[type]."'";
}
$q = "SELECT testimonial, city_name, name, web_address, image_name, membership_type
        FROM testimonials 
        INNER JOIN city ON city.city_id = testimonials.city_id
        ".$type."
        ORDER BY date_added DESC LIMIT $start, $display";

Of course make sure to check if the form was submitted.

edited Mar 06 '13 at 16:08

answered Mar 05 '13 at 13:52

SeanWM

16,789
7
51
83

I am going with your code. Now I have this error - An error occurred in script 'C:\wamp\www\website\testimonials.php' on line 92: Undefined index: type – TNK Mar 05 '13 at 14:09
I did as you said. Now no any errors but not filtering happen – TNK Mar 05 '13 at 18:25
What does `echo $q;` give you when you select a type? – SeanWM Mar 05 '13 at 18:58
where clause not display in my query. – TNK Mar 06 '13 at 14:10
problem is same. I updated with you code. But I changed values to this - ` ` – TNK Mar 08 '13 at 13:17

score 0 · Answer 4 · answered Mar 13 '13 at 11:56

0

add a class by the name to each of the testimonial divs

 <div class="student">testimonial goes here</div>

now use jquery to hide/show the categories at one shot.

 $('.student').show();

i hope it helps

answered Mar 13 '13 at 11:56

astroanu

3,901
2
36
50

I can't use show/hide function with this. Because I need to requery my query. – TNK Mar 13 '13 at 12:00
my idea was to do the filtering part on the browser without wasting bandwidth. :) – astroanu Mar 14 '13 at 06:35

Displaying database contents from drop down selection

4 Answers4