Pie chart based on checkbox array

PHP
1

Hi i would like to make a pie chart based on checkbox array submitted from the form where you can select more than 1 choice.
this is the form

<form method="post" action="" >
<input type="checkbox" value="ice cream" name=" food[] ">ice cream
<input type="checkbox" value="chocolate" name=" food[] ">chocolate
<input type="checkbox" value="milkshake" name=" food[] ">milkshake
<input type="checkbox" value="cake" name="food[] ">cake.
</form>

in the database the values are posted just fine and correctly.the only thing i need is how to present the choices that is how many times ice cream was chosen or cake. Please help me

2

How are the values being stored in your database?

3

if i choose two choices.in the database column it will be(cake,ice cream) if one of it it will be just be cake.
so the choices are separated by comma.

4

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@cielonovatus95 welcome to the forums. When you post code in the forum, you need to format it. To do so you can either select all the code and click the </> button, or type 3 backticks ``` on a separate line both before and after the code block.
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5

thank u

6

I suspected as much. This will make it more difficult to count the number of each. It can be done, but it’s a bit of a pain.

I would strongly recommend you take a look at Normalizing your database. It would make the process a lot easier.

If you insist on storing a comma separated string, you’ll have to retrieve every row of your database table and do some parsing to get the totals:

All code in this post is untested, i’m just pulling it from my head for example sake.

<?php
$foodstore = array();
$dbquery = mysqli_query($db,"SELECT food FROM mytable");
while($row = mysqli_fetch_array($dbquery)) {
  $food = explode(",",$row['food']);
 foreach($food AS $fin) {
   if(array_key_exists($fin,$foodstore) {
     $foodstore[$fin] += 1;
   }
   else {
     $foodstore[$fin] = 1;
  }
 }
echo $foodstore["cake"]

Whereas, if you had normalized the database, it goes a little simpler…and we can use some fancyness.

$foodstore = array();
$dbquery = mysqli_query($db,"SELECT COUNT(name) AS c, name FROM desserts GROUP BY name");
array_walk(mysqli_fetch_all($dbquery,MYSQL_ASSOC), function(v,i) { $foodstore[$v['name']] = $v['c']; });
echo $foodstore['cake'];
1 Like
7

@m_hutley Thank you so much it worked

8

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