CODEWITHSUNDEEP
phpjavascriptjquerymysql
the_boy_za
the_boy_za

Reputation: 287

How to automatically check checkboxes from mysql database

I have a problem to solve which im not sure if it can be solved in the way that i have in mind. It's about a form with checkboxes. The purpose of this form is to allow users to add or remove their favorite brands from the database by checking/unchecking the form checkboxes. The challege that i face now is that there may already be brands selected in the past from the user. And i would like the page when it loads to check in the database and automatically check the brands in the form which where found in the database. I checked to see if it was possible with jquery , but i got stuck. Here is my code:

/* == SQL CODE == */

CREATE TABLE IF NOT EXISTS `favoriteBrands` (
`pkFavoriteBrand` int(5) NOT NULL AUTO_INCREMENT,
`fkCompanyID` int(5) NOT NULL,
`Brands` varchar(30) NOT NULL,
PRIMARY KEY (`pkFavorietemerken`)
) ENGINE=MyISAM  DEFAULT CHARSET=latin1 AUTO_INCREMENT=81 ;


INSERT INTO `favoriteBrands` (`pkFavoriteBrand`, `fkCompanyID`, `Brands`) VALUES
(80, 7, 'Replay'),
(79, 7, 'Pepe Jeans'),
(71, 12, 'Nike'),
(70, 12, 'Le Coq Sportif'),

 /* == SQL CODE == */

 /* == PHP CODE == */
 <?php

 $find_fav_brands = " SELECT Brands FROM favoriteBrands WHERE fkCompanyID=$company_id";

 $get_brand        = mysql_query($find_fav_brands) or die ("No brand found");

 while($row = mysql_fetch_assoc($get_brand )){

$show_brand = $row["Brands"];

    echo"<p>$show_brand</p>";     
 }?>

 <form name="form" method="post" action="ready.php" >                       

<ul>

<li><input type="checkbox" name="merk[]" value="Adidas"/>Adidas</li>
<li><input type="checkbox" name="merk[]" value="Airforce"/>Airforce</li>
<li><input type="checkbox" name="merk[]" value="Armani"/>Armani</li>
<li><input type="checkbox" name="merk[]" value="Asics"/>Asics</li>
 </ul>

</form>
/* == PHP CODE == */

Upvotes: 3

Views: 4326

Answers (2)

JiangCat
JiangCat

Reputation: 125

1, I strongly do not suggest you to use 'string' as checkbox value. Try using IDs. It's much safer and easier to maintain your code.

2, Since you are using <li> as your input container, run a loop to do so with PHP makes it much easier, like while or for/foreach.

3, Your database only restored 'checked' values, but not unchecked. It is impossible to generate a complete list from it with all checkbox options. So, here are 2 solutions:

3.1 Either save all options in your table structure, for example: brand1=0, brand2=1, brand3=1 etc. So you can foreach all your results and print 'checked=checked' when brandX != 1;

3.2 Or, make an array of all possible options, and loop them, and put checked value if the data exists.

Upvotes: 1

John V.
John V.

Reputation: 4670

<?php

$find_fav_brands = " SELECT Brands FROM favoriteBrands WHERE fkCompanyID=$company_id";

$get_brand        = mysql_query($find_fav_brands) or die ("No brand found");
$brands = array();
while($row = mysql_fetch_assoc($get_brand )){
    $brands[]=$row["Brands"];
    $show_brand = $row["Brands"];
    echo"<p>$show_brand</p>";     
}
?>
<form name="form" method="post" action="ready.php">     

<ul>
<li><input type="checkbox" name="merk[]" value="Adidas"<?php echo in_array("Adidas",$brands)?" checked="checked"":""; ?> />Adidas</li>
<li><input type="checkbox" name="merk[]" value="Airforce"<?php echo in_array("Airforce",$brands)?" checked="checked"":""; ?> />Airforce</li>
<li><input type="checkbox" name="merk[]" value="Armani"<?php echo in_array("Armani",$brands)?" checked="checked"":""; ?> />Armani</li>
<li><input type="checkbox" name="merk[]" value="Asics"<?php echo in_array("Asics",$brands)?" checked="checked"":""; ?> />Asics</li>
</ul>

</form>

Upvotes: 4

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