CODEWITHSUNDEEP
phpsqlmysql
dede
dede

Reputation:

Inserting checkbox values into database

I have a form with several checkboxes which values are pulled from a database. I managed to display them in the form, assign an appropriate value to each, but cannot insert their values into other database.

Here's the code:

<form id="form1" name="form1" method="post" action="">
<?php
$info_id = $_GET['info_id'];
$kv_dodatoci = mysql_query("SELECT * FROM `dodatoci`") or die('ERROR DISPLAYING: ' . mysql_error());
while ($kol = mysql_fetch_array($kv_dodatoci)){
    $id_dodatoci = $kol['id_dodatoci'];
    $mk = $kol['mk'];


    echo '<input type="checkbox" name="id_dodatoci[]" id="id_dodatoci" value="' . $id_dodatoci . '" />';
    echo '<label for="' . $id_dodatoci.'">' . $mk . '</label><br />';
  }
?>
<input type="hidden" value="<?=$info_id?>" name="info_id" />
<input name="insert_info" type="submit" value="Insert Additional info" />
</form>
<?php
if (isset($_POST['insert_info']) && is_array($id_dodatoci)) {
    echo $id_dodatoci . '<br />';
    echo $mk . '<br />';

    // --- Guess here's the problem  ----- //
    foreach ($_POST['id_dodatoci'] as $dodatok) {
         $dodatok_kv = mysql_query("INSERT INTO `dodatoci_hotel` (id_dodatoci, info_id) VALUES ('$dodatok', '$info_id')") or die('ERROR INSERTING: '.mysql_error());
     }
}

My problem is to loop through all checkboxes, and for each checked, populate a separate record in a database. Actually I don't know how to recognize the which box is checked, and put the appropriate value in db.

Upvotes: 1

Views: 15033

Answers (6)

bastiandoeen
bastiandoeen

Reputation: 608

Well, as Eli wrote, the POST is not set, when a checkbox is not checked.

I sometimes use an additional hidden field (-array) to make sure, I have a list of all checkboxes on the page.

Example:

<input type="checkbox" name="my_checkbox[<?=$id_of_checkbox?>]">
<input type="hidden" name="array_checkboxes[<?=$id_of_checkbox?>]" value="is_on_page">

So I get in the $_POST:

array(2){
 array(1){"my_checkbox" => array(1){[123]=>"1"}}
 array(1){"array_checkboxes" => array(1){[123]=>"is_on_page"}}
}

I even get the second line, when the checkbox is NOT checked and I can loop through all checkboxes with something like this:

foreach ($_POST["array_checkboxes"] as $key => $value)
{
  if($value=="is_on_page")
  {
    $value_of_checkbox[$key] = $_POST["my_checkbox"][$key];
    //Save this value
  }
}

Upvotes: 0

dede
dede

Reputation: 2663

This is the loop that I needed. I realized that I need a loop through each key with the $i variable.

if(isset($_POST['id_dodatoci'])){
    $id_dodatoci=$_POST['id_dodatoci'];
    $arr_num=count($id_dodatoci);
    $i=0;
    while ($i < $arr_num)
    {
        $query="INSERT INTO `dodatoci_hotel`(id_dodatoci,info_id) 
            VALUES ('$id_dodatoci[$i]','$info_id')";
        $res=mysql_query($query) or die('ERROR INSERTING: '.mysql_error());
        $i++;
    }
}

Upvotes: 0

Bertrand Gorge
Bertrand Gorge

Reputation: 384

Also something that few people use but that is quite nice in HTML, is that you can have:

<input type="hidden" name="my_checkbox" value="N" />
<input type="checkbox" name="my_checkbox" value="Y" />

and voila! - default values for checkboxes...!

Upvotes: -1

Eli
Eli

Reputation: 99328

2nd Answer:

You might do something like this:

HTML:

echo '<input type="checkbox" name="id_dodatoci[]" value="'.$id_dodatoci.'" />';

PHP:

if ( !empty($_POST["id_dodatoci"]) ) {
    $id_dodatoci = $_POST["id_dodatoci"];
    print_r($id_dodatoci);
    // This should provide an array of all the checkboxes that were checked.
    // Any not checked will not be present.
} else {
    // None of the id_dodatoci checkboxes were checked.
}

This is because you are using the same name for all of the checkboxes, so their values will be passed to php as an array. If you used different names, then each would have it's own post key/value pair.

This might help too:

http://www.php-mysql-tutorial.com/php-tutorial/using-php-forms.php

Upvotes: 0

Ryan McCue
Ryan McCue

Reputation: 1658

As an extra note: You're using the wrong HTML syntax for IDs and <label>. <label>'s "for" attribute should point to an ID, not a value. You also need unique IDs for each element. The code you have posted would not validate.

Also, you're not validating your code at all. At the very least, do a htmlspecialchars() or htmlentities() on the input before you output it and a mysql_real_escape_string() before you insert data into the DB.

Upvotes: 0

Eli
Eli

Reputation: 99328

You can tell if a checkbox is selected because it will have a value. If it's not selected, it won't appear in the request/get/post in PHP at all.

What you may want to do is check for the value of it and work based on that. The value is the string 'on' by default, but can be changed by the value='' attribute in HTML.

Here are a couple snippets of code that may help (not exactly production quality, but it will help illustrate):

HTML:

<input type='checkbox' name='ShowCloseWindowLink' value='1'/> Show the 'Close Window' link at the bottom of the form.

PHP:

if (isset($_POST["ShowCloseWindowLink"])) {
    $ShowCloseWindowLink=1;
} else {
    $ShowCloseWindowLink=0;
}

        .....


$sql = "update table set ShowCloseWindowLink = ".mysql_real_escape_string($ShowCloseWindowLink)." where ..."

(assuming a table with a ShowCloseWindowLink column that will accept a 1 or 0)

Upvotes: 3

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