Insert checkbox data into database, based on if it is checked or not

Question

HTML:

<form action="add.php" method="get" id="form">
  <tr>
    <td>SALE ITEM</td>
    <td>
      <input type="checkbox" name="saleItem" value="n" <?php if(isset($_GET['saleItem'])) { ?> checked="checked" value="y" <?php } ?> />
    </td>
  </tr>
</form>

PHP (add.php)

$sql=("INSERT INTO table (sale) VALUES ('" . $_GET['saleItem'] . "')");
if (!(mysql_query($sql))) { die ("could not query: " . mysql_error()); }

It always inserts an N regardless if it is checked. I would love if it inserted y based on if the checkbox is checked. I also tried to put the value="n" into an else after the if, but it always put N regardless as well (i removed the value="n" previous to the if when I did this), here is what I meant:

<input type="checkbox" name="saleItem" <?php if(isset($_GET['saleItem'])) { ?> checked="checked" value="y" <?php } else { ?> value="n" <?php } ?> />

I want to it to send an N into my database if it is not checked, and a Y if it is checked. — eveo
– eveo, Commented Mar 16, 2012 at 20:48

nathanjosiah · Accepted Answer · 2012-03-16 20:52:02Z

2

You would change your html to:

<input type="checkbox" name="saleItem" value="1" <?php echo (isset($_GET['saleItem']) ? 'checked="checked"' : '')?> />

Then change the PHP to:

$is_checked = (isset($_GET['saleItem']) ? 'y' : 'n');

and then your SQL statement should look like:

$sql=("INSERT INTO table (sale) VALUES ('" . $is_checked . "')");

If a checkbox is not checked when the form is submitted, the browser doesn't send it's value to the server. So changing it's value to 1 and then checking server-side if it exists at all, is the best way to determine the state of the checkbox.

answered Mar 16, 2012 at 20:52

nathanjosiah

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2 Comments

Shattuck Over a year ago

And your sql statement to $sql=("INSERT INTO table (sale) VALUES ('" . $is_checked . "')");

nathanjosiah Over a year ago

Thank you, I added that to my answer.

Jazz · Accepted Answer · 2012-03-16 20:58:22Z

1

A checkbox can have only one value. Your attempt to change the value attribute depending on whether it is checked or not won't work.

Here's updated HTML for your checkbox:

<input type="checkbox" name="saleItem" value="Y" <?php if( isset( $_GET['saleItem'] ) ){ ?> checked="checked" <?php } ?> />

This will return a "Y" to the server if it is checked, but it will not return at all if it is not checked (this is the way the checkbox element is designed to work). So you need to have your PHP code explicitly check to see if anything was returned:

$saleItemValue = "N";
if( array_key_exists( 'saleItem', $_GET ) && $_GET['saleItem'] == 'Y' ){
    $saleItemValue = "Y";
}

$sql = "INSERT INTO table (sale) VALUES ( '$saleItemValue' )";
if( !mysql_query($sql) ){
    die( "could not query: " . mysql_error() );
}

This way, PHP will place an 'N' in the database if nothing comes back -- i.e. if the checkbox was not checked.

answered Mar 16, 2012 at 20:58

Jazz

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1 Comment

eveo Over a year ago

Absolutely perfect, worked. Thank you very much! I understand it now.

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