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 } ?> />

Answer 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.