php - 如何计算 while 循环中未选中的复选框

Question

$query = mysql_query("SELECT DISTINCT status, vendor FROM tbl_softwareinstalled WHERE vendor NOT LIKE ''");
$nums = mysql_num_rows($query);
echo "<form name = 'filter' action='softwarefilter.php' method='POST'>
$nums<br>

<table border = 1><tr><td>&nbsp;</td>
<td><strong>Software Vendor</strong></td></tr>";

$ctr1 = 1;
while($fetch = mysql_fetch_array($query)) {
$vendor = $fetch['vendor'];
$status = $fetch['status'];
    if(($ctr1 % 2)==1)
        { print "<tr bgcolor = 'white'>";}
        else 
        { print "<tr bgcolor = '#EEEEEE'>"; }
    print "<td><input name= 'chk[]' type='hidden' value='0'>
    <input type = 'checkbox' name = 'chk[]' value = '$vendor' ";




    if ($status == 'Enabled') {
    print "checked = 'checked'></td><td>$vendor</td></tr>"; }
    else {
    print "></td><td>$vendor</td></tr>"; }
    $ctr1++;
}
print "</table>";
print "<input type = 'submit' name = 'submit' value = 'Update Filter'>
</form> ";


$submit = $_POST['submit'];
if(isset($submit))
{ 
    $chk = $_POST['chk'];
    $count = count($chk);

    if (empty($chk)) {

    echo "qweqwe<br>";
    }

        for ($i=0; $i<$count; $i++) {
        $abc = $chk[$i];
        $query = mysql_query("UPDATE tbl_softwareinstalled SET status = 'Enabled' WHERE vendor = '$abc'");

        }

    echo '<meta http-equiv="refresh" content="0.5;url=/assets/softwarefilter.php">
            <script language="javascript">
            alert("Software filter updated.");
            </script>';
}

您好，如何使用这段代码计算 while 循环中的未选中复选框？ 比方说，我有 10 个复选框和 10 个值，用户在提交后选中了 3 个复选框(值:'abc'、'def'、'ghi')。它将计算所有未选中的复选框并回显它的值。不应回显 3 个选中的复选框的值。

Answer 1

浏览器只会告诉您哪些已被选中。从您的数据库中提取所有可能的框，然后删除浏览器告诉您已选中的框。