I have a isset($_REQUEST['form_submitted']) with these 4 entries (among other) in the form:
<div class="formLeft">
<input name="comic" type="checkbox" id="comic" value="1" <?php if (isset($auswahl)){if($dbcomic =='1'){ echo "checked";}} ?> />
</div>
<div class="formLeft">
<input name="comic" type="checkbox" id="comic" value="0" <?php if (isset($auswahl)){if($dbcomic =='0'){ echo "disabled";}} ?> />
</div>
<div class="formLeft">
<input name="onoff" type="checkbox" id="onoff" value="0" <?php if (isset($auswahl)){if($dbonoff =='0'){ echo "disabled";}} ?> />
</div>
<div class="formLeft">
<input name="onoff" type="checkbox" id="onoff" value="1" <?php if (isset($auswahl)){if($dbonoff =='1'){ echo "checked";}} ?> />
</div>
$auswahl is called before here:
$query = mysqli_query($link, "SELECT * from aktuelles where id = '$auswahl'");
while ($zeile = mysqli_fetch_assoc($query)) {
$dbonoff = $zeile['onoff'];
$dbcomic = $zeile['comic']; ... }
If the form is sent it uses:
if(isset($_POST['submit'])){$submit = $_POST['submit'];}
if(isset($_REQUEST['auswahl'])){$auswahl = $_REQUEST['auswahl'];}
and
if ($submit == 'aktualisieren'){
if (isset($_REQUEST['form_submitted'])) {
...
if(!empty($img)){$img = $img;} else {$img = $dbimg;}
mysqli_query($link, "UPDATE aktuelles set title = '$title',..., ablaufdatum = '$ablaufdatum', onoff = '$onoff', comic = '$comic' where id = $id");
}
It mainly works, but if just $dbonoff is checked and $dbcomic is disabled and not checked or the other way, I can not just disable $dbonoff as well, it turns into a white error. It works as soon I turn also $dbcomic as is checked or the other way. There must be an action on both sides.
Don't know how to solve this problem.
help for my problem and acknowledge...