[can I use a value="" in a form tag to select a DB table from $_POST?]

what about a hidden field specifying the table associated with the form?

[Total beginner plea for help - Formmail.php]

put this line just before the mail() line

 

ini_set("sendmail_from", "sending@from.com");

 

and see if that makes any change?

 

look into php.ini check values for:

 

SMTP =

sendmail_from =

 

[Hide video div, while audio continues]

first suggestion is try the javascript or ajax  forums.

 

but surely its not too hard to create a small script that on click, triggers hiding the video div, showing the form div, if they are positioned absolutely they should appear exactly where you specify. add a timer into the script that starts when the trigger is clicked, to switch back on timeout.

 

javascript isnt my friend, so I can't give any code, I'm still on my first cuppa of the day

[Card game "pairs" card sorting]

I vaguely recall doing something similar in asp years ago and I think I stored the 'picked cards' in an array and then each time i picked a new card, i looped through array to check it wasnt already picked.

 

 

[Total beginner plea for help - Formmail.php]

it is likely you need to set

ini_set("sendmail_from", $from);

 

try asking also in the 3rd party scripts section, you might get a better response.

[execute php code if javascript number = 0]

you might get more response in the javascript forum

[remove unexpected T_String error]

So the curly brackets around the $row['something'] variables are an alternative to the concatenation?

I not seen this before so am curious to how & why that works.

[Create a Dynamic Comment box for a web page]

woo hoo free praise 

 

drisate has given you a prefect solution, going to book mark it for myself too.

[Create a Dynamic Comment box for a web page]

it sounds like you need an identifing value associated to each page so comments can be tied to that 'pageid' or whatever you might want to call it

[Notice: Undefined index: name]

WHERE username=$username

 

 

should be

 

WHERE username='$username' ;

 

 

 

 

[Notice: Undefined index: name]

your form inputs are not inside the recommended  <form> </form>. move your inputs to echo inside those tags

[If Result >= 6 add link?]

I think at least... your last post was very confusing.

yeah it confused me

[If Result >= 6 add link?]

do you mean something like this?

 

foreach ($myarray as $k => $v) 
{
  if ($v >= 6) echo $link;
}

[Hi im Carlos]

welcome

 

I found this place great already and I only signed up yesterday

 

 

[If Result >= 6 add link?]

that will return the number of rows in your query, so yes, something like this perhaps

 

if(mysql_num_rows($result) >= 6) echo "<a href='somelink'>link</a>";

[Show results for one week?]

this is probably better asked in the mysql forum?

 

 

[Error]

got there in the end, thats the main thing, happy to help.

[Error]

hehe sure, sometimes I just overlook the simplest of errors, we are all human

$page='';
if (isset($_GET['page'])) $page = $_GET['page'];

switch ($page) {
    case "design":
         include ("includes/design.html");
break;
    case "gallery":
         include ("includes/gallery.html");
        break;
    case "contact":
         include ("includes/contact.html");
        break;
    default:
         include ("includes/home.html");

}

include("includes/footer.html");

[PHP Basics (help writing an Array)]

echo "thanks for the code markup advice";

[Error]

also move the include footer out of the switch statement, or it will only appear  in the homepage, not every page

[Error]

yeah sorry a small typo 

 

if (isset($_GET['page'])) $page = $_GET['page']

 

i only closed off one of the )

 

 

[partial code for adding pictures refreshes page and has done nothing]

 

Okay, i was'nt paying attention, because it was late in the evening,

 

The variable in the form was name="picAdd" en should have been addPic.

 

Topic solved.

 

Best regards,

 

Martijn

 

If I had a dollar for everytime I spent hours looking for a silly mistake like this, I'd be very rich by now

[Error]

page is undefined as its not a variable passed in the $_GET when you call the page first off.

wrap the whole if statement in an other if statement that has isset($_GET['page'])

 

personally I would do it this way:

$page;

if (isset($_GET['page']) $page = $_GET['page'];

 

switch ($page) {

    case "design":

        include ("includes/design.html");

break;

    case "gallery":

        include ("includes/gallery.html");

        break;

    case "contact":

        include ("includes/contact.html");

        break;

    default:

        include ("includes/home.html");

 

}

 

 

 

[Need help with listing]

this should get you started:

 

at the top of your <form>, add this line into the echo statement <input type='hidden' name='bookid' value='$bookid'>

 

then after the if($_POST) statement add this

$bookid = $_POST['bookid']

 

do the same for all variables you need to pass or you will see soon as you do this, $booktitle's value  is the last one in the recordset, not the one clicked on , highlighting the same error you have with bookid currently

 

that should be the quickest way to get the script working without a huge reworking of the code inplace. I would suggest though that you look at refining it somewhat in the long run.

[Need help with listing]

you are not passing a bookid in the <form> why not add a hidden input to pass the bookid, and because you dont pull that from the form, when you send the form, the value of $bookid is the one from the last row of record set, not from the form.

I hope that makes sense.

I would suggest either putting the form processing before the recordset loop or just not bother with the variables in the recordset, there is no need to do this

$bookauthor=$row['author']

<td>Author: <strong>$bookauthor</strong></td>

when you can just have

<td>Author: <strong>$row['author']</strong></td>

 

or even better post the form to a seperate page, then once processed, redirect back to book listing page.