Whats wrong with this line?

mattm591

Hi, Could anyone tell me what is wrong with this line

$field . $counter+1 = $_POST["field"] . $counter+1;

it gives me the error

Parse error: syntax error, unexpected '=' in /home/matt4617/public_html/cpcw/dostyle.php on line 7

I'm sure I've just got the syntax wrong or something. Any help appreciated.
Matt

Piranha

If you instead tell what you want it to do maybe someone can help you out. As it is now I have no idea what it should do, and then it is impossible to know how you can do it.

But there are several things that is strange. First PHP want one variable to the left of the = sign. Second I don't think it is possible to use the . the way you use it. And third I don't think it is possible to use $counter+1 on the left of the =.

mattm591

Piranha wrote:
If you instead tell what you want it to do maybe someone can help you out. As it is now I have no idea what it should do, and then it is impossible to know how you can do it.

But there are several things that is strange. First PHP want one variable to the left of the = sign. Second I don't think it is possible to use the . the way you use it. And third I don't think it is possible to use $counter+1 on the left of the =.

Ok then well what I want is for there to be several variables called field each followed by a number and then each receives data from the form. So the entire loops looks like this

$counter = 0;
while ($counter < $field){
$field . $counter+1 = $_POST["field"] . $counter+1;
$counter=$counter+1;
}

The line which isn't working needs to have the $field variable and then a number after it, so it creates variables $field1, $field2 and keeps creating them for the number it's been told and then on the right hand side it gets the data its had from field1 and then 2 etc. which is why I use $POST[$field] . $counter+1 although I think on the right it needs to be $POST[$field . ($counter+1)].

Piranha

The best way is probably to use an array:

$counter = 0;
while ($counter < $field){
$counter++;
$field[$counter] = $_POST["field"][$counter];
}

I made a few smaller adjustments. First I increment the counter in the start, then you don't need to add to it all the time. Second I used $counter++ that is equal to $counter = $counter + 1.

mattm591

Ok thanks a lot. I've never used an array before so I have a quick question on it. Does this create variables which are $field1, $field2 etc. so that if later on in the script I tell it to echo $field1 it will echo whatever was called from the $_POST["field"1].

mattm591

Sorry to be a pest but I've just developed a second problem as well!

$docnumdo=mysql_query("SELECT * FROM docs");
$docnum=mysql_num_rows($docnumdo);
$counter=0;
while ($counter < $docnum){
$docname=mysql_fetch_array($docnumdo,$counter);
echo "<option>" . $docname['title'] . " (" . $docname['year'] . ")" .  "</option>";
$counter++;
}

When I run this it only outputs half of the items in the table and puts a blank one in every other time. If I put the $counter++ before the $docname=mysql_etc. line it misses out the other half.

I don't know what I've done wrong with this one because it looks fine to me.

Piranha

To find out more about arrays go to www.php.net/array (I hope it works, I only get a Swedish page). The answer to your question is yes and no. It creates a variable with the name $field. That variable is an array, and the elements of the array is accessed with $field[0], $field[1] and so on. It also means that you can use it in loops without any problems.

Basically: Search for arrays and read up on it. You won't regret it since it will help out a lot.

mattm591

Thanks a lot. I've tried it out and it's working fine and yes the link works also. Thank you

Piranha

You can use another way to go through what you got from the database:

while ($docname = mysql_fetch_array($docnumdo)){
   // What you want done
}

By the way, $docname is an array and you use it just like any other array.

mattm591

Piranha wrote:
You can use another way to go through what you got from the database:
while ($docname = mysql_fetch_array($docnumdo)){
   // What you want done
}
By the way, $docname is an array and you use it just like any other array.

Ah. Thank you (again). I really need to get a book on this! My web learning doesn't seem to teach me things like this 😛

Thank you very much for all your help, most appreciated 🙂

mattm591

Sorry, I've stumbled upon another problem.

I have my select box like this

<select name="publications" size="10" multiple>
<?
$docnumdo=mysql_query("SELECT * FROM docs");
$docnum=mysql_num_rows($docnumdo);
$counter=0;
while ($docname = mysql_fetch_array($docnumdo)){ 
$counter++;
echo '<option value="' . $docname["id"] . '">' . $docname["title"] . ' (' . $docname["year"] . ')' .  '</option>';
}
?>
</select>

And thats all working great. But what I want is for the options selected to be submitted with the form.

So I tried what I normally do which is to include a

$selectedpubs=$_POST['publications'];

But then when I echo that I don't get anything. Now after googling and trying to make sense of php arrarys I think it might be because I need to get the data from the list box out into an array. But I can't do that! I'm not sure how I'd write an array which would be able to do that. It doesn't seem to be similar to the previous array problem.

mattm591

Sorted it using a foreach() command

Piranha

Just a tip for the future. Now you have started to ask first and look for answers after that. It will get people irritated that you don't even try before asking, and then you risk that noone will read help you out. We don't mind helping out, but we don't want to do your work for you. Read this to get the idea of how to post.

mattm591

I understand, but i did look first. I ended up getting help from a friend of mine who I hadn't been able to ask before. I'm sorry if it looks like I was simply being lazy, but I had looked and tried myself for quite a while before posting.