Does anyone know how to fix this problem? - Read Last Post

[j3rmain3]

I want to display information from several dropdown menus in a table and it would be too time consuming if i was to type in the name i have given to every option. So i have placed this:

echo "<td><p align=center><font face=Verdana size=-1>[b]".$row[filename]tech"[/b]</center></td>";
echo "<td><p align=center><font face=Verdana size=-1>[b]".$row[filename]fin"[/b]</center></td>";
echo "<td><p align=center><font face=Verdana size=-1>[b]".$row[filename]doc"[/b]</center></td>";

The names in the dropdown menu end in either TECH, FIN or DOC.

This is how the data in the dropdown menu has been setup:

<option value=1 name=$row[filename]tech>

This works fine because the name of the option is stored as 'segatech' like it is supposed to. But this only works becasue i have the whole thing within quotes.

I have placed the table like this so the data from the dropdown menu will appear in the cell. But i keep on getting an error which says i cannot have the $row[filename][b]tech[/b] bit.

Is there anyway i can work around this.  ???

Thanks,
J3rmain3

[blear]

The array cant be concatinated into HTML output that way.

[code]...size=-1>".$row[filename]tech"</cen.... [/code]

should become

[code] size=-1>" . $row[filename] . "tech</cen.... [/code]

[j3rmain3]

One more problem im having, i have created a form and a table which i want to display the data in. To move the data from the form to the table, i know that i have to place something like this in at the top:

$variable1 = $_POST['variable1'];
$variable2 = $_POST['variable2'];
$variable3 = $_POST['variable3'];

Is there anyway i can turn this into a loop so that variable4 will appear next without having to do it manually.

Thanks,
J3rmain3.

N.B. I have asked this question 3 times b4 and nobody as helped me so please (thats me pleading) somebody help me.

[Zane]

extract($_POST);

[blear]

j3r-

No, you cant assign $variable4, $variable5, $variable6, etc. inside a loop. That has to be done with an array. Instead of $variable4, you have $variable[4]. You will need to read up on arrays and their usage.

[Monkeymatt]

You [b]can[/b] assign $variable4, $variable5, $variable5, etc inside a loop:

[code]
$num=5; // number of variable(num) to extract
for ($i=1;$i<=$num;$i++) {
    $go="variable".$i;
    $$go=$_POST[$go];
}
[/code]

http://us2.php.net/manual/en/language.variables.variable.php

Monkeymatt

[Zane]

yeah your right, he can do that

but look at his keys for POST
$variable1 = $_POST['variable1'];
$variable2 = $_POST['variable2'];
$variable3 = $_POST['variable3'];

they're just the same as the name of the variables he wants
so he'd be just as well off using extract.

I don't know why you don't just use the POST vars themselves, directly

$variabel1 = $_POST['variable1'];
echo $variable1;

Does the same thing as

echo $_POST['variable1'];

nothing really different at all.

[j3rmain3]

[code]
$num=5; // number of variable(num) to extract
for ($i=1;$i<=$num;$i++) {
    $go="variable".$i;
    $$go=$_POST[$go];
}
[/code]

I have used the coding which MonkeyMat has explained. I looked up extract but did not understand it very well.

I have been able to use the foundations of the coding above and applied it to my table, but i am confused about something. To see if the coding comes out correct i used [b]echo[/b] >>>>> This is how i printed out the coding,

[code]

$x = "$";

echo "$x$go = ${$go}";

[/code]

the output to this is $variable1 =
But the $_POST[variable1] bit doesnt appear. I understand that it will not appear because its a reserved function (or something like that ) but how do i get this to be recognised by the server so that i can send information from the form to the table.

Thanks
J3rmain3

[j3rmain3]

I am trying to display numbers which have been sent from a form but i get the error message:

Parse error: parse error, unexpected T_CONSTANT_ENCAPSED_STRING, expecting ',' or ';' in C:\Apache\Apache2\PHPFiles\brainstorm\btable.php on line 42

This is the code:

while ($row=mysql_fetch_assoc($result)){

$num = $row;
for ($i=1;$i<=$row;$i++) {
$post = $row['filename']."tech";
$$post = $_POST[$post];
}
[b]42:[/b]  echo "<td><font face=Verdana size=-1>".${$post}"</td>";

Can anyone see what is wrong IN LINE 42? Isn't that the way i display something using variable variable

Thanks
J3RMAIN3