print failed

TheBeginner

Hi,

Got this file, anntyp.php :

anntyp.php

<?php
$typ=array(
"1"=>"Säljes",
"2"=>"Köpes",
"3"=>"Uthyres",
"4"=>"Sökes",
"5"=>"Erbjudes"
);
echo $typ[$typid];
?>

I will later in another file use above file to print the option like this

part of php

print (" directory/anntyp.php?typid=$number ");

I guess you can't do that with the print command or I use wrong syntax.
Please help!

rIkLuNd

What is the error? If there is no output, then maybe register_globals is set to off, which it should be! Then you must use the $_GET variable to get the passed value.

anntyp.php

<?php  

$typ=array(  

	"1"=>"Säljes",  

	"2"=>"Köpes",  

	"3"=>"Uthyres",  

	"4"=>"Sökes",  

	"5"=>"Erbjudes"  

);  

echo $typ[$_GET[ "typid" ]];  

?>

TheBeginner

he first part of the code is ok and the file behave like I want it to behave.

The problem seems to be:

print (" directory/anntyp.php?typid=$number ");

I don't see any output on the screen from this syntax above. I think it is the wrong way to use the print option and output from a file, but I can't find how to do it.

rIkLuNd

I tested this little piece of code on my testing server and it worked...

<?php

$number = 3;
print (" directory/anntyp.php?typid=$number ");

?>

TheBeginner

Ok that strange. I get:

directory/anntyp.php?typid=3

on the screen, and that is not what I want to get.

rIkLuNd

well, if you just pasted my code into your code, you will have to remove the line $number = 3;! By the way, where does the variable $number come from?

TheBeginner

Hi,

Sorry not answering. Was away in Copenhagen to see "The Boss", Springsteen.

The $number is just a variable comming from a HTML form. But anayway....if I use the code you gave me not changing it, I shuold get the same result everytime....right?

Still the print command don't gave me the right result from the function file. I just give me the phg code. I seems like it doesn't understand that it should be looking for a file in directory/anntyp.php?typid=3 and give an answer. I have no idea why.
If I go directly to the file with directory/anntyp.php?typid=3
it give me the print of the answer.

I have no clue anymore.

rIkLuNd

Well, if the form method is set to post, then try this:

<?php
print( "directory/anntyp.php?typid=" . $_POST[ "number" ] );
?>

If it's set to get, then change $POST to $GET..

TheBeginner

Thanks, but is doesn't matter.
I put this code in a own file and it does't work!:

php

<?php
$number =3;
print ("directory/anntyp.php?typid=3");
?>

It stills seems that if I use print and use "-sign around the funktion I want to give me the output I get on my screen only:

directory/anntyp.php?typid=3

and not the answer! It seems that it just don't go to my file to pick up the answer and the php thinks that it is only plain text within the "... ".

TheBeginner

I also tried:

php:

number = 3;

print (" directory/anntyp.php?typid=$number ");
print (" \"directory/anntyp.php?typid=$number\" ");
print (' directory/anntyp.php?typid=$number ');
print ([directory/anntyp.php?typid=$number ]);

Everything is failing. Is it that hard to get output from another php-file and then print it???

rIkLuNd

Try this code:

form.html

<form action="action.php" method="post">
<input type="submit" name="text" value="Press this button"/>
</form>

action.php

<?php
echo $_POST[ "text" ];
?>

Does this work?

TheBeginner

Yes it works fine.

But my old one still do not.......
I think is the problem with the print syntax, but I really cant't find the problem.

rIkLuNd

Try this:

print( "directory/anntyp.php?typid=" . $number );

TheBeginner

It gives me the answer I have got for 2 days now:

On the screen:

directory/anntyp.php?katid=3

It just don't want to recognize that it is a function it should execute!

rIkLuNd

Now I'm getting confused? Are you printing that on the screen to activate the other script?

TheBeginner

No, I hope that the expression in the print(" "); will be executed and give me the answer (the text) that my function anntyp.php gives as an output.

So this is what I want:

Have my php where I call anntyp.php?typid=3
anntyp.php gives an output like "some text"
My file take this answer and print it on the screen.

rIkLuNd

Maybe you can post some of the code here? Not just the one in anntyp.php but maybe some other code too. It's easier to help then...

TheBeginner

anntyp.php

<?php

$typ=array(
"1"=>"Säljes",
"2"=>"Köpes",
"3"=>"Uthyres",
"4"=>"Sökes",
"5"=>"Erbjudes"
);

//echo $typ[$typid];

echo $typ[$_GET[ "typid" ]];
?>

check.php

<?php
$number = 3;
print ("funktion/anntyp.php?typtid=$number");
?>

</BODY>
</HTML>

This code should print on the screen: "Uthyres", but
is print instead: funktion/anntyp.php?typtid=3

rIkLuNd

Well, not I understand what you're doing wrong. Try going to this link in your web browser and check the result:
http://www.your-domain.com/anntyp.php?typid=2

Then it will print the stuff you want it to. The [man]print()[/man] function will ONLY print stuff on the screen, it will not execute something!

TheBeginner

Yes the answer is then right. I knew that.

But how can I execute first the anntyp.php, get the answer and the print it from my php - file?