PHP Forum

**tirengarfio** · 01-09-2009, 03:58 AM

Hi,

im just trying to pass a variable from a page to another.

My problem:

im getting this error on the browser:

Code:


	Code:
	Not Found
The requested URL /red_social/mostrar_amigos.php was not found on this server.

These are the two pages:

PHP Code:


			
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01//EN" "http://www.w3.org/TR/html4/strict.dtd">
<html>
<head><title></title>

<style type="text/css" media=all>

</style>
</head>

<body>


<?php

//Apuntador de enlace
$link=mysql_connect("localhost", "root", "") OR die(mysql_error()); 

//Asociacion de enlace con base de datos
mysql_select_db("red_social",$link) OR die(mysql_error()); 

//Maneja todas las variables string de las consultas con comillas sencillas e indicale al query la variable de enlace para saber sobre que base de datos apuntar
$result = mysql_query("select * from amigos where email='$correo_electronico'",$link) OR die(mysql_error());  

// extraemos la primera fila de "amigos" y la pasamos a "$row".
$row = mysql_fetch_array($result);


if($row["password"]==$contrasena)

{

?>


<span id="saludo">

<span id="hola">
Hola
</span>

<?php
echo $row["nombre"];
?>

<span id= "elige..."> 
elige una de las siguiente opciones:
</span>

</div>

<div id="mostrar...">



<a href="mostrar_amigos.php?correo= <?php echo $correo_electronico; ?> ">Mostrar amigos</a><br>
Mostrar compañeros universidad<br>
Mostrar compañeros trabajo<br>
Buscar entre los usuarios registrados<br>



</div>

<?php

}

?>


</body>

</html>

PHP Code:


			
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01//EN" "http://www.w3.org/TR/html4/strict.dtd">
<html>
<head><title></title>





</head>

<body>

<?php


//Apuntador de enlace
$link=mysql_connect("localhost", "root", "") OR die(mysql_error()); 

//Asociacion de enlace con base de datos
mysql_select_db("red_social",$link) OR die(mysql_error()); 

//Maneja todas las variables string de las consultas con comillas sencillas e indicale al query la variable de enlace para saber sobre que base de datos apuntar
$result = mysql_query("select * from amigos where email='$correo_electronico'",$link) OR die(mysql_error());  

// extraemos la primera fila de "amigos" y la pasamos a "$row".
$row = mysql_fetch_array($result);


if($row["password"]==$contrasena)
{

?>

<span id="saludo">

<span id="hola">
Hola
</span>

<?php
echo $row["nombre"];
?>


<span id= "estos..."> 
estos son tus amigos:
</span>

</div>


<?php


echo $_GET['correo'];

?>

</body>

</html>

Any idea?

**stoot98** · 01-09-2009, 04:42 AM

That doesnt look like a PHP problem - it looks like the page you are linking to doesnt exist!

**tripy** · 01-09-2009, 05:10 AM

You might have a problem in the path...
Take care of the case, if your server is linux/unix based.
A page named "mostrar_amigos.php" is not the same as "Mostrar_amigos.php".

**tirengarfio** · 01-09-2009, 05:13 AM

Quote:

Originally Posted by stoot98

That doesnt look like a PHP problem - it looks like the page you are linking to doesnt exist!

I dont have any problem showing page (mostrar_amigos.php) when i delete the text which sends the variable (?correo= <?php echo $correo_electronico; ?>), so i think the page exists.

**tirengarfio** · 01-09-2009, 05:16 AM

Quote:

Originally Posted by tripy

You might have a problem in the path...
Take care of the case, if your server is linux/unix based.
A page named "mostrar_amigos.php" is not the same as "Mostrar_amigos.php".

"Mostrar_amigos.php" is not written any place.

**tripy** · 01-09-2009, 05:34 AM

Quote:

Originally Posted by tirengarfio

"Mostrar_amigos.php" is not written any place.

It was an example that you might have your file named with a different case than you refer in your file.
I cannot see your hdd content, so I cannot tell.

**tirengarfio** · 01-09-2009, 05:49 AM

Quote:

Originally Posted by tripy

It was an example that you might have your file named with a different case than you refer in your file.
I cannot see your hdd content, so I cannot tell.

In my hdd: "mostrar_amigos.php"

**jason_alan** · 01-09-2009, 06:02 AM

what's the url of the first page you're linking FROM? It looks to me like it is in the red_social folder and mostrar_amigos.php is not.

**tirengarfio** · 01-09-2009, 06:14 AM

Quote:

Originally Posted by jason_alan

what's the url of the first page you're linking FROM? It looks to me like it is in the red_social folder and mostrar_amigos.php is not.

This is the first code's path:

[root@andLinux red_social]# pwd comprobacion_logeado.php
/opt/lampp/htdocs/red_social

and this is second code's path:

[root@andLinux red_social]# pwd mostrar_amigos.php
/opt/lampp/htdocs/red_social

**tripy** · 06:41 AM

Ok. Then, are you sure that the apache root folder is /opt/lampp/htdocs/, and what are the permissions of the red_social folder ?
Usually, you must have the execution right for everybody on the folder for apache to access it, if it's not owned by the declared apache user.
do a

Code:

ls -la

in your htdocs folder to look at the red_social permissions.
They should looks something like this (an example from my own home server)

Code:

drwxrwxr-x  19 apache  webdev      4096 Nov 10 15:21 workspace

It's the 1st column that is important.
d means "directory", then it's 3 series of "rwx" meaning "read - write - execute".
The first series relate to the owner of the file (here: apache), the second to the members of the group (here: webdev), and the third is the permissions to everybody not in the 2 first cases.

In this case, apache user and members of the group webdev have full rights, and everybody else have execution and read rights only.

**tirengarfio** · 01-09-2009, 07:07 AM

Hi again,

i changed the permission of "mostrar_amigos.php" in this way:

Code:

sudo chmod 777 mostrar_amigos.php

and now it works ok.

Thanks!

**tripy** · 01-09-2009, 08:27 AM

Good, I don't know why, but I've seen that the bit "execute" should always be set to true.
Just "read" is not sufficient to have apache accessing it.
Or it's on the folder, I quite don't remember exactly, but there is something about this "execute" bit.

**anderswc** · 01-10-2009, 10:26 PM

well, PHP files are executed, so that would make sense

**tripy** · 01-11-2009, 08:47 AM

Quote:

Originally Posted by anderswc

well, PHP files are executed, so that would make sense

Nope, not in the sense of a linux execution though.
PHP files are resources that are used to give instructions to the php interpreter.

If you try to do an

Code:

./myScript.php

from the terminal, you will get back:

Code:

./dmpOrders.php
./dmpOrders.php: line 1: ?php: No such file or directory
./dmpOrders.php: line 2: =: command not found
./dmpOrders.php: line 3: syntax error near unexpected token `('
./dmpOrders.php: line 3: `$db= pg_connect($conn_string);'

because the shell tries to interpret the script itself.

I know that the execute bit on a folder means "permission to list the content".
If you have the read right but not the execution right on a folder, then you are blind.
If you specify the path and that the file exists, it'll work. But you cannot see which files are in the folder.

**anderswc** · 01-14-2009, 02:56 AM

What I meant was that in a sense the PHP interpreter "executes" the instructions in the PHP file. I've always assumed that PHP only interprets files which have the executable flag set to true. So to "execute" a PHP file, you have to make it CHMOD executable.

**jason_alan** · 01-14-2009, 03:13 AM

Well, no, the php interpreter reads the php file (as instructions), to interpret into bytecode. and just needs "read" permission to do this