RE: referencing external functions

"Bastien Koert" <bastien_k@xxxxxxxxxxx> · Tue, 15 Feb 2005 13:44:56 -0500

the error tells me that the files are not in the same dir...I have run into 
this one as well and it was always that case.

Your usage is off though...What you need to do is to include the 
functions.php page and then call a particular function..also any error 
handling should be with the function itself

I handle mine a little differently....try this:
<?php
//remote_conn.php
//check the password from the DB and get the access level = type
function conn($sql)
{
               $username = "root";
	$pwd      = "*****";
	$host   	= "localhost";
	$dbname 	= "apis";

	//echo "commnecing connection to local db<br>";

	if (!($conn=mysql_connect($host, $username, $pwd)))  {

	    printf("error connecting to DB by user = $username and pwd=$pwd");

	    exit;

	}

	$db3=mysql_select_db($dbname,$conn) or die("Unable to connect to local 
database");

	$result = mysql_query($sql) or die ("Can't connect because ". 
mysql_error());

	return $result;

}//end function
?>

Then I include the function like this

require("conn.php");

$sql = "select * from table";

$result = conn($sql); //runs the query and passes back the result

if ($result){
//do something here
}
?>

In your case...i changed your code somewhat....

<?php
function connect() {
//global $link;
$result = "";
	if (!$link = mysql_connect("localhost","username","password")){
	   $link = mysql_error();
	}
	 if(!mysql_select_db("database",$link))
	 {
	 $result = mysql_error();
	 }
	if ($result==""){
	  return $link; //pass the link without  globals
	}else{
	  die($result);
	}
}

function links() {
//global $link;
$sql = "select * from jozef_links";
$link = connect(); //call the connect function to create the link
$result = mysql_query($sql, $link) or die (mysql_error());
echo "<li>";
while ($array = mysql_fetch_array($result)) {
  $id = $array['link_name'];
  $url= $array['link_url'];
  echo "<ul><a href=\"$url\">$id</a></ul>\n";
}
echo "</li>";
}
?>

<?
include("functions.php");
links();  //call the links function
?>

bastien

From: "J. Connolly" <web@xxxxxxxxxx>
To: PHP list <php-db@xxxxxxxxxxxxx>
Subject:  referencing external functions
Date: Tue, 15 Feb 2005 12:45:29 -0500

Sorry if this is a duplicate. I accidentally sent to the listserve from the 
wrong address.

I am having trouble calling external functions. I am trying to use this to 
clean up the code in the display. I have this as the function.php

<?php
function connect() {
global $link;
$link = mysql_connect("localhost","username","password")
  or die ("Could not Connect to DB");
mysql_select_db("database",$link) or die (mysql_error());
}

function links() {
global $link;
$sql = "select * from jozef_links";
$result = mysql_query($sql, $link) or die (mysql_error());
echo "<li>";
while ($array = mysql_fetch_array($result)) {
  $id = $array['link_name'];
  $url= $array['link_url'];
  echo "<ul><a href=\"$url\">$id</a></ul>\n";
}
echo "</li>";
}
?>

and am trying to use it in this links3.php

<?php
include('functions.php') or die (mysql_error());
?>

but I keep getting the error message

*Warning*: main(1): failed to open stream: No such file or directory in 
*C:\Accounts\dbadream\wwwRoot\links3.php* on line *4*

*Warning*: main(): Failed opening '1' for inclusion 
(include_path='.;c:\php4\pear') in 
*C:\Accounts\dbadream\wwwRoot\links3.php* on line *4*

Can anyone explain what is wrong. Both files are in the same directory and 
and both have the normal permissions set to access.

Thanks,

jzf

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