RE: [PHP-DB] referencing external functions
| From: | Bastien Koert | Date: | Tue, 15 Feb 2005 18:44:56 +0000 |
| Subject: | RE: [PHP-DB] referencing external functions | ||
| References: | 1 | Groups: | php.db |
| Request: | Send a blank email to php-db+get-38585@lists.php.net to get a copy of this message | ||
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@jozef.info> To: PHP list <php-db@lists.php.net> Subject: [PHP-DB] 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 -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php