PHP - Hard Php Question, Regarding Files

i am trying to make a page to upload music to my website and add the information of the songs into a database, but i cannot get php to let me upload a file bigger than 2mb even after i went into the php.ini and changed the max_upload_size and a few others to 99mb.
So i was wondering if i could put my music in a folder on my website and pull the data from the items in the folder and make database entries(probably doubtful) or if i could get some help with getting php to upload bigger files?

thanks,
Adam

I'm a little confused with the use of absolute paths in PHP.  I've always used relative paths but it's becoming an issue with some of the apps I'm working with and absolute would solve the problem.

The thing is, I'm getting different results in different places and I'm a little confused.

With regular HTML I seem to be able to use / as the root of my site.  So if I do the following...

Code: [Select]
<img src="/images/name.jpg" />
That works fine no matter what page it's called from in the site and no matter what directory that page is in.  I need to be able to do that same thing with PHP.

Now, in my custom 404 page I have the following...

Code: [Select]
if($AddressURL == 'http://www.domain.com/blog/files/filename.zip')
{
header('HTTP/1.1 301 Moved Permanently');
header('Location: /download/filename.zip');
exit();
}

That actually works just fine. If I go to the /blog/files/filename.zip (which doesn't exist) in a web browser it winds up downloading the file at www.domain.com/download/filename.zip as expected. 

Now where I get confused is that include files don't seem to work like that with PHP.  If I do this...

Code: [Select]
require_once('/includes/config.php');

That winds up looking for an /includes folder all the way back at the local server root, not just the web root.  so it's looking for /var/includes instead of /var/www/includes and of course that doesn't end up working.

So, how come the include file paths work differently than regular HTML paths or header redirects in PHP?  Is there any simple way I can always refer to the SITE root no matter what server it's running from?  I've tried playing with $_SERVER['DOCUMENT_ROOT'] but it only seems to return the root of the current document, not the root of the whole site.  So if I use it within a php file located in /var/www/includes/test.php it considers /includes the root which is not what I want.

Any information would be greatly appreciated.  Thanks!

This may sound like a very stupid question, but I can't find a clear answer anywhere.

I'm making my first dive in OOP and all the tutorials only really show basic usage. They define a class, and show how to use it in a file.

My question is how does it work to use an object throughout a program? Is it available everywhere or do I store things in sessions and start a new object on each page?

For example: logging in a user. If I created an instance after confirming the credentials, would that same instance be available for use in say index.php, or would I just pass a session over and create a new instance on each page?

Again, sorry if this seems dumb to some people.

So say i have this:

Code: [Select]
<link rel="stylesheet" media="all" type="text/css" href="_css/styles.css" />
now the file that is linking to that is 3 directories deep, besides using a rel or absolute link, could i define the path by using something like this

Code: [Select]
<link rel="stylesheet" href="<?php require($_SERVER['DOCUMENT_ROOT'] . "/_css/portfolio-styles.css"); ?>" type="text/css" />
but this is actually opening the CSS in the header

thanks for any explanation on how to do this

I something like this to happen... A user has a text field and a button, he enters text in the field and clicks the button it will go to a URL. Once it goes to that URL it has a text field, this field would be automatically completed from what the user typed in at the start. On that URL there is also a button, that is automatically clicked once the text field has been automatically entered. BUT the user only sees the first page where he entered text in the field and clicked the button. Everything that happens after he clicked that button he cannot see.   Is there a particular phrase you would call this from what is happening? If so, what is it and how would I do it? This probably is hard to understand but I tried :/   Thanks

Get a hugh problem with a work in school!

We are suposed to optimize a website for moblie devices..

Just by using xhtml, css and php.

the problem is that I want the meny to just show at the front page for mobile devices..
I konw that I can choos pic with
< img >
< media="only screen and (min-width: 640px)" src="bild1.jpg" />
< media="only screen and (min-width: 800px)" src="bild2.jpg" />
</ img >

is it possible to do a if construct in any point with using media?

please help, Im lost.

I get the following error:
Quote

Parse error: syntax error, unexpected ';' in /hermes/web06/b1946/moo.rghollenbeck/index.php on line 35


But here is the only php code on the whole page:


. . .more html code above
<?php 

// PHP version: 5.2.12
// MySQL Version: 5.0.77

$link = mysql_connect('path', 'username', 'pword'); 
if (!$link) { 
die('Could not connect: ' . mysql_error()); 
} 
mysql_select_db(guestlist); 

$Users_IP_address = $_SERVER["REMOTE_ADDR"] ;
echo "<input type='hidden' name='ip' value=";
echo $Users_IP_address;
echo ">";

$first_name=mysql_query("SELECT 'fname' FROM tblMain; WHERE 'ip'=" . $Users_IP_address);

// pseudocode section:
// If $first_name !null{
// echo "Hello " . $first_name;
// }
// end pseudocode section:
// something like that
?>
more html code below. . .

Thank you.

Ok all I need some help with this. I have been stuck for two days. Basically I have a form that asks a registrant their gender and age, then my script spits out a list of race categories they qualify for. The situation calls for the registrant to be able to register for any category YOUNGER than them, but not older.

Screen Shot #1 shows my database setup (this can be changed if you have any better ideas...) that holds the race category information.

Here is the sql statement I am using to pull a female racer information, where $race_age is their submitted age.

"SELECT * from counts where female=1 and $race_age < age_end and $race_age > age_start"

This works, but doesn't pull any categories YOUNGER Than their age.

I'm stumpped Thanks for any help in advance.

Hey guys,

this is what I'm trying to do.
I want to print on the member's profile page, a list of the products he has bought so far.

I've come with two solutions:

A) add a new field to the products' table, with the ID of every clients that bought that product;
B) add a new field to the members table, with the ID of every products he bought;

These are pretty similar approaches, but I've no idea on how can this be done.

It would look like:

Bought: 2, 3, 4, 5,

I dont know how to write this data or how to read it.

Any idea?
Thanks!

hey guys

so I have this code:
$result = mysql_query("SELECT * FROM pec_mssgs  ORDER BY (y / m) DESC, d ASC");

while($row = mysql_fetch_array($result))
  {
  
  
echo "&bull;<a style='text-decoration: underline; font-weight: bold;' href='../index?id=".$row['id']."'>". $row['m']."/".$row['d']."/".$row['y'] . "  -  ".$row['title']."</a>";
echo "<br /><br />";


  }

in my database instead of doing date I have 3 rows d, m, y, which correlate to day, month, year.

now currently its echo out a list and sorting by year then month then day this way the events which are happening soonest will be at top.

the problem is it shows events that has already passed.

what do I do?

It will be way to much work to to back and recode the application to just use date.

is there a way I can create a var. using my d/m/y then compare that var. to the current date or something?

thanks for the help.

In ubuntu 18.10 ,  i created a folder name db in the external usb hard dirve changed the ownership to the www-data the apache2 server and set the permissions to 777 , i run this php code and get the permission error:

<php
 chdir('/media/user/e46cafba-2656-4e42-b1c9-6417f28839df/db/');
 system('./rs');
?>    

but the php code cannot change to the dir and get the error:

[21-Feb-2019 10:43:26 UTC] PHP Warning:  chdir(): Permission denied (errno 13) in /home/user/www/html/rse4db.php on line 5

 

I'd like to detect emails that hard bounce after i send out my newsletter, and then delete them from the table that stores them in my database.

Any ideas? Thank you.

Hi guys,

Here's my problem.  My company wants to use SSRS to generate reports.  The backend, what stores the data for the report is MS SQL 2005 (and no, I can't upgrade it to 2008, we don't have the licenses for it).

When I used one of their examples to login, I got an error stating that the URL, username or password might not be correct.  But when I do this, I connect just fine:
Code: [Select]
$client = new SoapClient("http://sampleURL/reportserver/reportexecution2005.asmx?wsdl", array('login' => "lmaster\SVCWEBDOCS", 'password' => "pdfconverter"));
Now, honestly, I don't know much about this and I'm still doing some research, has anyone tried to make this SDK spit out a report?  I'd appreciate some guidance/direction as to what to do next.  Also, if you could tell me about some of your experiences in getting to work, I'd really appreciate it.

This topic has been moved to MySQL Help.

http://www.phpfreaks.com/forums/index.php?topic=328677.0

Hi all,

I have created a successful registration and login page for what will eventually be a squash league website.

I am struggling with the following issue.

When a player logs in using their email address and password I want to populate some drop down boxes which are relevant to the person who has just logged in.

These drop downs should show all the leagues the logged in user belongs to and the other drop down should show all the players the currently logged in player has ever played against so that they could view a head to head.

So my code works but currently I have the id of the player who has logged in hard coded in my sql statements.

I'm wondering if anyone can help me so that I can pass in the id of the player who is logged in as obviously the whole site is not going to work with hard coded values.

The code is as follows:

Code: [Select]
<?php

include_once $_SERVER['DOCUMENT_ROOT'] . '/includes/magicquotes.inc.php';

require_once $_SERVER['DOCUMENT_ROOT'] . '/includes/access.inc.php';

if (!userIsLoggedIn())
{
 include 'login.html.php';
 exit();
}

if (!userHasRole('Squash Administrator'))
{
 $error = 'This page is only for Squash Administrators.';
 include 'accessdenied.html.php';
 exit();
}

if(isset($_REQUEST['email']))
{
 include $_SERVER['DOCUMENT_ROOT'] . '/includes/db.inc.php';
 
 $email = mysqli_real_escape_string($link, $_REQUEST['email']);
 $sql = "SELECT id, firstname, lastname, email FROM player WHERE email = '$email'";
 $result = mysqli_query($link, $sql);
 if(!$result)
 {
  $error = 'Unable to select players firstname from the database.' . mysqli_error($link);
  include 'error.html.php';
  exit();
 }
 
 while($row = mysqli_fetch_array($result))
 {
  $players[] = array('id' => $row['id'], 'firstname' => $row['firstname'], 'lastname' => $row['lastname'], 'email' => $row['email']);
 }

 //
 //POPULATE THE OPPONENTS DROP DOWN BOX TO SHOW ALL PREVIOUS OPPONENTS
 //

 $sqlid = "SELECT id FROM player WHERE email = '$email'";
 $playerid = mysqli_query($link, $sqlid);
 $sql = "SELECT DISTINCT player.firstname, game.player2_id AS opponent_id
 
FROM player INNER JOIN game
                
                ON player.id = game.player2_id

WHERE player1_id = '$row['id']'

UNION

SELECT DISTINCT player.firstname, game.player1_id AS opponent_id

FROM player INNER JOIN game
                
                ON player.id = game.player1_id

WHERE player2_id = '$row['id']'";

 $result = mysqli_query($link, $sql);
 if(!$result)
 {
  $error = 'Unable to populate the players dropdown from the database.' . mysqli_error($link);
  include 'error.html.php';
  exit();
 }
 
 while($row = mysqli_fetch_array($result))
 {
  $opponents[] = array('id' => $row['opponent_id'], 'firstname' => $row['firstname']);
 }

include 'squash.html.php';

}

?>
Thanks for your help and time in advance.