PHP - Totally New To Php. Need Urgent Help With Form.

I am creating a member site and I need help with coding the database. Here go the code

Code: [Select]
<?php
function connect(){
$DB_USER =  "root";
$DB_PASSWORD = "";
$DB_HOST = "localhost";
$dbc = mysql_connect ($DB_HOST, $DB_USER, $DB_PASSWORD) or $error = mysql_error();
mysql_select_db("membership") or $error = mysql_error();
mysql_query("SET NAMES `utf8`") or $error = mysql_error();

if($error){echo "<!-- $error -->";}}

function disconnect_data(){
@mysql_close($dbc);
@mysql_close();
}
?>I am suppose to fill it in with my database info but idk what to fill in! help me please! where do i enter the info!

<?php

$possibles[];
$alreadychosen[];
 
for ($i = 0; $i<20;$i++) $possibles =i+1;

for (int $i = 0; $i<8; $i++)
{
   $boolean= false;
   $randIndex=0;
   
   do
   {
      $randIndex = rand(1,20);
      for ($j = 0; $j< $i && !$hasInt; $j++)
      {
         if ($possibles[$randIndex] == $alreadyChosen[$j])
         {
            $hasInt = true;
         }
      }
   } while ($hasInt);
   {
   $alreadyChosen = $possibles[randIndex];
   echo($alreadyChosen);
    }
   
for ($i = 0; $i<8;$i++)
{
if($alreadyChosen<10)
{
echo ('<a href="/mp/info130'+$i+'.htm"><img src="/mp/p130'+$i+'-1.jpg"  height=180 width=120 align=left hspace=0 vspace=2 border=0 class=TopImages></a>');
}
else if($alreadyChosen>10)
{
echo ('<a href="/mp/info13'+$i+'.htm"><img src="/mp/p13'+$i+'-1.jpg"  height=180 width=120 align=left hspace=0 vspace=2 border=0 class=TopImages></a>');











it should select 8 images out of 20 but don't repeat please

hI for all
Am developping a script that allow to each user (seperatly ) upload or download his files and list all files in categories (such music or photos...) but i can 't achieve that the main problem is to upload files to server and store them in data base also i have to list for each user his files seperatly when he log in
any idea please
 

Hello dear friends  ,

I'm under attack huge attack from spam bots and i need your help

1) i've a form with input text and input submit
2) no way to add html cause using js banned any < or > ..ect
3) the spam attack now post html code but it stored into database after automatically  < and > removed as i mention in (2) okay

add to your note i can't add captcha cuase my website for kids 

Now please allow me to explain this idea and let me know if it ture or there is another way else more simple

1) i will add hidden input with value shows "IP Number Of Poster"

2) by that way i will know the IP of the spammer

3) i will use php and mysql to create database table and store the IPs of spamming i collect so that when that spammer hit my website again , will not be able to view input form.

What do you think ! is it right or bad solution or is there any any way

the bad idea is that spam is not sure human cause it is stupid and post something meaningless and not ture and fake urls like this

Code: [Select]
theuzgfnpbmi, [url=http://gdtuaemvkpif.com/]gdtuaemvkpif[/url], [link=http://rsxtnszcpceq.com/]rsxtnszcpceq[/link], http://awpfgwtvqkdm.com/" theuzgfnpbmi, [url=http://gdtuaemvkpif.com/]gdtuaemvkpif[/url], [link=http://rsxtnszcpceq.com/]rsxtnszcpceq[/link], http://awpfgwtvqkdm.com/"
please help if you have any suggestion or method how to stop such spammers

thank you so much 

This topic has been moved to Miscellaneous.

http://www.phpfreaks.com/forums/index.php?topic=321874.0

hi for all
am installing php security user mangement but i got some errors during install am using xampp 1.7.7 for windows but when i runed the script for the first time i got this message
Warning: require_once(C:/xampp/htdocs/xampp/psumthemes/shared.js.php) [function.require-once]: failed to open stream: No such file or directory in C:\xampp\htdocs\xampp\psum\index.php on line 20Fatal error: require_once() [function.require]: Failed opening required 'C:/xampp/htdocs/xampp/psumthemes/shared.js.php' (include_path='.;C:\xampp\php\PEAR') in C:\xampp\htdocs\xampp\psum\index.php on line 20()

any ideas please my document root is C:/xampp/htdocs

I have an array and I'd like to find the most popular value in the array. example.

$arr = array(1,1,1,1,2,2,3,4,5,5,5,5,5,5,5,3,4,198,10293,1,2);

I would like only display 5 please.

Okay so you may have noticed an earlier post from myself that problem have been fixed,

Some background info, The website is currently on my test server so not connected to the internet. I have only been learning PHP for the last 3-4 days so I am fairly new to most things so please keep your explanations simple, The finished code here is being designed for use on large scale websites with a lot of users around 100,000+ per 24 hour period and therefor will need to be secure. Yet I have no idea how to do that, The script once finished will also be freely distributed and contributors will be fairly credited.

I have some new problems that I need help fixing, So I am creating a PHP Blog connected to my MYSQL Database, Currently there is no admin backend that's a job for later but I do have the following files.

index.php - This displays all blog posts for all my categorys
news.php - This displays each blog post on there own
config.php - This contains my server connect code

I also have the following fields in my database
articleid - Unique ID number of each blog post
category - The category the blog post is in
title - The title of each blog post
body - The body of each post
author - The name of the author
date - date of posting

But now I need to add some comments so can anyone please help out, The comments should only appear below the article on the 'news.php' page so I was hoping someone could write me a script to post and display the comments, (No login system needed I got that covered surprisingly.) if you can also give me a .sql file to add the needed tables to my database that would be greatly appreciated,

Lastly would it be possible for someone to secure the script so it is invulnerable to SQL Injection attacks and any other form of attack that someone could pull off on the site.

So bring on the code,

Index.php
Code: [Select]
<?php include("config/config.php");
 $data = mysql_query("SELECT * FROM blog ORDER by date ASC") 
 or die(mysql_error()); 
while($row = mysql_fetch_array($data))
  {
echo "<table class='main'> <tr> <td> <a href='/news.php?articleid=" . $row['articleid'] . "' class='article_title'>" . $row['title'] . "</a>
<p>" . $row['introduction'] . "</p></td><tr><td ALIGN='RIGHT' class='small'> Posted by:" . $row['author'] . ", on " . $row['date'] .  ",</td></tr></table>";
  }
?>
Comments:I have removed all the junk out and left the basic script,

news.php
Code: [Select]
<?php include("config/config.php");
 $data = mysql_query("SELECT * FROM blog WHERE articleid = {$_GET['articleid']} ORDER by date ASC")
 or die(mysql_error()); 
while($row = mysql_fetch_array($data))
  {
echo "<table class='main'> <tr> <td> <a href='/news.php?articleid=" . $row['articleid'] . "' class='article_title'>" . $row['title'] . "</a>
<p>" . $row['introduction'] . "</p></td><tr><td ALIGN='RIGHT' class='small'> Posted by:" . $row['author'] . ", on " . $row['date'] .  ",</td></tr></table>";
  }
?>Comments:So here's the complex bit if someone could please add the comments below the article here,

Thanks again,
Blaze,
(Really bad PHP Programmer)

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

ive got a page that lists a users bank accounts...when the user chooses to update an account it goes to a new page where they can edit the details...when it goes to this new page it takes the user's ID and the account ID with it in the URL (http://localhost/oswp/bank/updateaccounts.php?id=34&cid=1)

so i know these variables are correctly being transferred.

now when the user has made their changes and clicked the update button it should update the database and then re-locate back to the original page listing all accounts...to get back to this page and show the correct accounts for that user i thaught i needed to take the ID from the URL...give it a variable...and use it in the URL back to the original page. so for example the variable and update sql and re-locate code would look like this.


$cid = $_GET['cid'];

//updating the table
$result=mysql_query("UPDATE accounts SET accno='$accno', pin='$pin', type='$type', name='$balance', balance='$balance', active='$active' WHERE accid=$id");

header("location:accounts.php?id=$cid");


however currently when the user clicks update and it tries to relocate back to the original page its not getting the variable in the URL.
(http://localhost/oswp/bank/accounts.php?id=)

and im getting the errors:

Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in C:\wamp\www\uni\oswp\assignment\bank\accounts.php on line 26
's bank accounts


Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in C:\wamp\www\uni\oswp\assignment\bank\accounts.php on line 30

basically saying its expected a number up in the URL and its not finding one...but i dont know why...any ideas please its quite urgent.

here is the relevant code for this section:

Accounts.php:
<a href="adminlogout.php">Logout</a><br />
<?php
$id = $_GET['id'];
if(!isset($_SESSION['myusername'])) {
header("location:adminlogin.php");
}
$dbhost = 'localhost';
$dbuser = 'root';
$dbpass = '';
$db = 'bank';

$conn = mysql_connect($dbhost,$dbuser,$dbpass);
mysql_select_db($db);
$result = mysql_query("SELECT * FROM customer WHERE cusid=$id");
$row2 = mysql_fetch_array($result);
echo $row2['name'] . "'s bank accounts" . "<br><br>";
$result2 = mysql_query("SELECT * FROM accounts WHERE cusid=$id");
echo "<form method='post' action='accounts.php?id=$id'>";
while($row = mysql_fetch_array($result2))
{
echo "<input type='hidden' value='$row[accid]' name='accid'>" . $row['accno'] . $row['name'] . $row['type'] . $row['balance'] . $row['active'] . "<a href='updateaccount.php?id=$row[accid]&cid=$id'>Update</a>" . "<br>";
}
echo "</form>";
?>


updateaccounts.php:
$id = $_GET['id'];
$cid = $_GET['cid'];
if(!isset($_SESSION['myusername'])) {
header("location:adminlogin.php");
}
$dbhost = 'localhost';
$dbuser = 'root';
$dbpass = '';
$db = 'bank';

$conn = mysql_connect($dbhost,$dbuser,$dbpass);
mysql_select_db($db);

if(isset($_POST['update']))
{
$accno=$_POST['accno'];
$pin=$_POST['pin'];
$type=$_POST['type'];
$name=$_POST['name'];
$balance=$_POST['balance'];
$active=$_POST['active'];
// checking empty fields
//if item field is empty
if(empty($accno))
{
echo "<font color='red'>account number field is empty.</font><br/>";
}
elseif(empty($pin))
{
echo "<font color='red'>pin field is empty.</font><br/>";
}
elseif(empty($type))
{
echo "<font color='red'>account type field is empty.</font><br/>";
}
elseif(empty($name))
{
echo "<font color='red'>account name field is empty.</font><br/>";
}
elseif(empty($balance))
{
echo "<font color='red'>balance field is empty.</font><br/>";
}
elseif(empty($active))
{
echo "<font color='red'>active field is empty.</font><br/>";
}

else
{
//updating the table
$result=mysql_query("UPDATE accounts SET accno='$accno', pin='$pin', type='$type', name='$balance', balance='$balance', active='$active' WHERE accid=$id");
header("location:accounts.php?id=$cid");
}
}

hello guys
I need little help bro,
I want to make application in which i want this
look,
I am from a company where we have over 12 companies license and all are those is renew after every year.
so i want to make a application which will inform us 1 month before expireing their license.
Mean if any company license is expire on 12 June, our application give us notification on 12 May.
Could you please help me how can I do this?
Thanks and regards

Hey, I need a bit of help replacing some variables in a string. I know its done with preg_replace.

Here is the string:

Code: [Select]
Header 1 <|header 2|> header 3 <b>header 4</b>
I would like to get ONLY <|header 2|> from the above string. But "header 2" is variable so please its useless if the preg_replace isn't flexible to find whatever word is in there regardless.

Thanks you.

What is all dependent on PHP? I mean lets say I removed echo from phps source code. Then I add function my_echo() which is the exact same code basically, what all would I need to update to allow my_echo() to be implemented correctly   Obviously I would need to modify every one of my php scripts on my webserver to replace echo with my_echo, but is there anything else?   I know i could just append my_echo to phps source, or modify the echo function itself, but this is conceptually and thats not the actual process I'm looking for right now.   Any insight would be greatly appreciated!