[[SOLVED] Select box copies it's value??]

Can you post some code? I'm not too good with javascript!

[[SOLVED] Select box copies it's value??]

I could go for any other approach if you have ideas! 

[PHP Newbie - Probably very simple answer - whats missing from code???]

OK, that is quite simple.

 

You can use PHP mail() function to send emails.

 

$to = "email@example.com"; //Here goes $_POST["email"];
$subject = "Test mail"; //Here goes some subject like "This is enqur. from web page"
$message = "Hello! This is a simple email message."; //Here goes like phone number etc.
$from = "someonelse@example.com";
$headers = "From: $from"; //Here goes you're email adress
mail($to,$subject,$message,$headers);
echo "Mail Sent.";

 

Also, DarkWater is right, go get new book. This is just causing problems for you.

[PHP Newbie - Probably very simple answer - whats missing from code???]

Well, you've got your opinion. 

 

And, you are getting OFF TOPIC!

 

We can open a discussion about this somewhere else!

[PHP Newbie - Probably very simple answer - whats missing from code???]

Well, now he will have much clearer picture of how to store data in a DB. I don't see any point in just displaying the values on the form.

 

If he is learning then I showed him much bigger picture and made the situation more understandable for him.

 

And, I'm sorry guys, but if he is learning PHP, then it's much better that he learns how to create a database and establish a link to it, first. Then he can start learning PHP.

 

It's like a building a house from the roof.

[PHP Newbie - Probably very simple answer - whats missing from code???]

You are not trying to get the values from form.

 

$1 = $_POST["name"];
$2 = $_POST["email"];
$3 = $_POST["company"];
$4 = $_POST["number"];

 

and then query to insert into DB

 

$sql = "INSERT INTO table_name (name_of_fields) VALUES ($variables)";
$result = mysql_query($sql) or die (mysql_error());

 

That's basiclly all the code you need to insert data in DB. just adjust it according to your needs and to match your DB fields.

 

P.S.

You know that you need a connection to DB??

[[SOLVED] Select box copies it's value??]

OK, I'm trying to rework this a little bit and I need your help. My knowledge of javascript is not to good.

 

This is what I have for now:

 

<SCRIPT LANGUAGE="JavaScript">
<!-- 	
// by Nannette Thacker
// http://www.shiningstar.net
// This script checks and unchecks boxes on a form
// Checks and unchecks unlimited number in the group...
// Pass the Checkbox group name...
// call buttons as so:
// <input type=button name="CheckAll"   value="Check All"
//onClick="checkAll(document.myform.list)">
// <input type=button name="UnCheckAll" value="Uncheck All"
//onClick="uncheckAll(document.myform.list)">
// -->

<!-- Begin
function checkAll(field)
{
for (i = 0; i < field.length; i++)
field[i].selected = true ;      ------ I have changed this from "checked" to "selected"
}

function uncheckAll(field)
{
for (i = 0; i < field.length; i++)
field[i].selected = false ;
}
//  End -->
</script>

 

And for the event I have done this:

 

onChange="checkAll(document.autoSumForm.list)

 

Couple of questions:

 

1. Is this possible?

2. document.autoSumForm.list -> list is a name of object on a form, is it possible to use ID instead of name????

[[SOLVED] Select box copies it's value??]

Hey guys,

 

Do you know of anyscript where user could choose a value in select box and then automaticlly 3 or more other select boxes would get the same value?

 

For instance:

User chooses in selectbox 1 value car

And the automaticly selectbox2 gets value car

And the automaticly selectbox3 gets value car

And the automaticly selectbox4 gets value car

And the automaticly selectbox5 gets value car

And the automaticly selectbox6 gets value car

 

Or you can point what to search on google!

[[SOLVED] Order By most recent dates from 2 tables]

You are not giving enough info. From the code you've posted we can't see anything.

 

Have you done anything in this matter. If not, post your DB structure so we could see the names of your date fields and explain which field is what.

[Header Location]

You are getting output in your db.info.inc file and then header is failing and reporting an error message.

["Topic solved" mod]

Here is a suggestion.

 

It would be nice to put a small icon next to the topic title for "Topic solved" on the forum index. It would be more visible and easier to access for a user that posted a topic to click that icon and mark topic as solved.

 

Right now, there is a lot users that don't click "Topic solved" and I think this could reduce number of those users.

 

Cheers.

["Topic Solved" Mod!]

Here is a suggestion.

 

It would be nice to put a small icon next to the topic title for "Topic solved" on the forum index. It would be more visible and easier to access for a user that posted a topic to click that icon and mark topic as solved.

 

Right now, there is a lot users that don't click "Topic solved" and I think this could reduce number of those users.

 

Cheers.

[[SOLVED] Calculating distances ...]

That's what I thought. I thought that maybe there is a way.

 

Thanks a lot for your help.

[[SOLVED] Calculating distances ...]

Here it is:

 

ID  From        To        Distance

1    Zagreb    Karlovac  52Km

2    Zagreb    Knin        274Km

3    Zagreb    Virovitica 154Km

4    Zagreb    Split        374Km

5    Karlovac  Pula        234Km

 

These are values from DB.

 

I'm not sure how to calculate distance (for example) from Karlovac to Knin??? And other combinations??

[problems with duplicate data]

OK, maybe you can do this:

 

$query = mysql_query("SELECT email FROM users WHERE email = '".$email."'");
while($row=mysql_fetch_array($query)){
$email1 = $row["email"];
}

if ($email1 == $email){
      if(@mysql_query($query)) $erroremail = true;
      
            $query2 = "INSERT INTO users SET
               username='$username',
               password='$password',
               email='$email'";

} else {

echo "Data is already in DB!";

}

[/code]

 

Here is antoher check, where we will check if that email is already stored in DB and if it is it will display an error message. Maybe you will need to adjust a code a litlle bit.

[[SOLVED] form handling]

Seems to me you will need to use javascript or AJAX for thing you are asking for.

 

There are plenty of simple scripts on Google.

 

Try Google for: "javascript form validation"

 

Cheers

[[SOLVED] Calculating distances ...]

Hey guys,

 

I'm creating a script where I should display distances between two cities. User can choose From - To cities and then distance is displayed

 

For example:

 

City 1    ->    City2    = 56km

City55    ->  City72  = 110km

...

 

In my DB I have this fields:

ID

date

from_city

to_city

distance

 

How do I calculate which is distance from one city to another???

 

[Using Sessions]

@iversonm:

 

$a != $b  Not equal  TRUE if $a is not equal to $b.

$a !== $b  Not identical  TRUE if $a is not equal to $b, or they are not of the same type. (introduced in PHP 4)

 

In both cases he will get the same result in his if function, because it's the same type.

 

Although, as looking now if

$_SESSION['token'] = uniqid(md5(microtime()), true);

is encoded with md5. Then

if ($_GET['token'] != $_SESSION['token']) 

is always FALSE unless decoded.

 

@lanmind:

 

Where do you get

$_SESSION['token']

for your if statment

if ($_GET['token'] != $_SESSION['token'])

?

[Using Sessions]

Try this:

 

<?php
session_start();


$_SESSION['token'] = uniqid(md5(microtime()), true);

if ($_GET['token'] != $_SESSION['token']) {
die('Invalid token');
}

$coin = $_SESSION['token'];

[[SOLVED] Print in landscape]

Thanks a lot guys!

 

I'm waiting for CSS3, until then I will print in Portrait with some adjustments.

 

Cheers.

[problems with duplicate data]

You already did it, but in a wrong order!

 

Here is a fix:

 

<?php
include 'db.php';
include 'function.php'; 

if($_POST['register']) { 

   
   //init error variables
   $errorusername = false;
   $erroremail = false;
   $erroremailconf = false;
   $errorpassword = false;
   $errorpassconf = false;
   
   $username = isset($_POST['username']) ? trim($_POST['username']) : '';
   $email = isset($_POST['email']) ? trim($_POST['email']) : '';
   $emailconf = isset($_POST['emailconf']) ? trim($_POST['emailconf']) : '';
   $password = isset($_POST['password']) ? trim($_POST['password']) : '';
   $passconf = isset($_POST['passconf']) ? trim($_POST['passconf']) : '';
   
   if(!eregi("^[_a-z0-9-]+(\.[_a-z0-9-]+)*@[a-z0-9-]+(\.[a-z0-9-]+)*(\.[a-z]{2,3})$", $email)) $erroremail = true;
   if(strlen($username) == '0') $errorusername = true;
   if(strlen($emailconf) == '0') $erroremailconf = true;
   if(strlen($password) == '0') $errorpassword = true;
   if(strlen($passconf) =='0') $errorpassconf = true;
   
   //display form again
   if($errorusername || $erroremail || $erroremailconf || $errorpassword || $errorpassconf) {
      showForm($errorusername,$erroremail,$erroremailconf,$errorpassword,$errorpassconf);
   
   } else {
      
      //check user email
      
      $query = mysql_query("SELECT email FROM users WHERE email = '".$email."'");
      if(@mysql_query($query)) $erroremail = true;
      
            $query2 = "INSERT INTO users SET
               username='$username',
               password='$password',
               email='$email'";
               
               
      if(@mysql_query($query2)) {
         echo 'Signup Completed';
         }else{
            echo 'failed to signup';
            }   
           
      }

} else {

echo "Nothing happens!"; //Leave this empty, I just putted it for test

}
?>

[[SOLVED] Print in landscape]

No, nothing is happening!

 

I don't see why. Should I use:

 

<div class="page">

 

And put everything inside that div element???

 

No, div element didn't help!!

[[SOLVED] Print in landscape]

Hey guys,

 

I can't work it out, so my page gets printed in landscape.

 

I'm using this peace of code:

 

@page {
size: landscape;
}

 

Web browser is Firefox 3.

 

I can't see what is wrong!!! Help!

[[SOLVED] query help]

$ranking = mysql_query("SELECT * FROM `" . DBPREFIX . "users` SUM(capture) as capture WHERE ClanID != '0' GROUP BY `ClanID` ORDER BY Capture DESC Limit $limit") or die(mysql_error());

[[SOLVED] Search a database via LIKE]

And your problem is...??