[Forms Validation]

You will need to use Javascript or AJAX.

 

Try to google for "javascript form validation".

[session error]

@wildteen88

 

Very nice and clean code!

[Server Processing]

You are on a good forum.

 

Try to google for basics of PHP and you will get your answer.

 

The thing you are asking are very basic steps of PHP. Learn it, post some code and will help you with your problems.

 

Good luck...

[[SOLVED] adding rows]

Of course, you can do it with PHP.

 

Use counter!! 

[[SOLVED] Select Box and PHP - Can this be done?]

Try Google for "Sticky forms" 

[[SOLVED] adding rows]

You can use SUM function in MYSQL

 

$sql = "SELECT *, SUM(amount) as amount WHERE id = '$id' GROUP BY amount";

 

Something like that.. Work with it...

[$row['cid']..need clarification!]

Well, this actually isn't making a lot of sence.

 

It's working as you asked...

 

There are simplier ways to do this.. 

[Parse error: syntax error, unexpected T_CONSTANT_ENCAPSED_STRING]

You have a problem with your if statment:

 

if $num is smaller then 0 show something?? Impossible...

 

 

This is fixed:

<?php
$sql = "SELECT * FROM `pilots` ORDER BY `date` DESC LIMIT 7";
$query = mysql_query($sql) or die(mysql_error());
$num = mysql_num_rows($query) or die(mysql_error());
?>

<table width=73% align="center">
    <tr bgcolor=#EEEEEE>
    <td width="10%">PID</td>
    <td width="15%">Name</td>
    <td width="17%">Hub</td>
    </tr>
   
<?php 
if($num>0) {   // <---- Bracket ONE opened                                                                  
    while ($data = mysql_fetch_assoc($query)) {  // <---- Bracket TWO opened ?>                   
        <tr bgcolor=#EEEEEE>
        <tr style="background-color:#EAEAEA;" onmouseover="this.style.backgroundColor='#0099FF'" onmouseout="this.style.backgroundColor='#EAEAEA'">
        <td><?php echo "{$data["login"]}"?></td>
        <td><?php echo "{$data["fname"]}","{$data["lname"]}"?></td>
        <td><?php echo "{$data["hub"]}"?></td>
        </tr>
        <?php
    } // <---- Bracket ONE closed 
        ?>                                                                               
    </table>
    <?php
} // <---- Bracket TWO closed                                                                                       
?>

[[SOLVED] avoiding double rows with same results]

$id is ID field form DB in table cancellation_train

 

As you wrote:

The table "cancellation_train" will have same fields plus id (primary key)

 

So, if you want info about some train you do:

 

$id = 4; //this is just an example
$sql = "SELECT * FROM  train_cancellation WHERE id = '4' ORDER BY id, day, month, year, hour, direction";

 

Hope it's more clear now...

 

 

[[SOLVED] avoiding double rows with same results]

??

 

 

$sql = "SELECT * FROM  train_cancellation WHERE id = '$id' ORDER BY id, day, month, year, hour, direction";

 

??

[Images in php]

Try this:

 

<td align=”left” valign=”top”><img src=”images/left_top.gif” border=”0”></td>

 

Are you sure that picture extension is .gif ???

 

Change slash \ to / for the picture path!

[cannot make the change in db...]

        
<?php  if($_POST["profile"])        {
          $username=$_GET["username"]; //Why are you using this???
          mysql_connect("localhost","root",""); //You need pass, although it's empty
          mysql_select_db("Plp");
          $query= mysql_query("UPDATE members WHERE pkid= '$pkid'") or die (mysql_error()); //Where do you get $pkid variable???
          $num=mysql_num_rows($query);
          if($num == 1)
          {
            echo "update successful";
          }
          else
          {
            $loginError= "didnt match with the database";
          echo "$loginError";
        }
       } 
      ?>

 

I assume that you took somebodys code and now it's not working. The code that you wrote is very bad and non functional, you had a lot of errors.

 

I don't know will this work in the way you wanted, becasue I don't know what is it supposed to do. I just corrected the code so it could work.

[Images in php]

Damn Kenrbnsn.

 

You've beat me for a second with your reply.  ;D

[cannot make the change in db...]

This is not the code we neeed to see. Send the code with query update.

 

Also, it's quite poorly written.

[[SOLVED] Something wrong in this query..has to be simple I just cant get it?]

Try this:

 

<?
require($_SERVER['DOCUMENT_ROOT'].'/includes/configs/sql_connect.php');
$sql = "SELECT * FROM products_description WHERE model_number=\'JT20-A8450\'";
$query = mysql_query($sql) or die ("MySQL Error: " . mysql_error());
while($row = mysql_fetch_array($query)){
$descriptions = $row['$id'];
echo $descriptions;
echo $row[$description];
}
?>

[[SOLVED] PDF create - getting fetal error]

Great!!!

 

Hit "Topic Solved"

[Inserting a php files script into another]

Yes, it will work!

[Inserting a php files script into another]

include("cookie.php");

 

[[SOLVED] PDF create - getting fetal error]

PDF_setfont() was deprecated since PDFlib version 5

 

http://uk3.php.net/manual/en/function.pdf-findfont.php

 

 

Check it out. That's you're problem!

[[SOLVED] PDF create - getting fetal error]

Try some other fonts ...

[PHP form won't work]

You're file "Update_ac.php" is not complete you are missing couple of things.

 

This is wrong:

<?php
$host="localhost";
$username="wowdream_domaine";
$password="mypassword";
$db_name="wowdream_domaine";
$tbl_name="members";

mysql_connect("$host", "$username", "$password")or die("cannot connect"); 
mysql_select_db("$db_name")or die("cannot select DB");

$sql="UPDATE $tbl_name SET username='$username', password='$password', access='$access' WHERE id='$id'";
$result=mysql_query($sql);

if($result){
echo "Successful";
echo "<BR>";
echo "<a href='editadmin.php'>View result[/url]";
}

else {
echo "ERROR";
}

?>

 

This is right:

 

<?php
$host="localhost";
$username="wowdream_domaine";
$password="mypassword";
$db_name="wowdream_domaine";
$tbl_name="members";

mysql_connect("$host", "$username", "$password") or die ("cannot connect"); 
mysql_select_db("$db_name") or die ("cannot select DB");

$id = $_POST["id"];
$username1 = $_POST["username"];
$password1 = $_POST["password"];
$access1 = $_POST["access"];

$sql="UPDATE $tbl_name SET username='$username1', password='$password1', access='$access1' WHERE id='$id'";
$result=mysql_query($sql) or die (mysql_error());
$num=mysql_num_rows($result);

if($num == 1){
echo "Successful";
echo "<br />";
echo "<a href='editadmin.php'>View result[/url]";
}
?>

[showing the table of current user]

Well you need something like ...

 

$result = mysql_query("SELECT * FROM regusers WHERE user_id='$id'")

 

But of course, your user needs to log in , otherwise you won't be able to know which users info to display.

[PHP form won't work]

Also on your Update_ac.php try this

 

$id = $_GET["id"];
echo "$id";

 

Maybe you are not getting an ID so the query can not be updated. Also try to echo you're query to see if you are getting all the values.

[Data in address bar and form]

Can you explain a little bit better and post some code so we could see what are you trying to explain??

[[SOLVED] form problem]

Well, I liked the idea from Acs ...

 

Try that one!

 

It seems good enough. Although I don't know what kind of site you are running, that would be helpfull to give you some ideas!