PHP - Function - Validate Ip Address And Return Errors

 

I have a basic form for collecting data, I have a function for collecting the ip address of the visitors unfortunately I cannot get the ip collection working with the rest of the form once I put the if($_POST['emailaddress']) {.........} in.

My aim is once the user has completed the form, data gone to MySQL database that the data will then be printed via the echo statements on to the form as the action sends it back to the same page.

Currently my code is


<?php  
/*
This script is a form handler, each section is commented as to what it does.
1. connects to the database (see connect.php).
2. it gets the users ip address.
3. Strips tags from the data entered so that no malicious code can be entered and corrupt/cause problems with the database or site.
4. Inserts the relevant data into the database in this case it is the ip, haveemail, emailaddress, browser, otherbrowser, resolution, otherresolution.
5. Sends the data just entered by the user back to the screen so they can see what they entered
*/

// 1. connection to MySQL  
require ("php/connect.php"); // if this script is unavailable then the rest of the code is pointless as need a connection to the database.  

// if fields are completed
if($_POST['emailaddress'])
{ // if this field has had data entered then process the data

// 2. collect ip address  
//$ip = getRealIpAddr(); 

// 3. Strips tags and POST variables from the form 
$haveemail = strip_tags($_POST['haveemail']);  
$emailaddress = strip_tags($_POST['emailaddress']);  
$browser = strip_tags($_POST['browser']);  
$otherbrowser = strip_tags($_POST['otherbrowser']);  
$resolution = strip_tags($_POST['resolution']);  
$otherresolution = strip_tags($_POST['otherresolution']);  


// 4. insert data to dbase  
$query="INSERT INTO datacollection1 (id, ip, haveemail, emailaddress, browser, otherbrowser, resolution, otherresolution) VALUES ('Null', '$ip', '$haveemail', '$emailaddress', '$browser', '$otherbrowser', '$resolution', '$otherresolution')";  

// Null is in the id field as this is added automatically in the database as it is set to auto increment upon an entry going in and is  primary key.  

// message to say if database has been updated  

mysql_query($query) or die (mysql_error());  

//echo "<b>Your IP address is: $ip</b> <br />"; 
  


/*function getRealIpAddr() 
{ 
    if (!empty($_SERVER['HTTP_CLIENT_IP']))   //check ip from share internet 
    { 
      $ip=$_SERVER['HTTP_CLIENT_IP']; 
    } 
    elseif (!empty($_SERVER['HTTP_X_FORWARDED_FOR']))   //to check ip is pass from proxy 
    { 
      $ip=$_SERVER['HTTP_X_FORWARDED_FOR']; 
    } 
    else 
    { 
      $ip=$_SERVER['REMOTE_ADDR']; 
    } 
    return $ip; 
} */

// 5. displays information on form to advise what the user just entered
echo '<br/>'."The database has just been updated with the following information: ".'<br/><br/>';
// echo "Your ip address is ".$ip.'<br/>';
echo "You answered ".$haveemail." to having an email address.".'<br/>';
echo "Your email address is ".$emailaddress.'<br/>';
echo "You use ".$browser." to browse the internet".'<br/>';
echo $otherbrowser.'<br/>';
echo "Your screen resolution is set at ".$resolution.'<br/>';
echo $otherresolution.'<br/>';
echo "Now you have completed this form, please follow onto the main form, this form seeks to get your valuable opinion regarding your likes and dislikes of the websites researched.";
}

mysql_close($db); // closes the database, this is good practice but the database will close once stopped running.


?>


I have tried moving the function outside of the if($_POST['emailaddress']) but it still didnt get the ip address.

Just as a side note everything else works, the data is written into the database bar the ip address.

I would appreciate some help 

Thanks in advance.

Hi, I've put together a callback request form that emails my client when someone fills it out.

the form is as follows

Code: [Select]
<form method="post" action="php/callbackform.php">
  Name: <input name="name" type="text" /><br />
  Phone number: <input name="phone" type="text" /><br />
  no: <input type="checkbox" name="no" value="This customer does not want their information to be used for marketing purposes" /><br />
  <input type="submit" />
</form>

it posts to this code

Code: [Select]
<?php
  $name = $_REQUEST['name'] ;
  $phone = $_REQUEST['phone'] ;
  $optout = $_REQUEST['no'] ;
  $message = $phone . "  " . $optout;

  mail( "client@example.com", "Call back form request",
    $message, "From: $name" );
  header( "Location: http://www.example.com/index.php" );
?>

The problem is that when the email is received, the email 'from' field shows my servers default email address.

So if someone called bob filled out the form the resulting email would show as;

From: bob<default@myserver.com>

My client doesn't want to add an email box to use as the from address so is there anyway i can hide my address being included in the email header?

thanks
Richard

let say i have a function

function checkme($x)
{
  if(strlen($x)>10)
 {
   return $x;
 }
 else
 {
   return false;
 }
}

$variable=checkme("5");

so two questions i wish to ask...
1. If i wish to get the empty string for the $variable, then is it correct to have return false there?
2. If it is true that return shall not be used twice in a function as it will somehow reduce bytes space or something like that??

thanks in advance

Hi, i got this function to get the wordCount that i store in a tabel:

Code: [Select]
function getWordCount($word) {
$query = "SELECT count FROM uniqueWords WHERE word='$word'";
$data = execQuery($query);
return $data;
}
Only it returns an array, what i want is to have it return a integer, how can i do that?

Hi all,

can you give me suggestion how to return value from function like this

Code: [Select]
function returnArray(){
for($i=1; $i<=3; $i++){
return $i; //how to return this as array??
}
}

when I code "echo" inside the function then call the function it give me output => 123
but when I code "return" inside the function, it will give output => 1
Can you suggest me how to output it as array?

Hi,Dear Frnds...........................I Hope U r all fine.
I am using php.Here is the problem in the Return array from function.
This is the code.
Code: [Select]
<?php
$store = array();

$test = get_my_links();
print_r($store);

function get_my_links() {

$my_array = array('e','f','g');

foreach($my_array as $key=> $list) {

$store[] = $list;

}
return $store;

}
?>

Suppose the array is maked dynamically.
please help me..THANKS...

Please help me fix these errors in my function.

Warning: mysql_result(): Unable to jump to row 0 on MySQL result index 10 in

Warning: mysql_num_rows() expects parameter 1 to be resource, boolean given in

Code: [Select]
function verify_user_fanned_company($uid, $cid)
{
$uid = (int)$uid;

$sql = "
SELECT
`company_id`
FROM `company_fans`
WHERE `user_id` = {$uid}
AND `company_id` = {$cid}
";

$query = mysql_query($sql);

$result = mysql_result($query, 0);

$num = mysql_num_rows($result);

if ($num === 1)
{
return true;
}
}

Code: [Select]
<?php

$is_fanned = verify_user_fanned_company($user_info['uid'], $info["companyid"]);

if ($is_fanned = true)
{

?>
do stuff
<?php

}

?>

Hi,
I'm trying to output some details from my database, along with a BLOB image.
At the moment I'm having some warnings and notices and I don't know how to solve them.

Here's the warnings and notices:
Notice: Undefined index: id in G:\xampp\htdocs\xampp\dsa\search_func.php on line 38

Notice: Undefined variable: terms in G:\xampp\htdocs\xampp\dsa\search_func.php on line 13

Warning: mysql_query() expects parameter 2 to be resource, object given in G:\xampp\htdocs\xampp\dsa\search_func.php on line 16

Warning: mysql_fetch_object(): supplied argument is not a valid MySQL result resource in G:\xampp\htdocs\xampp\dsa\search_func.php on line 22

Notice: Trying to get property of non-object in G:\xampp\htdocs\xampp\dsa\search_func.php on line 27

Notice: Trying to get property of non-object in G:\xampp\htdocs\xampp\dsa\search_func.php on line 27

Notice: Trying to get property of non-object in G:\xampp\htdocs\xampp\dsa\search_func.php on line 27

Notice: Trying to get property of non-object in G:\xampp\htdocs\xampp\dsa\search_func.php on line 27

Notice: Trying to get property of non-object in G:\xampp\htdocs\xampp\dsa\search_func.php on line 28

Notice: Trying to get property of non-object in G:\xampp\htdocs\xampp\dsa\search_func.php on line 28

Notice: Trying to get property of non-object in G:\xampp\htdocs\xampp\dsa\search_func.php on line 28

Here's the function code:
Code: [Select]
<?php
/*  return the details of an employee
     parameter 
         empno - employee number
         
     eg empHTML.php?empno=7521
    This is  very basic  e.g no error checking      
*/

function search($dbc, $id) { 

// define the query string 
    $query = "SELECT * FROM stick, location, identification WHERE make LIKE '%$terms%' AND stick.sid = location.lid AND identification.stickID = stick.sid ";
   
// execute the query and return a pointer to the resultant SQL relation  
    $dbresult = mysql_query($query,$dbc);
   
// return the result as a simple HTML table
$output = Array();
   
// get each tuple in the table as an object 
      $image = mysql_fetch_object($dbresult);

// interpolate the required employee values  -> is the equivalent of the java DOT notation

$output[] = '<ul>';
$output[] = '<li> Type: '.$image->make .'<br />Size: '.$image->size .$image->type .'<br />Colour: '.$image->colour .
'<br /> Street Missing in: ' .$image->street.'<br />Town missing in: '.$image->town .'<br />City missing in: '.$image->city.'</li>';
$output[] = '</ul>';

return join('',$output);

}
// connect
$dbc = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME);

// get the department name from the URL
    $id = $_REQUEST['id'];

// print the table of employees for this department
// put it ina  div HTML element with an emp class to support CSS styling
    print search($dbc, $id);
    print "<img src='getPhoto.php?id=$id'/>";

// close the database
    mysql_close($dbc);

This is my error:
Code: [Select]
Fatal error: Can't use function return value in write context in /home/a8152576/public_html/SecretSanta.html on line 41
confused about why.
This is the line 41:
if(isset($_POST('submit')){

this is the php of my page (without the mysql data):

<?php
class functions{
public static function chooseGiftee($SS){
     $done = false;
        while($done=false){
     $sqllink= mysql_connect($mysql_host,$mysql_user,$mysql_password);
     if(!$sqllink){
     echo "Could not connect! Please Try Again";
         mysql_close($sqllink);
     } else {
     $selectGiftee = "SELECT * FROM SecretSantas ORDER BY RAND() LIMIT 0,10;";
             while($row=mysql_fetch_array($result)){
                 $results = mysql_query($selectGiftee);
              if(!$row['Gifter']=''){
                  $done=false;
                 } else {
              $gifteeResults = $row['Name'];
                        $addSS = "INSERT INTO SecretSantas (Gifter) VALUES ('" .$SS. "')";
                        $done=true;
              return $gifteeResults;
                 }
             }
     }
            mysql_close($sqllink);
        }
    }
}

if(isset($_POST('submit')){
$echoName = functions::chooseGiftee($_POST['gifterName']);
echo $echoName;
}
?>

This is html:

<form id="form1" name="form1" method="post" action="">
  <p>
    <label>Your Name:
      <input type="text" name="gifterName" id="gifterName" />
    </label>
    <input type="submit" name="submit" id="submit" value="Submit" />
  </p>
  <p>&nbsp;</p>
</form>


Confused about the issue, first time using functions.
Thanks in advance,
Brandon

not sure if this is possible but i do not know how to return multiple values and read them where the call was made.

Code: [Select]
$string = "clause1";
func1($string);
// somehow read the response and use the return values seperately.
// both $string2 and $string3

Code: [Select]
function func1($input) {
if ($input == "clause1") { $string2= "value2"; $string3= "value6"; }
if ($input == "clause2") { $string2= "value3"; $string3= "value7"; }
if ($input == "clause3") { $string2= "value4"; $string3= "value8"; }
if ($input == "clause4") { $string2= "value5"; $string3= "value9"; }
return  ($requesttype, $messagetitle);
}

My tree gather function gives me multidimensional array. However i want it to be one dimensional . I was getting one dimensional earlier however i forgot how i got it. Can someone please tell me my mistake

Code
Code: [Select]
<?php
include'config.php';
 function tree_gather($node)   
  {
     
 $sql = "SELECT lchild,rchild FROM tree WHERE parent = '$node'";
    $execsql = mysql_query($sql);
    $array = mysql_fetch_array($execsql);
    
    if(!empty($array['lchild']) || !empty($array['rchild']))
    {
   
      $child[] = $array['lchild'];
      $child[] = $array['rchild'];
      $eh[] = tree_gather($array['lchild']);
      $child = array_merge($child, $eh);
      $eh1[] = tree_gather($array['rchild']);  
      $child =   array_merge($child, $eh1);
     
    } 
     return $child;
  }
$pair = tree_gather(1000000);
echo "<pre>";
print_r($pair);
echo "</pre>";  
?> 
Output
Code: [Select]
Array
(
    [0] => 3632873
    [1] => 5951538
    [2] => Array
        (
            [0] => 8930480
            [1] => 7563232
            [2] => Array
                (
                    [0] => 1144744
                    [1] => 4386810
                    [2] =>
                    [3] =>
                )

            [3] =>
        )

    [3] =>
)

Hi,
I have some strange issues with my code:

Code: [Select]
<?PHP
session_start();
$loginid = $_SESSION["valid_id"];
// First check if we are guest or user.
if (!$_SESSION["valid_email"])
{
$visitor = "yes";
}
else
{
$visitor = "no";
$email = $_SESSION['valid_email'];
$userid = $_SESSION['valid_id'];
}
//Load Header (blue menu)
require("./inc/header.php");
//Load Sub-acc (silver account menu)
require("./inc/sub-group.php");
//Load nav-group (Tabs)
require("./inc/nav-group.php");
//Load Config file
require("./inc/config.php");
//Set & get profile ID
$getid = $_GET["id"];
//check ID
$result = mysql_query("SELECT * FROM profiles WHERE id=('$getid')  LIMIT 1");
$row = mysql_fetch_array($result);

IF (mysql_num_rows($result) != 1)
{
 exit("Invalid ID");
}

//If we are guest, do we allow anon access to the profile
IF ($row["privacy"] <= 10 && $visitor = "yes")
{
exit("You may not view this profile as a visitor, due to the users privacy settings");
}
//Let's check if we are friends
ELSEIF ($visitor = "no")
{
$result2 = mysql_query("SELECT * FROM profiles_friends WHERE user=('$getid') AND target=('$loginid')  LIMIT 1");
$row2 = mysql_fetch_array($result2);
$friends = $row2["status"];
if (mysql_num_rows($result2) = 0)
{
$friends = "no";
}
}

//If we are friend, do we allow access to the profile
IF ($row["privacy"] >= 9 && $friends != 1)
{
exit("You may not view this profile because of the privacy settings.");
}

$row = mysql_fetch_array($result);
$memgroup = $row["group"];
IF ($row["activated"] != 1)
{
 exit("This account is suspended and cannot be viewed.");
}


//Check what group member is in.
$result2 = mysql_query("SELECT * FROM profiles_groups WHERE id=('$memgroup')  LIMIT 1");
$row2 = mysql_fetch_array($result);

?>

Alright, so the error: Fatal error: Can't use function return value in write context in C:\xampp\htdocs\prog\profile.php on line 45
Code: [Select]
42. $result2 = mysql_query("SELECT * FROM profiles_friends WHERE user=('$getid') AND target=('$loginid')  LIMIT 1");
43. $row2 = mysql_fetch_array($result2);
44. $friends = $row2["status"];
45. if (mysql_num_rows($result2) = 0)
Alright, this is one thing that bothers me, the other is:
Code: [Select]
//check ID
$result = mysql_query("SELECT * FROM profiles WHERE id=('$getid')  LIMIT 1");
$row = mysql_fetch_array($result);

IF (mysql_num_rows($result) != 1)
{
 exit("Invalid ID");
I tried to put an invalid ID, and already here the script should have died/exited before executing the parts of the code that doesn't work.
I tested my code on another page and it works flawlessly, perhaps this error is just generated before it actually exists i dunno..

Any way, any help is much appreciated
}

Hello everyone,

Pretty much what I am trying to do is run a query and based of the queries result, return a string - as of now it is turning up blank and cant seem to pass any strings through the function.

Current function:

function addNewBookmark($hash,$url,$title,$username){
     
      Code: [Select]
$query = "INSERT INTO bookmarks (hash,url,title,username) VALUES '".md5($hash)."', '".$url."', '".$title."', '".$username."')";
      $result = mysql_query(query, $this->connection);
      $message = '';
      if(mysql_affected_rows($this->connection)!=1) {

        $message = 'This URL already exists in the database!';
                return $message;
        }
        else
        $message = 'The URL was shared!';
        return $message;
       
           }
The above code seems to run fine, just cant seem to pass a string...read that you can possibly use a __toString function?  but that keeps throwing errors when I try to work with it.

Thanks all!

I hate array...   😞

So I had a block of code inside my photo-gallery.php script that took the path to my photos directory, and went to that directory, and then read all of the photo filenames into an array.  Then in my HTML, I iterate through this array to display all of the photos for my gallery.

Now I would like to move that code to an included file so multiple scripts can access it and always be working with the same array.

It seems to me that I need to encapsulate my code inside a function?

Then I could call my getPhotoFilesArray back to my callings cript, and use that array for whatever.

I haven't coded PHP in like 4 years and I am struggling to return the entire array back to my caling script.

This is what I have so far...

	function getPhotoFilesArray($photoPath){
	     $photoFiles = array();
	     <code to find corresponding files>
	     $photoFiles gets populated in a loop
	     return $photoFiles;
	}
	

 

Then in my calling script, I have...

	<?php
	     require_once('../../../secure_outside_webroot/config.php');
	     require_once(WEB_ROOT . 'utilities/functions.php');
	     getPhotoFilesArray($photoPath);
	     var_dump($photoFiles);
	

 

I get some error...

	Notice: Undefined variable: phtoFiles in photo-gallery.php line 133 (which is my var_dump).
	

 

<br>

Would appreciate help getting this to work!

 

 

Edited December 6, 2019 by SaranacLake