PHP - Generate A Unique Key

I'm not sure I'm in the right place, but I'm looking to create a unique URL that links to the same page every time I send an email after a form is submitted on my website and have it expire after 15 days.  After expiring, the link would redirect to a default page.  I'm still a beginner and realize this is probably going to be a tough task, but any resources or direction provided would be much appreciated.

Thanks for the help in advance!

hi!, is it possible to use rand() to generate unique number that will be saved in a database as a primary key? Since i dont want to have numbers like 00001 incrementing on the table. They dont look like real account numbers... i need something like 45642  or 95452 and the like.

thanks in advance.

I have a mysql table which will store users email addresses (each is unique and is the primary field) and a timestamp. 
I have added another column called `'unique_code' (varchar(64), utf8_unicode_ci)`.

What I would very much appreciate assistance with is;

a) Generating a 5 digit alphanumeric code, ie: 5ABH6 
b) Check all rows the 'unique_code' column to ensure it is unique, otherwise re-generate and check again 
c) Insert the uniquely generated 5 digit alphanumeric code into `'unique_code'` column, corresponding to the email address just entered. 
d) display the code on screen.

What code must I put and where?


**My current php is as follows:**

Code: [Select]
    require "includes/connect.php";
   
    $msg = '';
   
    if($_POST['email']){
   
    // Requested with AJAX:
    $ajax = ($_SERVER['HTTP_X_REQUESTED_WITH']  == 'XMLHttpRequest');
   
    try{
    if(!filter_input(INPUT_POST,'email',FILTER_VALIDATE_EMAIL)){
    throw new Exception('Invalid Email!');
    }
   
    $mysqli->query("INSERT INTO coming_soon_emails
    SET email='".$mysqli->real_escape_string($_POST['email'])."'");
   
    if($mysqli->affected_rows != 1){
    throw new Exception('You are already on the notification list.');
    }
   
    if($ajax){
    die('{"status":1}');
    }
   
    $msg = "Thank you!";
   
    }
    catch (Exception $e){
   
    if($ajax){
    die(json_encode(array('error'=>$e->getMessage())));
    }
   
    $msg = $e->getMessage();
    }
    }

Can this be done with php?
aaa
aab
aac
...
zzz

Hi, can someone help me to understand this? What i want to do is to write some information in a form and after i submit the form that data will be in a new php page.

Thanks in advance

<?php
if(isset($_POST['submit'])){

	//collect form data
	$location = $_POST['location'];
	$ID = $_POST['ID'];
        $section = $_POST['section'];

	//check name is set
	if($location ==''){
		$error[] = 'Name is required';
	}

    //if no errors carry on
    if(!isset($error)){

		# Title of the CSV
		$Content = "location, ID, section\n";

		//set the data of the CSV
		$Content .= "$location, $ID, $section\n";

    # set the file name and create CSV file
    $my_file = ("$location$ID$section.cvs");
    $handle = fopen("$my_file", "w") or die('Cannot open file:  '.$my_file);
    header('Content-Type: application/csv'); 
    header('Content-Disposition: attachment; filename="' . $FileName . '"'); 
    echo $Content;
    exit();
	}
}

//if their are errors display them
if(isset($error)){
	foreach($error as $error){
		echo "<p style='color:#ff0000'>$error</p>";
	}
}
?> 

<form action='' method='post'>
<p><label>Location:</label><br><input type='text' name='location' value=''></p> 
<p><label>ID:</label><br><input type='text' name='ID' value=''></p> 
<p><label>Section:</label><br><input type='text' name='section' value=''></p>
<p><input type='submit' name='submit' value='Submit'></p> 
</form>

I have wrote a .php file which is a form with 3 different fields, in these fields I want to write entries which lateron will be submitted into a .csv generating a csv file on my server with the name of the entries.   The issue is that the file is not being generated on my server. With the current code the entries are being recognized and being placed in the filename of the csv.   The point is not to make it downloadable sinds I want to have it on my webserver and keep editing information in it.   My code is top of this post.   My .php file replies that: Cannot open file: BOD10Buffer.csv   Pardon my sloppy code everyone, I am not really a person that codes but this "form" may save me a lot of time in the longrun, just difficult and I am asking for help on this. Is there anyone that may be kind enough to help me? and as to what I might be doing wrong?   All help is much appreciated!    

I have tried sitemap.org and a few others but have not found anything that will give the results I am after.

I have a blog site that has a URL and a Description for the URL and wish to have the description show as the anchor text which is linked to the URL, all data is taken from MySQL.

Does anyone have a code that will generate something like this ?

Hi guys, so i have this file upload script. When i upload a file it gets stored in /uploads and keeps the same file name. So if i upload a file "test.exe" the file will be available at uploads/test.exe What i want is that it generates a new file name like: "9daln292os.exe" so upload/9daln292os.exe   This is my code:

<?php

// Where the file is going to be placed
$target_path = "uploads/";

/* Add the original filename to our target path.
Result is "uploads/filename.extension" */
$target_path = $target_path . basename( $_FILES['uploadedfile']['name']);
$_FILES['uploadedfile']['tmp_name'];

?>


<?php


$file_type = $_FILES['userfile']['type'];
$file_name = $_FILES['userfile']['name'];
$file_ext = strtolower(substr($file_name,strrpos($file_name,".")));

if (!in_array($file_type, $FILE_MIMES) && !in_array($file_ext, $FILE_EXTS) )
$message = "Sorry, $file_name($file_type) is not allowed to be uploaded.";
else
$message = do_upload_function_here($upload_path_here, $upload_ur_upload_url_herel);
?>
<?php



$target_path = "uploads/";

$target_path = $target_path . basename( $_FILES['uploadedfile']['name']);

if(move_uploaded_file($_FILES['uploadedfile']['tmp_name'], $target_path)) {
echo "The file ". basename( $_FILES['uploadedfile']['name']).
" has been uploaded. Here is the link to your file: <a href=uploads/". basename( $_FILES['uploadedfile']['name']). ">". basename( $_FILES['uploadedfile']['name'])."</a>";

} else{
echo "There was an error uploading the file, please try again!";
}

?>

I don't know any basic php i really need someone to give me the code ready please. Thanks much appreciated.
Edited by darox, 21 July 2014 - 01:07 PM.

Hello PHP freaks,   my codes dont allow me to log-in to Student Home with unique id. It says "Invalid Login or Password" this is my form

<tr bgcolor="#E1E1E1" class="stylesmall">
              <td width="35%" align="left" class="style7 style1">Learner Id : </td>
                <td width="65%" align="left"><input name="learner_id" type="text" id="learner_id" action="Student_Home.php" method="post"></td>
              </tr>
            <tr bgcolor="#E1E1E1" class="stylesmall">
              <td align="left" class="style7 style1">Password:</td>
                <td align="left"><input name="student_password" type="password" id="student_password"><
/td>

and this is my handler.

<?php
	session_start();
	include 'Connect.php';
	$flag = "";
	$learner_id = $_POST['learner_id'];
	$student_id = $_POST['student_id'];
	$student_password = $_POST['student_password'];
	$query = "select last_login_date from student_information where student_id='$student_id' and student_password='$student_password'";
	$result = mysql_query($query,$link_id);
	if(mysql_error() != null){
		die(mysql_error());
	}
	if($date = mysql_fetch_array($result))
	{
        $lastdate = $date['last_login_date'];
	$date2 = date("d-m-Y h:i A",strtotime($lastdate));
	$_SESSION["student_id"] = $_POST["student_id"];
        $_SESSION["lastlogin"] =$date2;
	$_SESSION["type"] = "Student";
	mysql_query("UPDATE student_information SET last_login_date=now() where student_id='$student_id'",$link_id);
	if(mysql_error() != null){
	die(mysql_error());
	}
	header("location:  Student_Home.php?id={$student_id}");
		 die();
	}
	else
	{
		$flag = "invalid";
		header("location:Student_login.php?flag=$flag");
		die();		
	} 	
?>

PLease help me PHP friends to correct my codes.

I use this program on the server.  It does not work.

<?php
$pdf= pdf_new();

 ?>
This is the problem of the program or the problem of the server.

http://www.ptiimaging.ca/pdf.php

Hi, so i kinda planned to code a new project. But i'm not quite sure about this.   I could imagine you could pull the pages out.     So let me explain.       So i believe google is generating these numbers automatically and automatically linking it to a page. Is there a way to do that easily? Would help me alot.  30b3096a003f946c870b24e6f93538d7.png   4.72KB   0 downloads   So if i have 50 result per page. Thats the thing i want!   If 50 result = > then new page! And then it should create a new number down there.    =) Any questions? ASK ME!

Hi, I would like to generate a constant value that change from a website to website but have an identical value for a single website. For example 

$_SERVER['SERVER_ADDR']

doesn't change for a single website, but easy to guess. an other idea is 

realpath(dirname(__FILE__))

but this can change if the web application execute scripts located in sub directories of the main script that use this variable. So what other possibilities to get a constant value that doesn't change ?   Thank you.

Hi, I have a script to add images to my tabel and and upload them to my server, I have managed to get the upload script to rename the image with a 5 digit random number on the end to prevent overwriting, the only problem I can't seem to be able to add the same name to my tabel. Any ideas? Thanks in advance.

Code: [Select]
<?php

 $idir = "uploads/";   // Path To Images Directory


if (isset ($_FILES['fupload'])){

//upload the image to tmp directory
$url = $_FILES['fupload']['name'];   // Set $url To Equal The Filename For Later Use 
if ($_FILES['fupload']['type'] == "image/jpg" || $_FILES['fupload']['type'] == "image/jpeg" || $_FILES['fupload']['type'] == "image/pjpeg") { 
$file_ext = strrchr($_FILES['fupload']['name'], '.');   // Get The File Extention In The Format Of , For Instance, .jpg, .gif or .php 
$copy = copy($_FILES['fupload']['tmp_name'], $idir. basename($_FILES['fupload']['name'], $file_ext).rand(10000 , 99999).$file_ext);   // Move Image From Temporary Location To Perm 
}
}
if (isset ($_FILES['fupload2'])){

//upload the image to tmp directory
$url = $_FILES['fupload2']['name'];   // Set $url To Equal The Filename For Later Use 
if ($_FILES['fupload2']['type'] == "image/jpg" || $_FILES['fupload2']['type'] == "image/jpeg" || $_FILES['fupload2']['type'] == "image/pjpeg") { 
$file_ext = strrchr($_FILES['fupload2']['name'], '.');   // Get The File Extention In The Format Of , For Instance, .jpg, .gif or .php 
$copy = copy($_FILES['fupload2']['tmp_name'], "$idir" . $_FILES['fupload2']['name'] $file_ext).rand(10000 , 99999).$file_ext);    // Move Image From Temporary Location To Permanent Location 
}
}
if (isset ($_FILES['fupload3'])){

//upload the image to tmp directory
$url = $_FILES['fupload3']['name'];   // Set $url To Equal The Filename For Later Use 
if ($_FILES['fupload3']['type'] == "image/jpg" || $_FILES['fupload3']['type'] == "image/jpeg" || $_FILES['fupload3']['type'] == "image/pjpeg") { 
$file_ext = strrchr($_FILES['fupload3']['name'], '.');   // Get The File Extention In The Format Of , For Instance, .jpg, .gif or .php 
$copy = copy($_FILES['fupload3']['tmp_name'], "$idir" . $_FILES['fupload3']['name'] $file_ext).rand(10000 , 99999).$file_ext);      // Move Image From Temporary Location To Permanent Location 
}
}
if (isset ($_FILES['fupload4'])){

//upload the image to tmp directory
$url = $_FILES['fupload4']['name'];   // Set $url To Equal The Filename For Later Use 
if ($_FILES['fupload4']['type'] == "image/jpg" || $_FILES['fupload4']['type'] == "image/jpeg" || $_FILES['fupload4']['type'] == "image/pjpeg") { 
$file_ext = strrchr($_FILES['fupload4']['name'], '.');   // Get The File Extention In The Format Of , For Instance, .jpg, .gif or .php 
$copy = copy($_FILES['fupload4']['tmp_name'], "$idir" . $_FILES['fupload4']['name'] $file_ext).rand(10000 , 99999).$file_ext);      // Move Image From Temporary Location To Permanent Location 
}
}

 $usr = "user";
    $pwd = "pass";
    $db = "db";
    $host = "host";

    # connect to database
    $cid = mysql_connect($host,$usr,$pwd);
    if (!$cid) { echo("ERROR: " . mysql_error() . "\n");    }

    if ($_POST['submit']) {
$logo = mysql_real_escape_string("$idir" . $_FILES['fupload']['name']);
$image1 = mysql_real_escape_string("$idir" . $_FILES['fupload2']['name']);
$image2 = mysql_real_escape_string("$idir" . $_FILES['fupload3']['name']);
$image3 = mysql_real_escape_string("$idir" . $_FILES['fupload4']['name']);

$SQL = " INSERT INTO mhhire ";
        $SQL .= "  image1, image2, image3, logo) VALUES ";
        $SQL .= " ('$image1', '$image2', '$image3', '$logo') ";

$result = mysql_db_query($db,$SQL,$cid);
      $last=mysql_insert_id();

        if (!$result) { echo("ERROR: " . mysql_error() . "\n$SQL\n");    }

header("location:thanks.php");
exit();

    }
?>

Hi, im trying to apply a random css class to a div so that a random image is generated on each refresh. This is the code i am using to apply the class.

Code: [Select]
<div class="<?php echo($randomimage); ?>">
Say i have an array of $image1 $image2 $image3 etc, how can i make $randomimage = 1 of these at random on each refresh?

I've found lots of tutorials that generate a random number online but none that generate a random variable and am having trouble piecing this together.

Any help would be hugely appreciated.
Richard

I have a recursive sql query.  I'm doing this with an sql query as a function. 

Here is the function:

function select() {
  $sql="SELECT cat FROM $scat";
             if($result=$mysqli->query($sql)){
               if($result->num_rows>0){
                 $level =+1;
                 echo<<<HTML
                 <div class="menutitles">
                 onmouseover="document.getElementById('$scat').style.display='block';document.getElementById('$scat2').style.display='block';"
                 onmouseout="document.getElementById('$scat').style.display='none';document.getElementById('$scat2').style.display='block';"
                 >
                 <a class="topseltxt2" href="gem.php?cat=$scat"></a><br/>
                 <div class="contain" id="$scat">
                 <div class="menuitems" id="$scat2">
HTML;
               select();
               }else{$level=-1;
               if($level==1){echo"
               <a class='topseltxt' href='gem.php?cat=$scat'></a><br/>
               ";
               }else{
               While($row=$result->fetch_array()){
               $scat=replace($row[cat]);
               echo"
               <a class='topseltxt2' href='gem.php?cat=$scat'></a><br/>
               ";}
               }
        }
      }
}

and this is the database query with the function added to it:

$level=1;
$count=0;
    
    $sql = "SELECT cat FROM dczcat1 ORDER BY position";
    if ($result = $mysqli->query($sql)){
      if ($result->num_rows > 0){
        $count=+1;
         while ($row = $result->fetch_array()){
           $scat=replace($row[cat]);
           $scat2=replace($row[cat]) . 2;
           $cat=$row[cat];
           $sql="SELECT cat FROM $scat";
             if($result=$mysqli->query($sql)){
               $count=+1;
               if($result->num_rows>0){
                 $level =+1;
                 echo<<<HTML
                 <div class="menutitles">
                 onmouseover="document.getElementById('$scat').style.display='block';document.getElementById('$scat2').style.display='block';"
                 onmouseout="document.getElementById('$scat').style.display='none';document.getElementById('$scat2').style.display='block';"
                 >
                 <a class="topseltxt2" href="gem.php?cat=$scat"></a><br/>
                 <div class="contain" id="$scat">
                 <div class="menuitems" id="$scat2">
HTML;
               select();
               }else{$level=-1;
               if($level==1){echo"
               <a class='topseltxt' href='gem.php?cat=$scat'></a><br/>
               ";
               }else{
               While($row=$result . $count->fetch_array()){
               $scat=replace($row[cat]);
               echo"
               <a class='topseltxt2' href='gem.php?cat=$scat'></a><br/>
               ";}
               }
        }
      }echo"</div> </div> </div>";
    }
  }
}

I get the error:

Fatal error: Call to a member function fetch_array() on a non-object in /home/kangerc1/public_html/zleftsidebar.php on line 60

I think this is because I am using $result more than once and I need to call it $result1 $result2 $result3 etc. but how can I generate the name of the variable?