PHP - I Need Code That Reads From The Role Database And Displays The File According To It

I'm not totally new to php but I'm no guru either. I'm building an intranet and have three seperate user roles, user - manager - admin. There are some menu items I don't want to allow simple users to see. This will expand later to give them different views of a page as well (some can view/others can edit). As it stands the login validation is holding they're user level in $SESSION. To give you an idea take a look at what I'm trying to do:

Code: [Select]
function usermenu($usermenu) {
if($user_level=0)
echo ("<ul id="gooeymenu2" class="solidblockmenu">
<li><a href="main.php">Home</a></li>
<li><a href="forms.php">Forms</a></li>
<li><a href="/support/index.php" target="_new">Support</a></li>
<li><a href="documents.php">Documents</a></li>
<li><a href="admin/index.php">Admin</a></li>
<li><a href="logout.php">Logout</a></li>
</ul>
<script>
gooeymenu.setup({id:'gooeymenu2', selectitem:1, fx:'swing'})
</script>" ");

else($user_level=1,2)
echo ("<ul id="gooeymenu2" class="solidblockmenu">
<li><a href="main.php">Home</a></li>
<li><a href="forms.php">Forms</a></li>
<li><a href="/support/index.php" target="_new">Support</a></li>
<li><a href="documents.php">Documents</a></li>
<li><a href="new.php">New Adviser</a></li>
<li><a href="admin/index.php">Admin</a></li>
<li><a href="logout.php">Logout</a></li>
</ul>
<script>
gooeymenu.setup({id:'gooeymenu2', selectitem:1, fx:'swing'})
</script>" ");

Any help would be great, thanks in advance.

Jason

Hello all! So glad I found this forum. I would appreciate some assistance please.

I'm working on a filter for a Custom Post Type . I need it to filter the list depending on the user's role. The way this should work is the following...

* Users in roles "formusers1" and "formusers2" can post. Users can only see their own posts.
* Users in role "formchecker1" can see all posts assigned role "formusers1' and can approve each post.
* Users in role "formchecker2" can see all posts assigned role "formusers2' and can approve each post.
* Users in roles "formsupervisor" and "administrator" can see everyone's posts.

So far I can filter by roles "formusers1" and "formusers2" using  `$query->set('author', $current_user->ID);` . However, when try to filter the list for role "formchecker1" I see posts from all roles. What am I doing wrong? Here's the rest of the code. Thanks for checking out!

```
add_action('pre_get_posts', 'filter_posts_list');

function filter_posts_list($query) {
    //GLOBAL VARIABLES
    global $pagenow, $typenow;

    //MY VARIABLES
    global $current_user;
    get_currentuserinfo();

    //FILTERING
    if (current_user_can('formchecker1') && ('edit.php' == $pagenow) &&  $typenow == 'mycustomcpt' ) {
        $query->set('author', 'formusers1');
    }

    if (current_user_can('formchecker2') && ('edit.php' == $pagenow) &&  $typenow == 'mycustomcpt' ) {
        $query->set('author', 'formusers2');
    }

    if ((current_user_can('formusers1') || current_user_can('formusers2')) && ('edit.php' == $pagenow) &&  $typenow == 'mycustomcpt') {
        $query->set('author', $current_user->ID);
    }
}
```

 
Been scratching my head about this all day.  Cannot figure it out.  Any help would be appreciated.

Code: [Select]
<?php

$selectClause = 'SELECT * FROM idInfo WHERE';
$whereClause = '';

function addWhere($fxnName, $fxnValue) {
if (!$whereClause) {
return $fxnName . '=\'' . $fxnValue . '\'';
} else {
return ' AND ' . $fxnName . '=\'' . $fxnValue . '\'';
}
}
$whereClause.=addWhere("firstName", $firstName);
$whereClause.=addWhere("lastName", $lastName);
$whereClause.=addWhere("idNum", $idNum);

echo 'mysql_query("' . $selectClause . ' ' . $whereClause . '"); <br /><br />';
?>

This is meant to generate a query of the type:

Code: [Select]
SELECT * FROM idInfo WHERE firstName='firstNameINPUT' AND lastName='lastNameINPUT' AND idNum='idNumINPUT'
For some reason, though, the if statement always reads false.  No matter how I try to work the syntax Code: [Select]
if (!$whereClause)
if ($whereClause)
if ($whereClause == NULL)
if ($whereClause != NULL)
if (empty($whereClause)
etc.
I've tried all of those and more, experimenting with and without quotes.  They always return a FALSE for the existence of the variable, so that the string never contains the word "AND" as it should by the end.  I've tried these same if statements outside the function and they work fine.  Inside the function, it doesn't matter what the value of the variable is, the function runs as though it has no value.

Comments?
Suggestions?

I am trying to populate a custom field called "Customer Type" current user role.  The custom field is displayed on my checkout page.   I tried the below in my functions.php of my child theme and thought it would work but it does nothing.  Can anyone tell me what I might be doing wrong?  

$user = wp_get_current_user();
function onboarding_update_fields( $fields = array() ) {

       $fields['customertype'] = $user;

   return $fields;
}

add_filter( 'woocommerce_checkout_fields', 'onboarding_update_fields' );

Edited April 11 by JayX

Advance thank you.

Can you help please.

The error.....

Warning: mysql_fetch_assoc() expects parameter 1 to be resource, string given in C:\wamp\www\test_dabase.php on line 24

code.
Code: [Select]
<?php

//database connection.
$DB = mysql_connect("localhost","root") or die(mysql_error());

if($DB){

//database name.
$DB_NAME="mysql";

//select database and name.
$CON=mysql_select_db($DB_NAME,$DB)or die(mysql_error()."\nPlease change database name");


// if connection.
}if($CON){
    
    //show tables.
  $mysql_show="SHOW TABLES";
    
    //select show and show.
      $mysql_select2="mysql_query(".$mysql_show.") or die(mysql_error())";
    
    }
    
    //if allowed to show.
    if($mysql_select2){
        
        //while it and
    while($data=mysql_fetch_assoc($mysql_select2)){
        
        //show it.
        echo $data;
        
        }
    }    
    ?>

Hello Friends,

I need to convert the DOC (WordFile) in to PDF through PHP code in my web. According to requirement whenever the user upload the .Doc file code should convert that .Doc file into .PDF file at runtime.

I have tried by using PDFLib and FPDF functions but  when the .Doc file contains the images,tables etc. either FPDF or PdfLib being falied to generate PDF file

I am using Linux-PHP-MySQL Technolgoies for development.

I am expecting suggestions and help from expert friends who can help me out to achive this.

Thanks!

how do I use a connection file (connection.php) in multiple programs?  I think include?

How can I use PHP to export my database into a .sql file?

Hi PHP freaks!!   I have manage to make a script to upload file in database and its working. I will share to you the codes so for other viewers and readers to use also.

if($result){
			if($_FILES['LRCard']['name'] != ""){
				$filename = $_FILES['LRCard']['name'];
				$ext = strrchr($filename,".");
				$LRCardname = $student_id;
				$LRCardname .="_". $filename; 
				if($ext ==".jpg" || $ext ==".jpeg" || $ext ==".JPG" || $ext ==".JPEG" || $ext ==".gif" || $ext ==".GIF"){
					$size = $_FILES['LRCard']['size'];
					if($size > 0 && $size < 5000000){
						$archive_dir = "LRCards";
						$userfile_tmp_name = $_FILES['LRCard']['tmp_name'];
						if(move_uploaded_file($userfile_tmp_name, "$archive_dir/$LRCardname")){
							/*
								if LRC is successfully uploaded then LRC is stored in database.
							*/
							mysql_query("update student_information set LRCard='$LRCardname' where student_id='$student_id'", $link_id); 
							$flag = "success"; 
							if(mysql_error()!=null){
								die(mysql_error());
							}

						}
						else{
							if(file_exists('LRCard/' . $LRCardname)) {
								unlink('LRCards/' . $LRCardname); 
							}
							rollbackData();
						}
					}
					else{
						if(file_exists('LRCards/' . $LRCardname)) {
							unlink('LRCard/' . $LRCardname); 
						}
						rollbackData();
						die("You can upload LRCard of 5 MB size only. Please, try again.");
					}
				}
				else{
					if(file_exists('LRCards/' . $LRCardname)) {
						unlink('LRCards/' . $LRCardname); 
					}
					rollbackData();
					die("You can upload LRCard of .jpg, .jpeg, .gif extensions only. Please, try again. ");
				}
			} 	
		}
		else{
			$flag="error";
		}
		if($flag == "success"){
			mysql_query(" COMMIT ");
			$flag="success";
			if(mysql_error() != null){
				die(mysql_error());
			}
	    }

Now, my problem is how to make a Php script to DOWNLOAD this uploaded file considering the 1. file path 2. file name 3. file extension. 4. student_id (to determine the specific file from a specific student)   My plan is to make a download button which hyperlink to download.php and after clicking that button the specified file to a specific student id will automatically be downloaded by the browser.  

 Hey anybody can please guide me where I m wrong in this php code
please reply

here is the code
Only variables should be passed by reference on line 3
in that function area function get_file_extension($file_name)

main error I guess is the explode one

Aand yeah I checked adding
$file_name = $_FILES['fld']; before the function is started

<?php

require("include/dbconn.php");

    //$username = mysql_real_escape_string($_POST['username1']);



function get_file_extension($file_name) {
   return end(explode('.',$file_name));

}
function errors($error){
   if (!empty($error))
   {
         $i = 0;
         while ($i < count($error)){
         $showError.= '<div class="msg-error">'.$error[$i].'</div>';
         $i ++;}
         return $showError;
   }// close if empty errors
} // close function

if (isset($_POST['submit'])){

   $username = "Sunny";
   $date1 = date("d m y");
    mysql_select_db("deals") or die(mysql_error());

   if($username == "")
   {
   //header("location:newnewuser1.php");
   echo "Username empty";
    //header("location:newnewuser1.php");   
      
   }
   else
   {
   
   function createRandomPassword() {
    $chars = "abcdefghijkmnopqrstuvwxyz023456789";
    srand((double)microtime()*1000000);
    $i = 0;
    $pass = '' ;
    while ($i <= 7) {

        $num = rand() % 33;

        $tmp = substr($chars, $num, 1);

        $pass = $pass . $tmp;

        $i++;

    }
    return $pass;
}
// Usage

$password = createRandomPassword();

   //$password = mysql_real_escape_string($_POST['Password']);
$date = strtotime("+2hours");
$date1 =  date("d-m-y",$date);
 
if(get_file_extension($_FILES["fld"]["name"])!= 'csv')
{
$error[] = 'Only CSV files accepted!';
echo "Wrong Input";
}//get_file_extension
 
if (!$error)
{
 
$tot = 0;
$handle = fopen($_FILES["fld"]["tmp_name"], "r");

   
 while (($data = fgetcsv($handle, 1000, ",")) !== FALSE)
{
   
   for ($c=0; $c < 1; $c++) {
     
       try {
     mysql_query("INSERT INTO newusers (email,password,location,company,name,contact,sourcename,date)VALUES('".mysql_real_escape_string($data[0])."','$password','".mysql_real_escape_string($data[1])."',
   '".mysql_real_escape_string($data[2])."','".mysql_real_escape_string($data[3])."','".mysql_real_escape_string($data[4])."','$username1','$date1')");
    
    $tot++;
       
      }
      catch(Exception $e)
      {
      PRINT 'ERROR:' +$e;
     }   
   }//for



}//while
fclose($handle);


 
}// end no error
}//close if isset upfile
 



}

?>

and this is the form code

<form name="f1" action="" method="POST" enctype="multipart/form-data">
                             <table><tr><td><input type="file" value="Browse" name="fld" onselect="CheckExtension(fld)"><br/></td><td>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</td></tr><tr><td>Username :<input type="text" name="username1" value="" id="user" onfocus="validateForm1()"></td>
                      <td><input type="submit" name="submit" value="submit" id="submit"></td></tr></table>
                              </form>

Please help me I m stuck in this quite badly
Please


[attachment deleted by admin]

Hi, I use MySQL for real projects of course... but, I like to challenge myself and I was wondering how people make advance applications with Flat File databases... My questions are...

>Display
  >Reading The Flat File Data
  >Sorting it Alphabetically For Display
  >How to create a WHERE clause affect with the data

>Storage
  >Adding Records
  >Deleting Records
  >Editing Records

I went over a lot of this on my own, but I'm stuck on certain parts because google only has so much on Flat File Databases... also how would u secure a Flat File Database from someone just simply direct linking to it and viewing it?

I want to make a time limit in-between posts on my forum. Should I make a table in the database to store the seconds, or read it/write via a file?

Hello Everyone 

I am new to php and indeed Web Development.

After testing and Playing a bit, I can get the following code to work as two files, the form calling the *.php file to insert into the database, however, I am trying to create one html/php file that displays the form and then executes the php code to insert into the database once user clickes the button.

Please can you assist me with the code? I have something horribly wrong and I cannot find it.

Code: [Select]
<?php>
<html>
<head>
<title>Personal Details</title>
</head>
<body>
<form method="post" action="contactdetails.html"><font face="Arial">

Call Sign:<br>
  <input name="callsign" size="5" type="text"><br>

Surame:<br>
  <input name="surname" size="30" type="text"><br>

First Name:<br>
  <input name="firstnames" size="30" type="text"><br>
  
Known as:<br>
  <input name="knownas" size="30" type="text"><br>
  
RSA ID No.:<br>
  <input name="rsaid" size="13" type="text"><br>
  
Birth Date:<br>
  <input name="birthdate" size="12" type="text"><br>

<input name="Insert" value="Next" type="submit"></form>

</font><br>
</body>
</html>

//php to insert data into table

$callsign = $_POST['callsign'];
$surname = $_POST['surname'];
$firstnames = $_POST['firstnames'];
$knownas = $_POST['knownas'];
$rsaid = $_POST['rsaid'];
$birthdate = $_POST['birthdate'];

mysql_connect ("localhost", "jredpixm_testuse", "PHPDevelopment") or die ('I cannot connect to the database because: ' .mysql_error());
mysql_select_db ("jredpixm_test");

$query="INSERT INTO personal_details (callsign, surname, firstnames, knownas, rsaid, birthdate)Values ('$callsign', '$surname', '$firstnames', '$knownas', '$rsaid', '$birthdate')";

mysql_query($query) or die ('Error updating Database');

echo "<p>Thanks, your information has been added to the database.</p>";

?>


Regards

Allen

As per the client requirement he wants to read words from a file and want to insert them in database  one by one . ie a confirmation is required whether he wants to insert is or not. I have developed the thing but the problem is when i am retrieving the words it asks for all the words. i will show u wat the problem is. My main query is i want to pause the exectution until user's input.

Code: [Select]
<script type="text/javascript" src="jquery.js"></script>
<script type"text/javascript">
function show() {
document.getElementById('form#submit').fadeIn();
alert("Abhinav");
}
$(document).ready(function(){
$("form#submit").submit(function() {
// we want to store the values from the form input box, then send via ajax below
var word    = $('#word').attr('value');
var synonym     = $('#synonym').attr('value');

$.ajax({
type: "POST",
url: "ajax.php",
data: "word="+ word + "&synonym=" + synonym,
success: function(){
$('form#submit').hide();
//$('form#submit :input').val("");
$('div.success').fadeIn();
}
});
return false;
});
});
</script>

<?php 
$file=fopen("welcome.txt","r") or exit("Unable to open file!");
@$word="";
@$c="";
while (!feof($file))
{
  $c=fgetc($file);
if(strcmp($c," ")==0)
{ echo "".
 
"<form id='submit' method='post' name='submit' action='' >".  
        "<fieldset>  ".
            "<legend>Enter Information</legend>  ".
"            <label for='word'>Word:</label>  ".
"<input id='word' class='text' name='word' size='20' type='text' value='$word' >  ".
"            <label for='Synonym'>Synonym:</label>  ".
"<input id='synonym' class='text' name='synonym' size='20' type='text'>  ".
"            <button class='button positive' type='submit'>  Add Client </button>  ".
   " </fieldset>  ".
   " </form>  ".
"<div class='success' style='display:none;'>Client has been added.</div>  ".
"</div>";
 
$word="";
}
else
$word=$word.$c;
}
fclose($file);

?>



Now what i want is after user clicks add client then only next insertion box must be shown. please check the attached image to see the output. Guys pls help me over this. Thnx in advance

Hey all,
I've written a php search feature for a mysql database. The search returns a file name like sample.gif, how would I go about displaying the actual image instead? Since the images all have the same location could I have the search return the string into a variable then make the whole thing a link? Thanks.