PHP - Script To Display Array As A Tree

I have a database table with categories and subcategories, with the structure of ::
 categoryid, categoryname, parentid

I have found numerous examples of printing out a tree structure of the table, and it works, but I would like to get all the data into an array, ultimately to print the array to a listbox to select from.  The array structure would be categoryid => categoryname.  The basic query I have now is as follows ::

Code: [Select]
function createCatTree($thisParentCatID=0,$depth=0)  {
$catIDQ = mysql_query("SELECT * from category WHERE parentcatid = $thisParentCatID;");
while ( $catIDQRow = mysql_fetch_array($catIDQ) )   {
$thisCatName = $catIDQRow['categoryname'];
$thisCatID = $catIDQRow['categoryid'];
        for($i = 0; $i < $depth; $i++) $spcstr.= "&nbsp;&nbsp;&nbsp;";  // create spaces to mimic a tree structure
$value = $spcstr.$thisCatName;  //  add spaces
echo $value;
createCatTree($thisCatID,$depth+1);
}
}

I was hoping to send the categoryid and categoryname to an array, like
Code: [Select]
$arr[$thisCatID] = $value;then change my function call to be
Code: [Select]
createCatTree($thisCatID,$depth+1,$arr);to send the array as a variable back to the function to hold the data going forward. Also, I added a section at the top of the function to see if the array needed to be initialized, like
Code: [Select]
if ($arr == "") {
$arr = array();
}
but this does not work.  Is there any way to use this basic recursive code and save all the data to an array to be worked on at a later time?

Hello all,

I need to make a rotating banner system for a client. They want about 5 banners to rotate with Javascript. However they want the visitor to sort of step through each one on each refresh.

So if I have 5 banners and I go to the site, I see banner 1 and then the javascript rotates through each one.

When I go back to the site, they want it to start with banner 2 and then the javascript will rotate through each one. Then on and on. Go back and see banner 3, then step through again.

I have a nice Javascript image slider I'd like to use. It basically just requires the images in order. Like:

<img src="1.jpg"/>
<img src="2.jpg"/>
<img src="3.jpg"/>
<img src="4.jpg"/>
<img src="5.jpg"/>

I've thought about it and I think the best way to do this would just be using PHP and a cookie. So when the viewer first goes to the site, PHP checks for the cookie. If it exists, it pulls the starting image #. If it doesn't, it starts at 0 and sets the cookie.

As for the images part, I figured an associate array would work well. The key would be what the cookie logic is based on, and then the image HTML would be in the array.

So you visit the site without a cookie it would be...

0 => <img src="1.jpg"/>
1 => <img src="2.jpg"/>
2 => <img src="3.jpg"/>
3 => <img src="4.jpg"/>
4 => <img src="5.jpg"/>

Then when you go back to the site, I need the array to be re-ordered like this:

0 => <img src="2.jpg"/>
1 => <img src="3.jpg"/>
2 => <img src="4.jpg"/>
3 => <img src="5.jpg"/>
4 => <img src="1.jpg"/>

What's the best way to go about this array logic? My experience with manipulating data inside arrays is rather limited. Basically I need it to say "Start with key # and then reorder array based on that"

Hi,

I'm hoping someone can help me with a simply array problem.

I have a table with two columns: "ID" and "Name" and 100 Names.

I also have an array called $myIDList containing 10 IDs. I want to display 10 Names from the table that correspond with the IDs from my array.

Originally, I tried to implode the array, adding commas and inserted into my MySQL query using:

$myQuery = "SELECT ID, Name FROM namesTable WHERE ID IN ($myImplodedIDList)";

The problem was that duplicated names (i.e. the same Name, but assiged to different IDs) would only be displayed once.

So now I'm trying to use the original array to loop through a second array and display Names where the ID matches an ID from my first array. I have succesffuly listed the table like this:

$myQuery = "SELECT ID, Name FROM myTable ORDER BY ID";

$myResult = mysql_query($myQuery) or die(mysql_error());

    while($row=mysql_fetch_array($myResult)) {
         echo $row['ID'];
         echo $row['Name'];
    }

But I want to do something like:

    while($row=mysql_fetch_array($myResult)) {
      while $row['ID'] = $myIDList {
      echo $row['Name']
      }
    }

Can someone shed some light onto this for me please?

Thank you!
Bryan

Well the subject line is pretty explicit.  I found this script that uploads a picture onto a folder on the server called images, then inserts the the path of the image on the images folder onto a VACHAR field in a database table. 
Code: [Select]




<?php

//This file inserts the main image into the images table.

//address error handling

ini_set ('display_errors', 1);
error_reporting (E_ALL & ~E_NOTICE);





//authenticate user
//Start session
session_start();

       
        //Connect to database
require ('config.php');

//Check whether the session variable id is present or not. If not, deny access.
if(!isset($_SESSION['id']) || (trim($_SESSION['id']) == '')) {
header("location: access_denied.php");
exit();
}

        else{ 







// Check to see if the type of file uploaded is a valid image type

function is_valid_type($file)

{

     // This is an array that holds all the valid image MIME types

     $valid_types = array("image/jpg", "image/jpeg", "image/bmp", "image/gif");

 

     if (in_array($file['type'], $valid_types))

         return 1;

     return 0;

}

 

// Just a short function that prints out the contents of an array in a manner that's easy to read

// I used this function during debugging but it serves no purpose at run time for this example

function showContents($array)

{

     echo "<pre>";

     print_r($array);

     echo "</pre>";

}

 

// Set some constants

 

// This variable is the path to the image folder where all the images are going to be stored

// Note that there is a trailing forward slash

$TARGET_PATH = "images/";

 

// Get our POSTed variable

$image = $_FILES['image'];

 

// Sanitize our input

$image['name'] = mysql_real_escape_string($image['name']);

 

// Build our target path full string.  This is where the file will be moved to

// i.e.  images/picture.jpg

$TARGET_PATH .= $image['name'];

 

// Make sure all the fields from the form have inputs

if ( $image['name'] == "" )

{

     $_SESSION['error'] = "All fields are required";

     header("Location: member.php");

     exit;

}

 

// Check to make sure that our file is actually an image

// You check the file type instead of the extension because the extension can easily be faked

if (!is_valid_type($image))

{

     $_SESSION['error'] = "You must upload a jpeg, gif, or bmp";

     header("Location: member.php");

     exit;

}

 

// Here we check to see if a file with that name already exists

// You could get past filename problems by appending a timestamp to the filename and then continuing

if (file_exists($TARGET_PATH))

{

     $_SESSION['error'] = "A file with that name already exists";

     header("Location: member.php");

     exit;

}

 

// Lets attempt to move the file from its temporary directory to its new home

if (move_uploaded_file($image['tmp_name'], $TARGET_PATH))

{

     // NOTE: This is where a lot of people make mistakes.

     // We are *not* putting the image into the database; we are putting a reference to the file's location on the server

     $sql = "insert into images (member_id, image_cartegory, image_date, image) values ('{$_SESSION['id']}', 'main', NOW(), '" . $image['name'] . "')";

     $result = mysql_query($sql) or die ("Could not insert data into DB: " . mysql_error());

     header("Location: images.php");

echo "File uploaded";

     exit;

}

else

{

     // A common cause of file moving failures is because of bad permissions on the directory attempting to be written to

     // Make sure you chmod the directory to be writeable

     $_SESSION['error'] = "Could not upload file.  Check read/write persmissions on the directory";

     header("Location: member.php");

     exit;

}







        } //End of if session variable id is not present.
 


?> 


The script seems to work fine because I managed to upload a picture which was successfully inserted into my images folder and into the database.
Now the problem is, I can't figure out exactly how to write the script that displays the image on an html page. I used the following script which didn't work.


Code: [Select]

//authenticate user
//Start session
session_start();

       
        //Connect to database
require ('config.php');

$sql = mysql_query("SELECT* FROM images WHERE member_id = '".$_SESSION['id']."' AND image_cartegory = 'main' ");
$row = mysql_fetch_assoc($sql);
$imagebytes = $row['image'];
header("Content-type: image/jpeg");
print $imagebytes;







Seems to me like I need to alter some variables to match the variables used in the insert script, just can't figure out which.  Can anyone help??

Hello all,
I am new to php and was wondering if i could get some guidance here. I am using phpAdmin 2.6.0 and running Mysql 4.1.21. here is my situation....

I have a script that allows us to upload a new product name, product code, category and a PDF file to the data base. There is also a folder on the server that has the PDF files in it. I think I deleted the code on the page (library.php) that displays the files for the client to download. My goal here is after I upload everything I want it to then be displayed on the page with a link to the PDF file. Here is the page that has the links on it. I hope that I explained this correctly. I am not a programmer but do have some idea and have been reading up on php to try and figure this out. I am looking to create the script that would display the links on the library.php page. Any help would be great. The other link is to the script that allows us to upload.

www.pennstateind.com/library.php

www.pennstateind.com/lib-admin.php

Hi,

Is it possible to display the whole array of voteID in this url?

https://falling-skies-tv.loopiasecure.com/vote_result.php?voteID=292&comp=falling-skies-danmark

Is there a script/website/code I can use?

Thank you!

my user pm table has a touserarray field which is a serialized array that looks like this:
a:1:{s:2:"cc";a:1:{i:15773;s:14:"testusername";}}
i know I can unserialize it with the unserialize() function.
But I need to extract the value corresponding to 'i' field and the 's' field that has the username.

So  from the above array, i need to retrieve the value '15773' and 'testusername'.
Can someone tell me how?
 

$color=array("01 orange","02 blue","03 red");

If my array was structured this way where the first two characters represented a code, how would I be able to display the code and color distinctly?

I had posted a similar post  a few days back, about an display script which is supposed to retrieve a stored image from my database and displays it on an html page, but fails to do so.  Someone suggested that I  upload my image onto a folder on the server and then save the image path name in my database.  I found this script(below)  which does just that.  It uploads the image into a folder on the server called images and stores the image path into a table in my database called images and I know this script works because i saw the file saved in the images folder and the path name inserted into the images table.
Code: [Select]


<?php

//This file inserts the main image into the images table.

//address error handling

ini_set ('display_errors', 1);
error_reporting (E_ALL & ~E_NOTICE);





//authenticate user
//Start session
session_start();

       
        //Connect to database
require ('config.php');

//Check whether the session variable id is present or not. If not, deny access.
if(!isset($_SESSION['id']) || (trim($_SESSION['id']) == '')) {
header("location: access_denied.php");
exit();
}

        else{ 







// Check to see if the type of file uploaded is a valid image type

function is_valid_type($file)

{

     // This is an array that holds all the valid image MIME types

     $valid_types = array("image/jpg", "image/jpeg", "image/bmp", "image/gif");

 

     if (in_array($file['type'], $valid_types))

         return 1;

     return 0;

}

 

// Just a short function that prints out the contents of an array in a manner that's easy to read

// I used this function during debugging but it serves no purpose at run time for this example

function showContents($array)

{

     echo "<pre>";

     print_r($array);

     echo "</pre>";

}

 

// Set some constants

 

// This variable is the path to the image folder where all the images are going to be stored

// Note that there is a trailing forward slash

$TARGET_PATH = "images/";

 

// Get our POSTed variable

$image = $_FILES['image'];

 

// Sanitize our input

$image['name'] = mysql_real_escape_string($image['name']);

 

// Build our target path full string.  This is where the file will be moved to

// i.e.  images/picture.jpg

$TARGET_PATH .= $image['name'];

 

// Make sure all the fields from the form have inputs

if ( $image['name'] == "" )

{

     $_SESSION['error'] = "All fields are required";

     header("Location: member.php");

     exit;

}

 

// Check to make sure that our file is actually an image

// You check the file type instead of the extension because the extension can easily be faked

if (!is_valid_type($image))

{

     $_SESSION['error'] = "You must upload a jpeg, gif, or bmp";

     header("Location: member.php");

     exit;

}

 

// Here we check to see if a file with that name already exists

// You could get past filename problems by appending a timestamp to the filename and then continuing

if (file_exists($TARGET_PATH))

{

     $_SESSION['error'] = "A file with that name already exists";

     header("Location: member.php");

     exit;

}

 

// Lets attempt to move the file from its temporary directory to its new home

if (move_uploaded_file($image['tmp_name'], $TARGET_PATH))

{

     // NOTE: This is where a lot of people make mistakes.

     // We are *not* putting the image into the database; we are putting a reference to the file's location on the server

     $sql = "insert into images (member_id, image_cartegory, image_date, image) values ('{$_SESSION['id']}', 'main', NOW(), '" . $image['name'] . "')";

     $result = mysql_query($sql) or die ("Could not insert data into DB: " . mysql_error());

     header("Location: images.php");

echo "File uploaded";

     exit;

}

else

{

     // A common cause of file moving failures is because of bad permissions on the directory attempting to be written to

     // Make sure you chmod the directory to be writeable

     $_SESSION['error'] = "Could not upload file.  Check read/write persmissions on the directory";

     header("Location: member.php");

     exit;

}







        } //End of if session variable id is not present.
 


?> 











Now the display image script accompanying this insert image script is shown below. Code: [Select]
<?php

// Get our database connector

require("config.php");

?>

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN"

      "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">

 

<html xmlns="http://www.w3.org/1999/xhtml">

    <head>

        <title>Dream in code tutorial - List of Images</title>

    </head>

 

    <body>

     

        <div>

 

            <?php   

                // Grab the data from our people table

                $sql = "SELECT * from images  WHERE  image_cartegory = 'main'";

                $result = mysql_query($sql) or die ("Could not access DB: " . mysql_error());

 

                while ($row = mysql_fetch_assoc($result))

                {

                    echo "<div class=\"picture\">";

                    echo "<p>";

 

                    // Note that we are building our src string using the filename from the database

                     echo "<img src=\"images/ " . $row['filename'] . " \" alt=\"\" />";

                   

                    echo "</p>";

                    echo "</div>";

                }



            ?>

         

        </div>

    </body>

</html>



Well, needless mentioning, the picture isn't displayed.  Just a tiny jpeg icon is displayed at the top left hand corner of the page.  Figuring out that the problem might be the lack of a header that declares the image type, I add this line
header("Content-type: image/jpeg");
but all it does is display a blank white page, without the tiny icon this time.  What am I missing? Any clues??

I came up with the following script for displaying an uploaded picture on a webpage but for some reason I can't pinpoint, the picture won't be displayed.  I checked my database from php myadmin and sure enough, the picture was successfully uploaded but somehow, the picture won't be displayed.  So here is the display script.  I named it display_pic.php

Code: [Select]

<?php


//address error handling

ini_set ('display_errors', 1);
error_reporting (E_ALL & ~E_NOTICE);





//authenticate user
//Start session
session_start();

       
        //Connect to database
require ('config.php');

                             //address error handling

                             ini_set ('display_errors', 1);
                             error_reporting (E_ALL & ~E_NOTICE);

                             //include the config file
                              require('config.php');

                             $image = stripslashes($_REQUEST[imname]);
                             $rs = mysql_query("SELECT* FROM images WHERE member_id = '".$_SESSION['id']."' AND image_cartegory = 'main' ");
                             $row = mysql_fetch_assoc($rs);
                             $imagebytes = $row[image];
                             header("Content-type: image/jpeg");
                             print $imagebytes;

                            
 ?>


The tag on the html page that's supposed to display the picture reads something like this
<img src="display_pic.php"   width="140" height="140">



And just in case this might help, I will include the image upload script below, which I think worked just fine because my values were successfully inserted into the database.
Code: [Select]
<?php

//This file inserts the main image into the images table.

//address error handling

ini_set ('display_errors', 1);
error_reporting (E_ALL & ~E_NOTICE);





//authenticate user
//Start session
session_start();

       
        //Connect to database
require ('config.php');

//Check whether the session variable id is present or not. If not, deny access.
if(!isset($_SESSION['id']) || (trim($_SESSION['id']) == '')) {
header("location: access_denied.php");
exit();
}

        else{ 
// Make sure the user actually 
// selected and uploaded a file
if (isset($_FILES['image']) && $_FILES['image']['size'] > 0) { 

      // Temporary file name stored on the server
      $tmpName  = $_FILES['image']['tmp_name'];  
       
      // Read the file 
      $fp      = fopen($tmpName, 'r');
      $data = fread($fp, filesize($tmpName));
      $data = addslashes($data);
      fclose($fp);
      

      // Create the query and insert
      // into our database.
      $query = "INSERT INTO images (member_id, image_cartegory, image_date, image) VALUES ('{$_SESSION['id']}', 'main', NOW(), '$data')";
      $results = mysql_query($query);
      
      // Print results
      print "Thank you, your file has been uploaded.";
      
}
else {
   print "No image selected/uploaded";
}

// Close our MySQL Link
mysql_close();


} //End of if statmemnt.
 


?> 



So any insights as to why the script fails to display the image? Any help is appreciated.

Hello all, I have followed the stop on this page to what I thought was a tee: http://www.inkplant.com/code/pull-tw...-your-site.php but I guess I was wrong because I am getting errors. I have setup the database and table properly and created all files, but when I do step 4 and place the code on my site I am getting this error Code: [Select]
Fatal error: Call to undefined function dbQuery()
The api pull went fine and it stored the tweets in the database, but it can't display them. Is there a missing step about connecting to the database somewhere? Thanks for any help

Example: i have an array of names
array{
=> john [1] = ted
[2] = jeff
}

and i display them like this:

john, ted, jeff

but jeff is currently logged in so i want his name to display first(i have my reasons lol)

how would i go about this?
i got as far as:


if(in_array($user->user_name, $likes))
            {
                $user_key = array_search($user->user_name, $likes);
            }


then hit a brain freeze.

Thanks

Here is my current code to display files from the end to beginning : Code: [Select]
<?php
function func() {
$date = date("Y-m-d");
$picarray = array();
$userarray = array();
        $handle = opendir ('pqimages/'.$date.'/');
        while (false !==($file = readdir($handle))) {
            if($file != "." && $file != "..") {
                $picarray[] = "http://blah.com/pqimages/".$date."/".$file."";
                $userarray[]= "pqimages/".$date."/".$file."";
            }
        }
        sort($picarray);
        sort($userarray);
        closedir($handle);
$json_array = json_encode($picarray); 
echo $json_array;  
}
func();
?>

How could I change this from end to beginning  for ($picarray)?

Cheers,
George

Hi, 

It's my first time here. I'm not a coder but I sometimes need to find solutions by myself so here I am...

I think this one will be easy for all of you.

I generate a PDF displaying data from fields on a DB

They all display very well except the multiple choices.

Here is an exerpt of my code; I tried the implode code to display my values separated by commas but in the document it stil displays "Array" (screenshot enclosed) instead of something like: "Sante mentale, Reussite scolaire, Environnement sain "

And I obviously don't know what's wrong...

<tr><td><?php echo $fields['prenomact']['value']; ?> <?php echo $fields['nomact']['value']; ?></td><td><?php echo $fields['titreact']['value']; ?></td></tr>
<tr><td><?php echo $fields['courrielact']['value']; ?></td><td><?php echo $fields['telact']['value']; ?></td></tr>
<tr><td>Enjeux visés: <?php $fields['enjeux']=implode(",", $array); echo $fields['enjeux']['value'];?></td><td>territoi <?php echo $fields['territoire']['value']; ?></td></tr>
<tr><td>Participation des enfants: <?php echo $fields['niveauparticipation']['value']; ?></td><td>Nombre d'enfants participants : <?php echo $fields['nbenfants']['value']; ?></td></tr>

Thanks in advance for your help,

Valérie

 

Hi, i'm getting an error when I load my php code in a browser. Here's my code snippet:
<?php
  mysql_connect("localhost","root");
  mysql_select_db("something");
  $bellProductsArray=array(1=>"Apple_iPhone3GS.jpg",2=>"Apple_iPhone4.jpg");
 
  $result=mysql_query("SELECT Name, Manufacturer FROM bellProducts WHERE ID=1");
  $row=mysql_fetch_assoc($result);
  print "Name: {$row['Name']} Manufacturer: {$row['Manufacturer'] }";
 
?>

<img src=<?php array_values($bellProductsArray); ?>  alt="Apple iPhone 3GS" title="Apple iPhone 3GS" />

**I can't get it to display the first image (eg: Apple_iPhone3GS.jpg), I made sure that I have the image in same directory. please help!1!