PHP - Creating Multipul Menus From A Single File

This topic has been moved to Third Party PHP Scripts.

http://www.phpfreaks.com/forums/index.php?topic=358810.0

Hello.

I want to make a simple website, where I can upload a zip file and download it from a URL with just HTTP GET request.

When I download the file, the file will then be deleted on the server.

Can't find any examples. Maybe, I've searched wrong.

Has somebody some written project links or some tips for me how can I achieve this?

 

Hi everyone! Nice to meet/see/read you.

This is my first post and it will probably be a lame one so apologies. I'm a pretty good frontend designer and I've always struggled to learn PHP and my new job is forcing it on me so I'm happy in a "throw me in the deep end" kinda way.

I am really stuck! I have a single page that is using jquery to scroll my content left and right via the menu (example attached). What I am trying to do is when on the homepage, don't show the "main logo". When on any other link, show it.

I immediately thought of an if/else statement but realised I don't know what or how exactly to target since everything is on the same page and the URL doesn't change. Could I target the image title that is the only image on the homepage? So, something like:

<?PHP
$image_title = 'home';

if (XXXXX($image_title)) {
echo "IMAGE";
} else {
echo "NO IMAGE";
}

?>
I know that's wrong, I put XXX where I feel like something possibly helpful should go. Heh.   

I'm not asking for someone to write code for me, I'm happy to learn but but I would be very grateful if someone could just guide me to what I need to do and off to Google I will go. I haven't been able to find anything that is like: "if image title equals".

Hmm... I hope I can do this with PHP.

Thanks again,
Linda

I know it needs a for loop, but i don't know where in the code i should be putting it?

Code: [Select]
function check_input($value)
{
// Stripslashes
if (get_magic_quotes_gpc())
  {
  $value = stripslashes($value);
  }
// Quote if not a number
if (!is_numeric($value))
  {
  $value = "'" . mysql_real_escape_string($value) . "'";
  }
return $value;
}

$_POST = array_map('check_input', $_POST);

$sql="INSERT INTO testimonials (CustomerName, Town, Testimonial, SortOrder, Images)
VALUES
({$_POST['customername']}, {$_POST['town']}, {$_POST['testimonial']}, {$_POST['sort_order']}, '$imgname'

)";

}
if (!mysql_query($sql,$con))
  {
  die("<br>Query: $sql<br>Error: " . mysql_error() . '<br>');
  }
echo "<p align=center><b>1 testimonial added</b></p>";

mysql_close($con);


Thanks in Advance,
Steve

Hey all,

I am building a simple cms. I have a posts table and I have an images table. A post has many images and images has a foreign key to the posts table. So when a user edits, updates, creates, and deletes a post, they affect the images related to the post. Sometimes a post can have more than one image, like three images. Hence I rendered this in the view (note that I am using a datamapper that converts tables to objects and fields to key/value pairs):

Code: [Select]
foreach($records as $post){
echo form_open_multipart("homes/update/$post->id"); //File uploads require a multipart form. Default form handling uses the application/x-www-form-urlencoded content type. Multipart forms use the multipart/form-data encoding. //this is critical to pass the id as part of the action attribute of the form, so we can use our params hash to target the id to update the specific record
 
  echo label('Update Title');
echo form_input('title',$post->title);
echo label('Update Body');
echo form_textarea('body',$post->body);

  $images = $post->images->include_join_fields()->get();
if(!is_null($images->image_file_name)){ 
  echo label('Update Images');
foreach($images as $image){
echo form_upload('image_file_name',$image->image_file_name); 
}
}

}
echo form_submit('submit','Update');
The above line of code will render a few input file types. The problem occurs during posting to my update method. It is looking for one parameter from the input file field and so if I upload three different images, it will only look for one and write only one to database:

Code: [Select]
$field_name = 'image_file_name';
  if ( ! $this->upload->do_upload($field_name)){
  $error = array('error' => $this->upload->display_errors());
echo $error['error'];
// redirect('homes/edit');
  }
else {
  $data = array('upload_data' => $this->upload->data());
$image_file_name = $data['upload_data']['file_name'];

Is it possible to do wht I am trying to do? Should I only have on input file type per form submission or is that I need to fix the code to accomodate for multiple submissions by creating an array of sorts?

Thanks for response.

I'm ok with PHP but probably not half as good as some of you guys on here.

I am basically trying to find a way to grab a line from a huge and I mean huge text file.... its basically a list of keywords I want to call by line number but without preferably going through them all before I get to that line.....otherwise couldmcrash my server obviously.

At the moment im using this

Code: [Select]
$lines = file('http://www.mysite.com/keywords.txt');

// Loop through our array, show HTML source as HTML source; and line numbers too.
foreach ($lines as $line_num => $line) {
echo "$line_num";
}
This works but im sure theres gotta be a better way of doing to save on usuage because this is putting the whole file into the memory and if I can simply say to php give me line number 97, would umm RULE....

Hope you guys can come up with a solution as your much smarter than me ty

I am retrieving values in form of array from a database table and out of the I have to build multidimensional array depending upon the parent->child relationship in the same array.
The result is as
 
    => Array
                (
                    => Array
                        (
                            [label] => Shirt
                            [uri] => #
                        )

                    [1] => Array
                        (
                            [label] => Jeans
                            [uri] => #
                        )

                    [2] => Array
                        (
                            [label] => Lowers
                            [uri] => #
                        )

                    [3] => Array
                        (
                            [label] => T-Shirts
                            [uri] => #
                        )

                    [4] => Array
                        (
                            [label] => cotton shirt
                            [uri] => #
                        )

                    [5] => Array
                        (
                            [label] => Trousers
                            [uri] => #
                        )

                )
each is related to some parent-id.
Can any body me any idea.

PHP script return 20 UL LIST values like,

< ul >

A
B
C
D
E
F
G
H
I
J
K
L
M
N
O
P
Q
R
S
T

< /ul >

How to display UL LIST into row wise 5 columns like

A B C D
E F G H
I J K L
M N O P
Q R S T (20 ITEMS)

Edited December 7, 2019 by aveeva

Hi all

I am having some issues

basically

Code: [Select]
$sftp->put("server.properties",
"allow-nether=".$nether."
level-name=".$lname."");

is not printing the variables. $sftp is part of the phpseclib and the fuction is in the same format as fwrite(), I have tried and failed using fwrite as well! I can echo out the 1st variable correctly, but as soon as I add the second I just get

Code: [Select]
allow-nether=
level-name=

The variables $nether and $lname exist and I have echo'd them to check.

I guess this is a formatting problem, just cant work it out!

I am new to Php and the linux environment.  I have written a php program which sames a file to folder using xampp in Windows but I am having a problem with the pathway using my Ubuntu server and LAMP.  My code I used in Windows was:

$fileName = "guessbook.txt";
$file = fopen ($fileName, "ab");
if(!$file)
{
    echo "ERROR! did not create the file! Exiting.<br />;
    exit();
}

This file was saved correctly using xampp into the htdoc folder and worked fine when I called it in my browser.  I tried to substitute a Linux pathway ie: /var/www/assignment2/guessbook.txt, where guessbook is in the code but still nothing.  Can someone explain how to write the correct pathway to save this file?

Thank you

Paul

Hello
I was trying from days to create an array from a txt file by i can't make it happen. I am so desperate now and i want your help guys

The code create a txt file from a textarea feild after it's submited each on a line. After that the php code need collect these data from the txt file and create an array from them.

Example: if i type in the textarea:
test
test1
test2

I want these usernames to be inserted in the array so it would be like this:
$names = array('test','test1','test2');

Anyone can help me 

This is the code


<html>
  <head>
   <title></title>
   <style type="text/css">
    body{
    background-color:#1166aa;
    }
   #mainbox{
   width:inherit;
   background-color:#FFFFFF;
   padding:14px;
   border:solid 3px #000000;
   float:left;
   }
   #mainbox2{
   background-color:#CCCCCC;
   border:solid 2px #000000;
   padding:11px;
   }
   #contaner{
   border:solid 0px #00FF00;
   }
   h1:hover{
   color:#0000CC;
   cursor:pointer;
   }
   #msg{
   cursor:pointer;
   }
   table td{
   border-top:outset 1px #999999;
   }
   #amount{
   color:#FF0000;
   cursor:crosshair;
   }
   #to{
   color:#3300FF;
   }
   #from{
   color:#3300FF;
   }
   #subject{
   color:#3300FF;
   }
   </style>
  </head>

<body alink="#003399" vlink="#003399">


<?php
/////////////////////////////////////////////////
function create_list(){

     $list= ($_GET ['name_list']);
     $file= "list.txt"; 
     $file_pointer = fopen($file ,'w');
     fwrite($file_pointer , $list);
     fclose($file_pointer);
     }
                
//////////////////////////////////////////////////
     $list= "";
 create_list();
//////////////////////////////////////////////////
function explodeRows($data) {
  $rowsArr = explode("\n", $data);
  return $rowsArr;
}

$filename = "list.txt";
$handle   = fopen($filename, 'r');
$data     = fread($handle, filesize($filename));
$rowsArr  = explodeRows($data);

for($i=0;$i<count($rowsArr);$i++) {
  $name = explodeRows($rowsArr[$i]);



function name(){



$names = array('the usernames submited to be inserted here','the usernames submited to be inserted here',.......);



}

 }
////////////////////////////////////////////////////
 
 if ($_GET ['submit'] == "submit"){
     create_list();
 name();
 }
 

?>
<script type="text/javascript">
function foo(textarea,limit){
var val=textarea.value.replace(/\r/g,'').split('\n');
if(val.length>limit){
alert('You can not enter\nmore than '+limit+' lines');
textarea.value=val.slice(0,-1).join('\n')
}
}
</script>

<table align="center" id="contaner">
<tr><td>
<form action="<?php $_SERVER [SERVER_NAME]; ?>" method="get">
<div id="mainbox">
  <div id="mainbox2">
  <table>
  <tr>
     <td>name List :</td>
    </tr>
  </table>
<textarea id="name_list" name="name_list" cols="60" rows="9" onkeyup="foo(this,50)"></textarea>
  </div>
<tr>
<td><center><input name="submit" type="submit" value="submit"/></center></td>
</tr>
</div>
</form>
</tr></td>
</table>
</body>
</html>

This code works: 

// create object
$zip = new ZipArchive();

// open archive 
if ($zip->open('./download/zipfile.zip', ZIPARCHIVE::CREATE) !== TRUE) {
    die ("Could not open archive");
}


but, I want to create the name of the zip file dynamically.  The following code does not work ... doesn't give me an error, just doesn't create the zip file:

// create object
$zip = new ZipArchive();
$thisTime = time();

// open archive 
$thisZip = './download/'.$thisTime.'_download.zip';
$thisZip_b = $thisTime.'_download.zip';
if ($zip->open($thisZip, ZIPARCHIVE::CREATE) !== TRUE) {
    die ("Could not open archive");
} 



thanks in advance oh wise ones.

mrsburnside

I have the code below , and when I run it on my website , it returns a XML Parsing Error: not well-formed error , which says it has a problem reading the & i think   please see http://www.jamesflow...om/php/xml2.php   I think it has something to wiht the str_replace as when I change the $row['customerName'] . "</Name>\n"; to $row['customerNumber'] . "</Name>\n"; as a test it works fine , I understand XML cant handle special characters , how would prevent this happening?   code as below       $query = "SELECT * FROM customers";
$resultID = mysql_query($query, $linkID) or die("Data not found.");

$xml_output = "<?xml version=\"1.0\"?>\n";
$xml_output .= "<entries>\n";

for($x = 0 ; $x < mysql_num_rows($resultID) ; $x++){
    $row = mysql_fetch_assoc($resultID);
    $xml_output .= "\t<entry>\n";
    $xml_output .= "\t\t<Name>" . $row['customerName'] . "</Name>\n";
        // Escaping illegal characters
        $row['text'] = str_replace("&", "&", $row['text']);
        $row['text'] = str_replace("<", "&lt", $row['text']);
        $row['text'] = str_replace(">", "&gt;", $row['text']);
        $row['text'] = str_replace("\"", "&quot;", $row['text']);
    $xml_output.= "\t\t<Number>" . $row['customerNumber'] . "</Number>\n";
    $xml_output.= "\t</entry>\n";
}

$xml_output .= "</entries>";

echo $xml_output;     regards   James

Hi, I've been scratching my head for a while now about how to do this, I'm relatively new to php and mysql and perhaps foolishly taking on creating a user area for a website.  I have everything else working, all of my register account functions and confirmations and all of the login scripts etc. 

I have created a profile page which returns various information to the user (this bit works fine) and I've got some nice show/hide toggles running with some javascript/css but my intention is to allow the user to change thier information (e-mail address, contact phone number and also whether they are subscribed to the e-mail list), it also displays any support tickets or messages.

So after the long intro, here's what I'm struggling with...  I have a form in a visibility toggled <div> which submits a 'change_email' script, so a user wants to change their e-mail, clicks on change, the <div> appears, they bang in the new e-mail and hit submit.

My php script appears to work (because it doesn't throw up any errors), until you realise that actually it's not updated the record in the db...

I'm using mysql_query("UPDATE users SET email='$new_email' WHERE username='$user'");

Do I need to setup variables for all of the information in the db (name, username, password, email, contno etc etc) and include them in the command to get it to work or should that just pick the correct record and then update it? If that is the case is there a way I can include 'blank' variables so I don't have to set them all up...

e.g.  mysql_query("UPDATE users SET user='',password='',email='$new_email', etc WHERE username='$user'");

Many thanks in anticipation

Hey guys

i've been trying to find out how to create and and write in a file, so that you can for example add a nav button to your website by form (you know like a admin panel). It tried some things but it doesnt work cause it wont write to the file... here is the code i've used:

$ourFileName = $_POST['sitename'];
$ourFileHandle = fopen($ourFileName, 'w') or die("can't open file");
fclose($ourFileHandle);

$myFile = $_POST['sitename'];
$fh = fopen($myFile, 'w') or die("can't open file");
$stringData = "test\n";
fwrite($fh, $stringData);
$stringData = "test\n";
fwrite($fh, $stringData);
fclose($fh);

the "sitename" is the id from the form where you choose what the menu tab should be called.
I would really glad if you could help me out with this one thanks!

MinG