PHP - Moved: Php Single File Help

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?

 

I have created a css menu for the top of my site that is included on each page with a php include.  What I am wanting to do now is create the side menu options for each page from the one file.

Each page will have different links side links based on what section of the site you are in however I want to administer them all from one php file if that is possible.  I'm quite new to php and am not sure if I can even do this.  Any ideas on how this can be done?

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 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

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

This topic has been moved to PHP Applications.

http://www.phpfreaks.com/forums/index.php?topic=318539.0

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

This topic has been moved to JavaScript Help.

http://www.phpfreaks.com/forums/index.php?topic=359179.0

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

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