PHP - How Do I Go About Displaying This File?

I'm trying to determine why when I select Upload, on the html page I see a message "My extension is:" only, instead of My extension is: webM" for example.Here's the code, I look forward to any assistance:

<?php
foreach (array('video', 'audio') as $type) {
    if (isset($_FILES["${type}-blob"])) {

        $fileName = $_POST["${type}-filename"];
        $uploadDirectory = 'uploads/' . $fileName;

            // make sure that one can upload only allowed audio/video files
            $allowed = array(
                'webm',
                'wav',
                'mp4',
                'mov'
    );

    $extension = pathinfo($filePath, PATHINFO_EXTENSION);
    die("My extension is: " . $extension);

   if (!$extension || empty($extension) || !in_array($extension, $allowed)) {
        echo 'Invalid file extension: '.$extension;
       return;
 }
        if (!move_uploaded_file($_FILES["${type}-blob"]["tmp_name"], $uploadDirectory)) {
            echo (" problem moving uploaded file");
        }
    }
}
?>

 

Hello

So in my index.php code i include('somefile.html');. In the html file itself there is an <img src="/images/picture.jpg">. When looking at index.php on a browser, the html looks fine everything loaded fine except for the <img> There is a little broken image link displayed on the page. i checked through the browser if i mistyped the img src link but somefile.html in the browser displays the image fine. But if i try to display the img in the html file through the php file, its a broken link. Any reason why this is happening? Am I missing something here? Thanks in advance.

Ok, my php skills are such that I can customize an existing script to some extend, but I can't write one.

My problem is the following:
I'm using a php-based forum script in a friends website. I load the 'forum.php' script  in an iframe of a static page.

When a user gets a response to a post that he wrote on the forum, he receives a mail with a link which - if clicked - gets him to the response he got ie. http://someurl.com/forum.php?todo=viewtopic&tid=6#t2

The problem is now that if he clicks the link, the file of course doesn't open in the iframe of the static page (http://someurl.com/forum.html) where normally this forum.php file opens.

I guess I will have to make the static page also into a php-file and then use parts of the clicked url (forum.php?todo=viewtopic&tid=6#t2) and insert it into the iframe as the file to load ... would appreciate a lot if anyone can advice me how to do that!

Thanks in advance,
K.

I am using the following php script to display all the files that are uploaded in a directory on a page. There are two files in the folder I do not want displayed: .htaccess & index.php.

Can someone help me out and provide a little guidance about what I need to do to print all the file names EXCEPT for those two. I imagine some sort of if statement.

CODE:

Top of page: Code: [Select]
<?php
$dirname = ".";
$dir = opendir($dirname);
?>
Body:
Code: [Select]
<?php
while(false != ($file = readdir($dir)))
{
if(($file != ".") and ($file != ".."))
{
echo("<a href='$dirname$file'>$file</a> <br />");
}
}
?>
Thanks so much!

So, I'm trying to get this read my text document "news.txt" and display the contents.

My code:
Code: [Select]
div.newscontent{
padding:0px;
margin-top:60px;
margin-bottom:0px;
margin-left:400px;
margin-right:0px;
position:absolute;
width:457px;
height:330px;
text-align:center;
color:#FFFFFF;
}
Code: [Select]

<div class='newscontent'>
<?PHP

$filename = "news.txt";
$arr = file($filename);
foreach($arr as $line){
    print $line . "<br/>";
}

?>
</div>
This is the output I get:
Code: [Select]
"; } ?>
Whats wrong?

Hi!
I have 2 files based on the Model View Arhitecture.

The file login.php and the config file containing the db connection are in the model folder and the login_view is in view folder.

I am trying to make a login form am display some error messages to the user, but I don't get the error messages.

Can you help me figure it out?

PS: the connection to the db works fine!

Thx!

LOGIN.PHP

<?php
 //start the session
 session_start();
 foreach (glob("model/*.php") as $filename) {
    include $filename;
}

 //form defaults
 $error["alert"]= "";
 $error["user"]= "";
 $error["pass"]= "";
 $input["user"]= "";
 $input["pass"]= "";

if(isset($_POST["submit"]))
{
    //process form
    if($_POST["username"] == "" || $_POST["password"] == ""){
        //display errors
        if($_POST["username"] == ""){$error["user"] = "required";}
        if($_POST["password"] == ""){$error["pass"] = "required";}
        $error["alert"] = "Please fill in the required fields";

        //get the values for the username and password
        $input["user"]= $_POST["username"];
        $input["pass"]= $_POST["password"];
   
        include("../views/login_view.php");
    }
}else{
    include("../views/login_view.php");
}

LOGIN_VIEW.PHP

<?php
    session_start();
    foreach (glob("model/*.php") as $filename) {
        include $filename;
    }
?>
<link href="//maxcdn.bootstrapcdn.com/bootstrap/4.0.0/css/bootstrap.min.css" rel="stylesheet" id="bootstrap-css">
<script src="//maxcdn.bootstrapcdn.com/bootstrap/4.0.0/js/bootstrap.min.js"></script>
<script src="//cdnjs.cloudflare.com/ajax/libs/jquery/3.2.1/jquery.min.js"></script>
<!------ Include the above in your HEAD tag ---------->

<!--author:starttemplate-->
<!--reference site : starttemplate.com-->
<!DOCTYPE html>
<html lang="en">
<head>
    <meta charset="utf-8">
    <meta name="viewport" content="width=device-width, initial-scale=1, shrink-to-fit=no">
    <meta name="keywords"
          content="unique login form,leamug login form,boostrap login form,responsive login form,free css html login form,download login form">
    <meta name="author" content="leamug">
    <title>Unique Login Form | Bootstrap Templates</title>
    <link href="/styles/style.css" rel="stylesheet" id="style" type="text/css">
    <!-- Bootstrap core Library -->
    <link href="//netdna.bootstrapcdn.com/bootstrap/3.2.0/css/bootstrap.min.css" rel="stylesheet" id="bootstrap-css">
    <script src="//netdna.bootstrapcdn.com/bootstrap/3.2.0/js/bootstrap.min.js"></script>
    <script src="//code.jquery.com/jquery-1.11.1.min.js"></script>
    <!-- Google font -->
    <link href="https://fonts.googleapis.com/css?family=Dancing+Script" rel="stylesheet">
    <!-- Font Awesome-->
    <link href="//maxcdn.bootstrapcdn.com/font-awesome/4.2.0/css/font-awesome.min.css" rel="stylesheet">
</head>
<body>

<!-- Page Content -->
<div class="container">
    <div class="row">
        <div class="col-md-offset-5 col-md-4 text-center">
            <h1 class='text-white'>Login Form</h1>
            <form action ="" method="POST">
              <div class="form-login"></br>
                </br>
                <?php
                    if($error["alert"] != ''){
                        echo "<div class='alert'>".$error["alert"]."</div>";
                    }
                ?>
                <div>
                    <input type="text" name ="username" id="userName" class="form-control input-sm chat-input" placeholder="Username" value= "<?php echo $input['user'] ?>" required/>
                    <?php echo $error['user'] ?>
                </div>
                </br></br>
                <input type="text" name ="password" id="userPassword" class="form-control input-sm chat-input" placeholder="Password" value= "<?php echo $input['password'] ?>" required/>
                <div class="error"> <?php echo  $error['pass'] ?>
                </br></br>
                <div class="wrapper">
                <input type="submit" class="btn btn-primary" value="submit">
                 
                </div>
            </div>
        </div>
    </form>
    </div>
</div>
</body>
</html>

 

Edited October 28, 2020 by Alexa

Hi,

I have a separate form created using Dreamweaver that calls a separate php script when the Submit button is clicked. Currently I am able display form validation messages in a new html page. If the user leaves fields blank, I would like the messages to appear on the form itself instead of in a new page.

How do you make the messages from the php form validation code display into the calling html form? I know I could just do this with Dreamweaver but I would like to learn to do this using php.

It might be easier to embed the code within the html page but I was thinking that using the separate script would be more secure.

My form can be found here. Validation is working but opens a new page:
http://www.tallfirshoa.com/adform.htm

Thanks!
Rob

Hi,

i have export the search result to excel file, but arabic characters not displaying properly in excel file, eventhough i mention the header content type as utf-8

here my code...

Code: [Select]
<?php
include("global.php");
// Original PHP code by Chirp Internet: www.chirp.com.au 
// Please acknowledge use of this code by including this header. 

function cleanData(&$str) 
{
 $str = preg_replace("/\t/", "\\t", $str); $str = preg_replace("/\r?\n/", "\\n", $str); 
 if(strstr($str, '"')) $str = '"' . str_replace('"', '""', $str) . '"'; 
} 

$sql = $_SESSION['emp_search_sql'];
// file name for download 
$filename = "employees_" . date('Ymd') . ".xls";
header("Content-Disposition: attachment; filename=\"$filename\"");
header("Content-Type: application/vnd.ms-excel; charset=UTF-8"); 
header("Pragma: no-cache";


$flag = false;
 $result = $DB_site->query($sql) or die('Query failed!');
while(false !== ($row = mysql_fetch_assoc($result)))
{
if(!$flag) 
{
// display field/column names as first row 

echo implode("\t", array_keys($row)) . "\n"; $flag = true; 
} 
array_walk($row, 'cleanData'); 
echo implode("\t", array_values($row)) . "\n";
}

?>


Thanks,

Hey everyone. Currently I have a site set up where I manually edit the txt file and the site pulls it from the text file and displays via php. I wanted to automate it by creating one file which contains the sunset times for the whole year, and have php determine todays date and then display the corresponding sunset time.

Can someone please point me in the right direction?

maybe it can be set up in a way where each line in the txt file represents the day of the year. ex: Jan 1 would be line 0, Feb 1's sunset time would be on line 31 etc.

Thanks

I currently am working on a project where I code a "simple" telephone directory.

There are three main tasks that it needs to do:
1. Directory.php(index page) has a "First Name" and "Last Name" field and a search button. When a name is searched from the directory.txt file, it displays First Name, Last Name, Address, City, State, Zip and phone in findinfo.php in designated text boxes...first name, last name, etc.

2. From the findinfo.php, like previously stated, the users information is listed in the appropriate text boxes. From there, there is an update button that will overwrite the user's information to directory.txt if that button is selected. It will then say the write was sucessful.

3. (completed this step) From the index page, there is a link that will take you to addnew.php where you enter First Name, Last Name, Address, City, State, Zip and phone in a web form and write it to directory.txt.

 This is the php code for the third step:

<?php
$newentryfile = fopen("directory.txt", "a+");
$firstname = $_POST['fname'];
$lastname = $_POST['lname'];
$address = $_POST['address'];
$city = $_POST['city'];
$state = $_POST['state'];
$zip = $_POST['zip'];
$phone = $_POST['phone'];
$newentry = "$firstname $lastname\n\r $address\n\r $city, $state $zip\n\r $phone\n\r";
if (flock($newentryfile, LOCK_EX)) {
if (fwrite($newentryfile, $newentry) > 0)
echo "<p>" . stripslashes($firstname) . " " .
 stripslashes($lastname) . " has been added to the directory.</p>";
else
echo "<p>Registration error!</p>";
flock($newentryfile, LOCK_UN);
}
else
echo "<p>Cannot write to the file. Please try again later</p>";
fclose($newentryfile);

if(empty($firstname) || empty($lastname) || empty($address) || empty($city) || empty ($state) || empty($zip) || empty($phone))
{
echo "<p>Please go back and fill out all fields.</p>";
}
?>


So to sum it all up, what would be my best approach? I am totally stumped and not sure which function to use. Should I work my way from step 1 to step 2? I see it as when I do the search for the name from directory.php, it takes me to findinfo.php, listing the users information in the text boxes. From there, if I needed to, having the user's information already listed I could hit the update button to overwrite the new information to directory.txt. Doing the update when then tell me that the write was successful.

I have literally been scouring the internet for hours. What would be the best function to do this? I hope I was clear enough. Please help me out and thank you for your time.

how to get the name of the file including a file from the included file,

This one has me mixed up a bit.. I am trying to record site activity information from a common.php file using a user object. But since the file is included into different php files based on different situations I need a dynamic way of finding the file name that is including it. I could be over complicating things but right now this seems like the best solution other wise I'll have to rewrite the code on every page i write. Is there a function for doing this? Or if someone gets what I'm trying to do if they could point me to the direction of some more information on it.

Thanks.

Hey everyone, I'm pretty new to this, not my full time job, but just something I thought I'd give a shot...

I have a database, in postgres, in which I make my query and I go fetch all the info I need.
What I need to do next is the tricky part for me.

For each line of results received, I have to output to a text file (.txt) but I have to respect a format that was given to me from the person requesting the info.

Example:
Character 1 must be G or N.
Character 2 to to 11 will be a result from my database search, which is a 10 digit string.
Character 12 to 14 must be spaces.

Another rule is:
start at character 78: input a value from my database search but it must not exceed 20 characters and if it is less then 20 character, fill the remaining with spaces.


If anyways has a code I can copy and work off of I would really appreciate it....thanks!

I'm using an Upload Script, and I've been working on trying to style the 'Choose File Button' (input file button)

In this multi file upload form, choose three images, click submit and preview the images on the preview page. If the user wishes to delete or replace an image, click edit and the form will go back to the previous page. Select the replace radio button for example on one of the three images and select a new image from the file input prompt and click submit. The form will go to the preview page again to display the images. During this process the image names are being input into a table and the images are being moved to a directory.

The table is

  `id`  AUTO_INCREMENT,
  `image0`
  `image1`
  `image2`
  `status`

So input name='image[image0]' can be directed to table `image0` and so on.

The code for keep and delete work fine, but how do I replace an image?
I have two foreach blocks. The first one deletes the image file from the directory and deletes the image name from the table, but the second foreach dose not move the new image file into the directory.
Thanks.

<input type='radio' name='image[image0]' value='keep' checked='checked'/>
<input type='radio' name='image[image0]' value='delete' />
<input type='radio' name='image[image0]' value='replace' />
<input type="file" name="image[]" />

<input type='radio' name='image[image1]' value='keep' checked='checked'/>
<input type='radio' name='image[image1]' value='delete' />
<input type='radio' name='image[image1]' value='replace' />
<input type="file" name="image[]" />

<input type='radio' name='image[image2]' value='keep' checked='checked'/>
<input type='radio' name='image[image2]' value='delete' />
<input type='radio' name='image[image2]' value='replace' />
<input type="file" name="image[]" />

<?php
if (isset($_POST['status'])) {

$status = $_POST['status'];
$confirm_code = $status;

#--------------------------- replace --------------------------------------------

if (isset($_POST['submitted']) &&
($image = $_POST['image'])) {

foreach($image as $imageKey => $imageValue) {
if ($imageValue == 'replace') {
$query = "SELECT $imageKey FROM table WHERE status = '$status' ";
if($result = $db->query( $query )){
$row = $result->fetch_array();
}

unlink( UPLOAD_DIR.$row[0] );

$query = "UPDATE table SET $imageKey = '' WHERE status = '$status' ";
}
}

foreach($image as $imageKey => $imageValue) {
if ($imageValue == 'replace') {

$filenm = $_FILES['image']['name']; 
$file = $_FILES['image']['tmp_name'];

move_uploaded_file($file, UPLOAD_DIR . $filenm);
$filename[] = $filenm; 

$query = "INSERT INTO table VALUES ('','$filename[0]','$filename[1]','$filename[2]','$confirm_code')";
}
}
}
}
?>

Hiya,

Firstly, I'm a complete novice, apologies! But I have got my upload.php working which is nice. I will post the code below.

However, I would now like to restrict the file size and file type to only word documents.

I currently have a restriction of 200KB but it's not working - no idea why as I've looked at other similar codes and they look the same.

Also, just to complicate things - can I stop files overwriting each other when uploaded? At the moment, if 2 people upload files with the same name one will overwrite the other.

Is this too many questions in 1?

Any help is very much appreciated!

Code below:

Code: [Select]
<form enctype="multipart/form-data" action="careers.php" method="POST">
        Please choose a file: <input name="uploaded" type="file" /><br />
<input type="submit" value="Upload" />
</form>

<?php 
 $target = "upload/"; 
 $target = $target . basename( $_FILES['uploaded']['name']) ; 
 $ok=1; 
 
 //This is our size condition 
 if ($uploaded_size > 200) 
 { 
 echo "Your file is too large.<br>"; 
 $ok=0; 
 } 
 
 //This is our limit file type condition 
 if ($uploaded_type =="text/php") 
 { 
 echo "No PHP files<br>"; 
 $ok=0; 
 } 
 
 //Here we check that $ok was not set to 0 by an error 
 if ($ok==0) 
 { 
 Echo "Sorry your file was not uploaded"; 
 } 
 
 //If everything is ok we try to upload it 
 else 
 { 
 if(move_uploaded_file($_FILES['uploaded']['tmp_name'], $target)) 
 { 
 echo "Your file ". basename( $_FILES['uploadedfile']['name']). " has been uploaded."; 
 } 
 else 
 { 
 echo "Sorry, there was a problem uploading your file."; 
 } 
 } 
 ?>