PHP - (help) Php To Get Random Line From Text File For Video Embed

Hi,

Have a text file that is being read. I want to print the contents however I want each line of the text file to appear on a new line in the browser.

So far
$str = file_get_contents('records.txt');
$lines = explode("\n", $str);
echo $lines[0];

Help appreciated

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

The title says it all,..well as all u can see, there are many methods to accomplish this done but not before undergoing through a heavy tasking process. For instance,
1. Open the file
2. Reads the content of the file into array.
3. Find out the desire line of text that we want to replace.
4. Make that line of text as variable and assigns with the new data(that we want to replace) as its values.
5.open another file.
6.Write the whole contents to the file opened in 5.
7. Put the contents back to the file of original.

Can anyone contributes a better way to get this done? Like replace/delete a specific line without affecting other characters or anything therein?
Thanx.

P.S
reading and using str_replace to make the alteration always ends up in giving some kind of result beyond desired after the first round of code execution.

Here's what I'm trying to do, and I am having trouble getting started with this.  It's a very simple process, but I didn't want to spend the next 6 hours in frustration, so some help getting started would be great.

Here's the purpose of the script:
1. Allow user to add a text file to a form.
2. Take the text file, add HTML code to the beginning and end of each paragraph (a single line of text, usually paragraphs would be separated by a line return)
3. Send the user an email with the HTML file attached and thank them or whatever.
4. Allow the system to throttle itself (one-at-a-time) so that many people using the site won't bog it down.  These files will probably be anywhere from 100 KB to 1,000 KB in size, usually hitting in the 300-500KB range.

Here's what I can do very easily:
1. Allow user to add a text file - very simple and straightforward.
2. Take the text file, add HTML... - this is what I need a little help figuring out.  Each paragraph needs to have <p> at the beginning and </p> at the end, and the script will also search for keywords on certain lines (section headers) and add a <align="center"> tag to that, and so forth.  I can handle the formatting rules, but making sure the loop runs correctly could be a problem.
3. Send the user an email... - very easy, I can do that myself.
4. Allow the system to throttle itself... - this could be tricky.  I was thinking a database with a TINYINT field, 0 for not processed yet, 1 for processing, 2 for processed.  Cron job checks the next one on the list to see if it needs to send it to the processor, if the file is already being processed, or can be sent to a different database (completed entries) and removed from the current queue.  The cron job would also be responsible for triggering the "Your file is converted!" email and the attachment.

Any/all help would be greatly appreciated on this.  I am going to work on the parts that I can do myself, and I'll be checking back for the discussion - in between Mountain Dew runs.

I have a PHP file that is executed via batch file very frequently for live updating. This is a sort of "sync" file for reading/writing to a MySQL database. I am able to get it functioning absolutely fine when executed manually via a web browser (and my echo debug lines output the expected information), yet when I run the same PHP file via the command line, it seems to just not be capable of opening any file for reading, so the equivalent variables that are correct when executed on a web browser are blank when echoed through the command line.

Is there a different way of handling reading of files when running a php script via the command line, or should it function exactly the same as when run via a browser?

For instance:

$k = "0";
$line = file("examplefile.log")[$k];
echo $line;

Im trying to figure out the best way to random a line from a text file and after the random delete that line from the file. This way the same username cannot be picked twice thanks for any input

I have an excel file that I want to show in PHP page. I dont want this to first download and then view. I want it something like gmail where you just click on the file name and it opens up. Can you please give me an idea as how to do this.

Regards,

Faisa

I have a script that reads a .gz file into an array and prints the name of each record but will not work on larger files. Is there a way to read 1 line at a time?
Here is the code I have so far.
Code: [Select]
<?php

if ($handle = opendir('.')) {

print "<ol>";
    
    while (false !== ($file = readdir($handle))) {
if($file != '..' && $file!="." && $file!="start_update.php" && $file!="sharons_dbinfo.inc.php" && $file!="root.php" && $file!="read_directory.php" && $file!="read_dir.php" && $file!="new_category.php" && $file!="index.php" && $file!="file_count.php" && $file!="dir_loop2.php" && $file!="dir_loop1.php" && $file!=".htaccess" && $file!="Answer.txt" && $file!="Crucial_Technology-Crucial_US_Product_Catalog_Data_Feed.txt"){
$filename = $file;
$go = filesize($filename);
if($go >= 1){
$filename2 = explode("-", $filename);
$filename2 = $filename2[0];
echo str_replace("_"," ",$filename2) . ' | Filesize is: ' . filesize($filename) . ' bytes<br>';
$gz = gzopen($filename, 'r');
$lines = gzfile($filename,10000);
foreach ($lines as $line) {
$line2 = explode(",", $line);
$line2 = str_replace("," , "-" , $line2);
    echo "<li>".str_replace("," , "-" , $line2[4])."</li><br>";
}

}
}
    }

    closedir($handle);
}
?>
</ol>

How would i do a random file name for my upload script

Code: [Select]
$type = $_FILES['uploadedfile']['type'];
// Where the file is going to be placed
$target_path ="/***/***/public_html/****/lofslidernews/images/";

/* Add the original filename to our target path. 
Result is "/images/uploads/filename.extension" */
$target_path = $target_path . basename( $_FILES['uploadedfile']['name']);

$uploadedfile2 = $_FILES['uploadedfile']['name'];

if (empty($quicktitle) || empty($maintitle ) || empty($description ) || empty($uploadedfile))
   {
       $_SESSION["promoerror"] = "Please Select an image and enter a Quick Title, Main Title and Description!" ;
       header("Location: promo.php"); //This sets the redirection information
       //echo "Going to login.php, username or password is empty";
       exit(); //Ends the script and redirects to above
   }


$query = "INSERT INTO promotion (title, maintitle, description, image) VALUES ('$quicktitle','$maintitle','$description', '$uploadedfile2')";



$result = mysql_query($query) or die ("Error in query: $query. ".mysql_error());


if ( ( $type != "image/jpeg") && ($type != "image/gif") )

{
die ("That format is not allowed");
}
else
{
    move_uploaded_file($_FILES['uploadedfile']['tmp_name'], $target_path);
header("Location: promo.php");
$_SESSION["promoerror"] = "Added Successfully!";

}

Hi Guys,

I have a php script to upload a photo to a mysql table. This works great, until someone uploads a file with the same name, then it replaces the old one. Bad times!

So Ive researched it and I have put a timestamp on the filename and it uploads with the correct timestamp in the images folder.

However the data that it uploads to the table is just the original filename, ie it doesnt stamp the filename in the table therefore they dont match...

Any ideas...

<?php 

include('config.php'); 
if (isset($_GET['Ter']) ) { 
$ter = (int) $_GET['Ter']; 
if (isset($_POST['submitted'])) {

//Photo Upload

//This is the directory where images will be saved
$name=time();
$target = "images/"; 
$target = $target .$name. basename( $_FILES['photo']['name']);
//This gets all the other information from the form 
$photo = ($_FILES['photo']['name']);
//Pause Photo Upload


foreach($_POST AS $key => $value) { $_POST[$key] = mysql_real_escape_string($value); } 
$sql= "UPDATE `ter` SET `Ter` =  '{$_POST['Ter']}' , `BranchName` = '{$_POST['BranchName']}' ,  `BranchAddress` =  '{$_POST['BranchAddress']}' ,  `BranchTel` =  '{$_POST['BranchTel']}' , `BranchEmail` =  '{$_POST['BranchEmail']}' , `BranchLink` =  '{$_POST['BranchLink']}' , `Theme` =  '{$_POST['Theme']}' , `LocalInfo` =  '{$_POST['LocalInfo']}' , `BranchInfo` =  '{$_POST['BranchInfo']}' , `photo` =  '{$_FILES['photo']['name']}' WHERE `Ter` = '$ter' "; 

mysql_query($sql) or die(mysql_error()); 


//Unpause Photo Upload
//Writes the photo to the server 
if(move_uploaded_file($_FILES['photo']['tmp_name'], $target))
{
//Tells you if its all ok
echo "<br />The file ". basename( $_FILES['photo']['name']). " has been uploaded. <br />";
}
else {
//Gives and error if its not
echo "";
} //End of Photo Upload

echo (mysql_affected_rows()) ? "<br />Edited Branch.<br />" : "<br />Nothing changed. <br />";  
} 
$row = mysql_fetch_array ( mysql_query("SELECT * FROM `ter` WHERE `Ter` = '$ter' "));
?>

Hi. I am trying to create a simple video file server. I finished the skeleton of it but when it came to creating the multitude of pages I realized that there could be an easier way to do it.   I started out basically getting the folder names in the server then printing them on the page.

<?php
$dirs = glob("*", GLOB_ONLYDIR);
echo '<ul>';
foreach($dirs as $dir)
{
	$forbidden_folders = array("not4u", "ignore", Styles);
	$filename = str_replace($forbidden_folders, '', $dir);
	if (!empty($filename))
	{
    echo '<li><a href="'.$dir.'">'.$dir.'</a></li>';
	}
}
echo '</ul>'
?>

Then I created in each of the folders with files a php file with this code:

<?php

function date_sort_desc($a, $b)
{
  preg_match('/\w+ \d{4}/', $a, $matches_a);
  preg_match('/\w+ \d{4}/', $b, $matches_b);
  $timestamp_a = strtotime($matches_a[0]);
  $timestamp_b = strtotime($matches_b[0]);
  if ($timestamp_a == $timestamp_b) return 0;
  return $timestamp_a < $timestamp_b;
}

$files = array();
$dir = opendir('.');
while(false != ($file = readdir($dir))) {
        if(($file != ".") and ($file != "..") and ($file != "index.php")) {
                $files[] = $file;
        }   
}

natsort($files);

$i = 0;
foreach($files as $file) {
$i++;
$string = str_replace("TV Show Name S1 E$i - ", '' , $file);
        echo '<div><a href="../Discriptions/'.$file.'.php">'.basename($string, '.m4v').'</a></div><br>';
}
?>

This opens up a php file with the same file name as the file to be played containing the episode's thumbnail and description. That file then contains a link pointing back to the real file. The problem here is that I'd have to make a new php file for every file in my collection. I'm wondering if there's somehow a way to simplify all of this.
Edited by chrisyroid, 11 August 2014 - 12:13 AM.

Have page#1 with thumbs view_thumbs.php , when a thumb is clicked thumb link goes to page#2 which is the video_play.php page(which must play embed codes from xml sheet)
Using GET method to identify incoming link from page#1.
QUESTION is how can i retrieve file(embed code)  from xml to play on page#2 with incoming GET info from page#1?
(embeds not you only from you tube, so no youtube solutions only, just retrieve file)

I somehow have to fill play video div on page#2/video_play.php  according to GET data from page#1/view_thumbs.php

Dear all,
Can somebody help me please?
Instead of ‘username’ and 'password', I need random values from CSV file.
Can somebody show me how can I do this?

My CSV file looks like this:

user001,userpass001
user002,userpass002
user003,userpass003

My script looks like this:

<?php
 $t= new post();
 $t->username='username';
 $t->password='password';
 $res = $t->update('This is some text.');
?>

Thank you very much advance.

Anyone know of a script that will allow you to upload a file and converts it to a specified file format?

Folder layout:
script/
script/videos/mp4/                                       }
script/videos/ogg/                                        }    folders for converted files from uploaded files
script/videos/webm/                                     }
script/videos/swf/                                        }

Video player:
Code: [Select]
<video width="320" height="240" controls="controls">
  <source src="script/videos/mp4/<? echo $movie; ?>.mp4" type="video/mp4" />
  <source src="script/videos/ogg/ <? echo $movie; ?> .ogg" type="video/ogg" />
  <source src="script/videos/webm/ <? echo $movie; ?> .webm" type="video/webm" />
<object data="script/videos/mp4/ <? echo $movie; ?> .mp4" width="320" height="240">
<embed src="script/videos/swf/ <? echo $movie; ?> .swf" width="320" height="240">
Your browser does not support video
</embed>
</object>
</video>

I have searched and searched the web for an answer and cannot find one pleas help?

This is the php file code to a video jukebox add-on script, for a popular web video script.
I'd like to modify it so the "Click Here" text font and color can be changed. Can you help or suggest what I can do to accomplish this?

Thanks

Code: [Select]
<?php include("../classes/config.php");?>
<artworkinfo>
<?
$query = "select * from videos where viewtime<>'0000-00-00 00:00:00' and public_private='public' AND approved ='yes' AND promoted ='yes'  limit 10"; // Max 10!";
$db=mysql_connect ($dbhost,$dbusername,$dbpassword) or die ('I cannot connect to the database because: ' . mysql_error());
mysql_select_db ($dbname);

$num_rows = mysql_num_rows(mysql_query("select * from videos WHERE public_private = 'public' AND promoted = 'yes' AND approved ='yes'"));

$result = mysql_query($query) or die ('Query Error: ' . mysql_error());
while ($results = mysql_fetch_array($result)) {
?>

<albuminfo>
<artLocation>../uploading/thumbs/<?=$results['video_id']; ?>.jpg</artLocation>
                <artist>Click Here!!</artist>
<albumName><?=$results['title']; ?></albumName>
<artistLink>../play.php?vid=<?=$results['indexer']; ?>&src=jukebox</artistLink>
<albumLink>../play.php?vid=<?=$results['indexer']; ?>&src=jukebox</albumLink>
</albuminfo>
<?
}
mysql_close();

?>

</artworkinfo>