PHP - Trouble Calling Multiple Pages Into My Php File

I had Tom Cat Java installed on my Linux sever and I need to call a java file, which isn't working. Someone told me to verify Java is in fact installed on my server using this command:

Code: [Select]
echo exec('which java');

but when I run it, nothing displays!

The java command I'm trying to run is this:

Code: [Select]
exec('java -Xmx1024m -jar "/home/account/public_html/java -cp b4a_c2dm.jar anywheresoftware.b4a.c2dm.C2DM send  $php_device_id \"test\""');

The .jar file is there.

My path to java is: /usr/local/jdk/bin/java

Thanks for any help...

Hi,

I wrote this to retrieve a file based on a POST value.
For some reason it will not retrieve the file content and display it at all.

I get this message:
Warning: fopen(18.txt) [function.fopen]: failed to open stream: No such file or directory in /home/content/x/y/z/path/html/includes/different.php

The 18 is the correct POST value.  So that is a positive

There is a file named 18.txt in /home/content/x/y/z/path/html/includes/18.txt

Thanks.

<?php

$id5 = $_POST[id];
 //Display Text File
 $myFile = $id5;
 $myFile .= ".txt";
 $fh = fopen($myFile, 'r');
 $theData = fgets($fh);
 fclose($fh);
 echo $theData;
      
?>   

HY

I have index.php and pictures.php.
In index.php I have 3 columns:
- left (for menu)
- right (for advertising)
- center (where I want to include pictures.php)

What is the best way to include pictures.php in center of index.php

Code: [Select]
if($_get[pictures]){
include ("pictures.php");
}

I ask this because I have multiple variable like "pictures" and I will have multiple "If".

Or to include in DB all this variables "pictures" and just add ".php" extension. It is secure in this way ?

Thanks

Hey guys, how is everyone today?

I am needing a little help with this one. I have a database that updates itself and the display page for it with every form submission on the website. I am wondering how I go about getting it to display across multiple pages? For example, once it hits 25 submissions, it would make a page 2, once page two hits 25 submissions, it creates page 3, etc. etc. Much like you see on something like a forum, it hits so many replies before it creates a new page, to limit one long page of results.

Thanks in advance!! 

Hello.

I'm coding myself an small webpage, In the internet you can see that there is pages like index.php?id=223923 <- for example or index.php?=news.
 So, I'm trying to create similar to that myself. I tried googling and searching youtube how to do this but didn't really find anything.

I figured it out that it needs some database etc.

I tried myself doing some table in my mysql db. And in the table some 'id, title, content' and in the id would be the url, (index.php?='id') the title would be the <title> </title> and the content would be all the code inside the webpage. I got no idea how to link these to an php or whatever it should be done So would anyone kindly tell me howto do this or give some link to an tutorial?

I've read through a few examples of how to paginate query results, but with my currently php skills I don't understand them, or they do not fit into my code.  I wondered if I could make my own, and the logic behind this code makes sense to me, but it's still not working.  Here is the code:

<?php

if (!$Start) {
$Start = 0;
}
else {
     $Start=$_POST['Start'];
}

$crittercount = 120;

$numpages = $crittercount / 30;
$pagei = 1;
$pagereit = 1;

echo '<p><center>
<FORM method=POST action=testpagination.php>
<SELECT NAME="Start">
        <OPTION>Page 1';
        while ($pagereit < $numpages) {
        $pagei = $pagei + 30;
        echo '<OPTION value='.$pagei.'>';
        $pagereit = $pagereit + 1;
        echo "Page: " .$pagereit;
        }
        echo '</SELECT>';
echo'<input type=submit value=Go></FORM><p>';

echo $Start;

?>

The test site is he
http://lab-lib.com/felishorns/Felishorns/websitetest/testpagination.php

$crittercount is a variable that changes depending on the main $query.  This query is invisible, and is used to pull a count for all the results in total.  I am modifying it with $Start to create a second query which the viewer sees.

$queryF = $query . " LIMIT " . $Start . ", 30";

The idea is that $Start changes depending on the value selected, thereby changing the query.  In this test code, I've simply echoed the $Start value of that page, to see if the code is working, and it's not changing.  What's wrong with this code?  Is it not possible to have a form lead to the same link of the page it's on?  Does the variable not become updated?

Thanks in advance.

Hi guys,

I hope you may be able to shed some light on a problem I am having. I am fairly new to PHP although do understand bits and peices such as login system.

Basically - I have a website with lets say 10 users (more like 500 but that will do for now!) - Each user has their own page wich is password protected.

Each month - I want to be able to upload PDF files to the server - which CAN ONLY BE ACCESSED BY THE RELEVANT user. They must not be able to see each others PDF files.

To do this I have been advised to have a non-web accessible folder on my server to put the PDF's in and then use PHP to handle the operation. Therefore my path would be:

1/ User logs into page

2/ User clicks the PDF link

3/ PF link goes to PHP page that checks they are logged in and then the PDF they want before delivering to the server.

I do actually understand the theory but my PHP is not at the stage where I can just write the code that will handle the operation. I am also very confused over how to access the non-web accessible files!

Could anyone please give me an example of the code I will need to use to a) check the user is logged in (I guess I can use the same code I used for the login) and then b) call the relevant PDF and display it? I have been given a path of c:\blahblahblah to access my PDF files but don't even know how to begin implimenting this!

Here is the code I use for my login system. Could anyone show me how to adapt it to get what I need?

Thank so much for anyone that can help - hopefully I will be in a position to give back one day!

--------------------------------------------
if(isset($_SESSION['loggedin']))
{
    header("Location:" . strtolower($username) . ".php");

if(isset($_POST['submit']))
{
   $username = mysql_real_escape_string($_POST['username']);
   $password = mysql_real_escape_string($_POST['password']);
   $mysql = mysql_query("SELECT * FROM mydb WHERE username = '{$username}' AND password = '{$password}'");
   if(mysql_num_rows($mysql) < 1)
   {
     die("Password or Username incorrect! Please <a href='login.php'>click here</a> to try again");
   }   $_SESSION['loggedin'] = "YES";
   $_SESSION['username'] = $username;
   header("Location:" . strtolower($username) . ".php");

AND THEN IN THE HEADER OF THE PAGES

<?php
session_start();
if(!isset($_SESSION['loggedin'])) {
  header('Location: /login.php');
} elseif ($_SESSION['username'] . '.php' != basename($_SERVER['SCRIPT_FILENAME']) ) {
  // Logged in user attempting to view someone else's page
  header("Location:" . strtolower($_SESSION['username']) . ".php");
  exit;
}
?>

Again - any help would be truly appreciated. I will say now that the last person I asked said "Use the open() function - that'll work!" ...... answers like that are a bit lost on me at the moment and leave me even more confused!

Cheers in advance
 

 

Help!
HTML5 Submit button not calling PHD file

My HTML:

<html>
<head>
</head>
<body>
<form action="my.php" method="POST" name="process_form">
<Input type="submit" name="submit" value="Submit" formmethod="post" formaction="my.php">
</form>
</body>
</html>

PHP

if(isset ($_POST["submit"])){
   Do something; 
   Do something; 
} else {
   Do something; 
   Do something;
}

I click the button and nothing happens.
What am I missing?

I am creating a webpage for my golf league and some of the information(variables) such as Name, Address and email, must be reused across multiple pages for different outputs.  May question is "Can I, or is there a way to reUse variables across multiple pages.   "Do I need to create a session and make everything session variables"?  "Should I link to a database and call the information when needed"? 

Any explanation would be greatly appreciated. 

Hello Everyone was wondering if I could get some help with the following code? I am querying a database for results of listings that are in a database these listings are displayed on the page in a form. I am wanting each listing to be on a different page. Below is my code.
Code: [Select]


$lim=1;
if (!isset($s) || $s < 1 || !is_numeric($s)) {
            $s = 1;
        }


    $start = ($s - 1) * $lim;
     $sql = "select id,bussimg,imagewidth,imageheight,email,usridm,company,businesscategory,address1,address2,state,city,zip,website,email,repname,description,phonenumber,country,status from $approvecheckbusinesses where usridm='$user_id'";
    $result=db_query($sql);
     



        $countpages = $sql;
        $sql = $sql . " order by id asc limit $start, $lim";
       $result=db_query($sql);
        $pages = ceil(mysql_num_rows(mysql_query($countpages)) / $lim);
     
   


$result=db_query($sql);
for ($i = 0; $i < mysql_num_rows($result); $i++) {
$Listid= mysql_result($result, $i, "id");

$usridm= mysql_result($result, $i, "usridm");
$CompanyName= mysql_result($result, $i, "company");
$realname= mysql_result($result, $i, "repname");
$email= mysql_result($result, $i, "email");
$BusinessCategory= mysql_result($result, $i, "businesscategory");
$status= mysql_result($result, $i, "status");
   



echo ("FORM IS TO BE DISPLAYED HERE");

}

if ($pages > 1) {
            echo("<p align=left style='font-size: 85% color=white'>");
            for ($i = 1; $i <= $pages; $i++) {
                echo("[");
                if ($i == $s) {echo("<b>");}
                else {echo("<a id=home_offerLink href='index.html?EditMemberListing&user_id=$user_id&s=$i'>");}
                echo("Page $i");
                if ($i == $s) {echo("</b>");}
                else {echo("</a>");}
                echo("]  ");
            }
            echo("</p>")


Page Numbers here using the above code.. The problem I seem to be running into is that it only displats the first record. The page numbers show up page 1 page 3
page 2 and three are blank there is no mysql error or anything for some reason I only get that first result out of three

Hi guys,
 I have a php file that will go to a site and scrape the data I need. However this site is setup to use pagination so when I try to scrape all the players names I have to do separate queries to search each page. Is there a way to find out by using code how many pages there are and query all the pages at same time?

I use  this code
Code: [Select]
<?php //first page

//turn error reporting on
libxml_use_internal_errors(true);

//get data from this page
$dom = new DOMDocument;
$dom->loadHTMLFile('http://www.gametracker.com/server_info/76.73.3.42:1716/top_players/?searchipp=50#search');
$xpath = new DOMXPath($dom);

// Get the total player count
$rows2 = $xpath->query('//div[@class="block774"]/div');

// Get the rows from the search list
$rows = $xpath->query('//table[@class="table_lst table_lst_spn"]/tr');


for ($i=1; $i<$rows->length-1; $i++) {
    $row = $rows->item($i);

    // Get the columns for a row
    $cols = $row->getElementsByTagName('td');

// Get the player rank (1st column)
    echo 'Rank:'.trim($cols->item(0)->textContent).PHP_EOL;

    // Get the player name (2nd column)
    echo 'Name:'.trim($cols->item(1)->textContent).PHP_EOL;

// Get the player score (3rd column, actually 4th but number 3 is hidden)
    echo 'Sco '.trim($cols->item(3)->textContent).PHP_EOL;

echo "<br />";
}
?>





<?php //secondpage

//turn error reporting on
libxml_use_internal_errors(true);

//get data from this page
$dom = new DOMDocument;
$dom->loadHTMLFile('http://www.gametracker.com/server_info/76.73.3.42:1716/top_players/?searchipp=50&searchpge=2#search');
$xpath = new DOMXPath($dom);

// Get the rows from the search list
$rows = $xpath->query('//table[@class="table_lst table_lst_spn"]/tr');


for ($i=1; $i<$rows->length-1; $i++) {
    $row = $rows->item($i);

    // Get the columns for a row
    $cols = $row->getElementsByTagName('td');

// Get the player rank (1st column)
    echo 'Rank:'.trim($cols->item(0)->textContent).PHP_EOL;

    // Get the player name (2nd column)
    echo 'Name:'.trim($cols->item(1)->textContent).PHP_EOL;

// Get the player score (3rd column)
    echo 'Sco '.trim($cols->item(3)->textContent).PHP_EOL;
echo "<br />";
}

?>
I also have to go to that website first to see how many pages there are so I can have enough queries.

Hey Im always trying to remove code and cut corners to reduce work in the long run, soIim wondering how I could link my menu bar from say a template to ALL my php pages for my site so I don't have to write/change links on every page when I need to.

Thanks

I have an ordering process that consists of 3 pages:

1) Page1: Enter name/address
2) Page2: Based on address, determine availability of products, and let user choose which products, how many, etc.
3) Page3: Enter credit card info to send to the processor

What is the best way to store information fro page 1 to page 2, then from page 2 to page 3? I thought about just using hidden fields. Is this OK? Is there another preferred way?

Thanks!

Hi,

I have some code which scrapes data from a page. However there are around 1200 product pages on the site I need to scrape, when I attempt to loop through all the pages I get a server timeout. I can only get to around 40 without timeout. Has anyone else had this problem?