PHP - Help With Writing Basic Code

am total newbie to programming, apart from knowing SQL, the thing is i have been given a MYSQL database containing various information about kids diseases and a web interface written in php to create reports from the database that can be accessed via the interface. there are almost 25 different variables that need to be computed in the report, i have written sql queries to compute all these values, but i don't know anything about PHP, if i have all these queries isn't there a way for me to combine all these sql queries to be display results on a webpage and come up with this report without writing PHP code?

Thanks for your help

very sorry if this is too basic

Hi again....

I'm working on a project that creates a profile page for the user (i.e. 'username.php') when they register. Because there are two ways to register, through Facebook and through the website itself, there has to be an if statement in this page that it crates as to which ID to use for that user. Because Facebook usernames are not unique, we must identify a user through their OAuth User ID which is a 10 digit number. When a user registers through the website itself, usernames are unique so their profile page can be ID'd by their username.

Here's the code that creates that profile page:
function createProfile($user) {
        $userFile = 'users/'.$result['oauth_uid'].'.php';
        $fh = fopen($userFile, 'w') or die("can't open file");
        $stringData = "<?php\n" . '$pageowner = "' . $result['oauth_uid'] . '";' . "\n" . 'include "profile.php";' . "\n?>";
        fwrite($fh, $stringData);
        fclose($fh);
}

$user is the username passed to the function when it's called.

Currently that is set to create a line of code in the new profile page ('userid'.php) that looks like this:
<?php
$pageowner = "100001745088506";
include "profile.php";
?>

I want to add an IF to that page that follows this structu
if oauth_provider == facebook { $pageowner = "oauth_uid" } else { $pageowner = username }

But I don't know how to write that he
 $stringData = "<?php\n" . '$pageowner = "' . $result['oauth_uid'] . '";' . "\n" . 'include "profile.php";' . "\n?>";

to make it show up in the page it creates. I want that if statement to be written IN the page that is created.

I just don't know the syntax well enough yet to do that. Would somebody help me out or point me in the right direction?

Thanks!

for timthumb in my wordpress i use this code to
function first_image() {
  global $post, $posts;
  $first_img = '';
  ob_start();
  ob_end_clean();
  $output = preg_match_all('/<img.+src=[\'"]([^\'"]+)[\'"].*>/i', $post->post_content, $matches);
  $first_img = $matches [1] [0];
 
  if(empty($first_img)){ //Defines a default image
    $first_img = "images/default.gif";
  }
  return $first_img;
}

it works really well, then i thought why not to make it show random images if timthumb dont find any image?

so the 1st googling i did came up with this function:
<?php 
function getRandomFromArray($ar) { 
    mt_srand( (double)microtime() * 1000000 ); 
    $num = array_rand($ar); 
    return $ar[$num]; 
} 

function getImagesFromDir($path) { 
    $images = array(); 
    if ( $img_dir = @opendir($path) ) { 
        while ( false !== ($img_file = readdir($img_dir)) ) { 
            // checks for gif, jpg, png 
            if ( preg_match("/(\.gif|\.jpg|\.png)$/", $img_file) ) { 
                $images[] = $img_file; 
            } 
        } 
        closedir($img_dir); 
    } 
    return $images; 
} 

$root = ''; 
// If images not in sub directory of current directory specify root  
//$root = $_SERVER['DOCUMENT_ROOT']; 

$path = 'images/'; 

// Obtain list of images from directory  
$imgList = getImagesFromDir($root . $path); 

$img = getRandomFromArray($imgList); 

?>

and to show random image i Place the following where i wish the random image to appear:

Code: [Select]
<img src="<?php echo $path . $img ?>" alt="" />
my question is how to place this code in my function first_image??
i need to do something like this, but it doesnt make any sense:
if(empty($first_img)){ //Defines a default image
    $first_img = " <img src="<?php echo $path . $img ?>" alt="" />
this was my thought, if i can mix those 2 normally it should work, right?

or maybe there is another way to show random image in the Code: [Select]
[i] if(empty($first_img)){ //Defines a default image[/i]
  [b]  $first_img = ?

Hi all, 

Hi guys,

I've just started using xampp for php. My friend sent me the files i've attached below to see if apache is working for me, but i don't understand why they won't work together. Basically the .html file opens a simple login which works fine. When the password is typed in it should direct me to some links, but for some reason it doesn't. It does however look like the apache server is working fine - it could be something with the code i'm really not to sure. When i try to log in these are one set of the errors it gives me:

Warning: file_get_contents(listlinks.array) [function.file-get-contents]: failed to open stream: No such file or directory in /Applications/XAMPP/xamppfiles/htdocs/test1/userlinks.php on line 8

Warning: sqlite_open() [function.sqlite-open]: unable to open database: /Applications/XAMPP/xamppfiles/htdocs/Test1/userlinks.sqlite in /Applications/XAMPP/xamppfiles/htdocs/test1/userlinks.php on line 16

Warning: sqlite_query() expects parameter 1 to be resource, string given in /Applications/XAMPP/xamppfiles/htdocs/test1/userlinks.php on line 20

Warning: sqlite_fetch_array() expects parameter 1 to be resource, null given in /Applications/XAMPP/xamppfiles/htdocs/test1/userlinks.php on line 21

Warning: Cannot modify header information - headers already sent by (output started at /Applications/XAMPP/xamppfiles/htdocs/test1/userlinks.php: in /Applications/XAMPP/xamppfiles/htdocs/test1/userlinks.php on line 25

Warning: sqlite_close() expects parameter 1 to be resource, boolean given in /Applications/XAMPP/xamppfiles/htdocs/test1/userlinks.php on line 61

I've literally just started learning web scripting so i apologise for over-looking anything simple.

The login.php file is also supposed to be .html aswel - i just changed it to upload.

Thanks in advance, Josh.

Hi All,

I'm having a major mind blank, and can't find anything in the previous posts resolving what I'm after.

I'm setting

$searchtext = $_POST['searchtext'];

I want to check $seachtext is not null.

I've seen isset($searchtext) but it doesn't solve my problem.

Basiclaly; I want an if statement to say if(isset($searchtext)) {......}

Thoughts?

I have a code for writing a log file...
The code is working fine but it inserts the new details at the bottom of the file but i want to insert in the top...

So what function should i use... Here is the file my_log.php

Code: [Select]
<?php

$usname = $_SESSION['usname'];
date_default_timezone_set('Asia/Calcutta');
$date = date("l dS \of F Y h:i:s A");
$file = "log.php"; 
$open = fopen($file, "a+"); 
fseek($open,289);
fwrite($open "<b><br/>USER NAME:</b>              ".$usname . "<br/>"); 
fwrite($open, "<b>Date & Time:</b>            ".$date. "<br/>"); 
fwrite($open, "<b>What have they done :</b>   ".$reason . "<br/><br/>"); 
fclose($open); 

?>

and here is my log.php file :

Code: [Select]
<?php
session_start();
if($_SESSION['stage']!=1 || $_SESSION['stage2']!=2)
{header('location:index.php');
die(" ");
}
?>
<?php
if($_GET['valu']=="view_log")
{
$reason=" ".$_SESSION['usname']." Viewed the LOG";
include('my_log.php');
header('location:log.php');
die("");
}
// bytes till here are 289

//LINE 1 : Log details have to insert here



I want to insert every detail from the top of the page but just below the php code..
What should i do...

ANy helo would be appreciated

Pranshu Agrawal
pranshu.a.11@gmail.com

I am using a cache script, well it writes a array to a file like this:

Code: [Select]
fwrite($fh, '<?php'."\n\n".'define(\'PUN_LOTTERY_LOADED\', 1);'."\n\n".'$lottery = '.var_export($output2, true).';'."\n\n".'?>');
output2 is

Code: [Select]
$result2 = $db->query('MY QUERY ');
$output2 = array();
while ($cur_donors = $db->fetch_assoc($result2))
$output2[] = $cur_donors;

Now, I want to ditch the mysql and I want to use this script with 7 variables that I already have loaded, so I dont need to use the mysql, how do I add my 7 variables to my var_export function instead of using mysql to loop them?

I have a web app hosted on Just Host that I have nearly finished writing but still needs to have coded the ability to write a text file to user's computer. the user should be able to specify where on their computer they would like the file to be stored. Is this possible ? And can you tell me how it's done, or point me to a souce that can tell me 

I rarely ever ask for help regarding programming, but this has flumoxed me.


If it is not possible to do this, then would I have to generate this file on the Host's server (Just Host in my case), then download it ? If this can be done then can you please tell me how, as any info I have found related to file downloads seems a bit obscure 

thanks in advance 

Ok, I know how to write to a file, but what I'm looking for is to check if a line of code exists, and if it is, don't recreate it. It would also be nice to not recreate the file either.

Code: [Select]
$file = fopen("index.html", "w");
fwrite($file,"This is a line of text");
fclose($file);

im trying to make a program that changes some files from 0 to 1,but im having some trouble...

heres my code,
Code: [Select]
$id = $_GET["id"];

$file1 = "http://mysite.co.cc/users/".$id."/file1.txt";

$fh = fopen($file1, 'w');
fwrite($fh, "1");
fclose($fh);

I followed this tut: http://www.tizag.com/phpT/filewrite.php

I cant find anything wrong with the code,but it will not make the file have 1 in it.

I'm trying to write to a file with the following code.

$fh = fopen("../inc/config.php", "a") or die("\r\nCan't open file.");
$write = "\$config['database_host'] = {$mysqlh};
\$config['database_user'] = {$mysqlu};
\$config['datbase_pass'] = {$mysqlp};
\$config['datbase_name'] = {$dbn};
\$config['table_prefix'] = {$tp};";

fwrite($fh, $write);
fclose($fh);


It is printing out "Can't open file", which I guess means it can't find the file or I've given the wrong path. The file that is executing this code is /install and the file I'm trying to write to is /inc/config.php. I thought .. put you up a directory so if I do .. I will be at the root and then do /inc I will be in inc.

Can someone please guide me on what I'm doing wrong?

Thanks.

Greetings,   I am new to these forums, I am working on this assignment, and these are the current issues I am running into.   Notice: Undefined variable: year in G:\EasyPHP-5.3.2i\www\PHP_Projects\ChineseZodiacs\zodiac_year_switch.php on line 77 ie. $year = validateInput($year,"Birth Year"); Notice: Undefined variable: year_count in G:\EasyPHP-5.3.2i\www\PHP_Projects\ChineseZodiacs\zodiac_year_switch.php on line 138 ie. echo "<p>You are person " . $year_count . "to enter " . $year . "</p>\n";   Honestly, I believe they are linked, because what should be happening, as the user enters the year, and hits submit, it should create a file called counts/$year.txt - $year should equal the entered data in the textbox, any help would be appreciated.   Thank you for your help.

<!DOCTYPE html>
<head>
<title>Write to and From a File</title>
<meta http-equiv="content-type" content="text/html; charset=iso-8859-1" />

<?php

$dir = "counts";
if ( !file_exists($dir)) {
	mkdir ($dir, 0777);
	}

function validateInput($year, $fieldname) {
	global $errorCount;
	if (empty($year)) {
		echo "\"$fieldname\" is a required field.<br />\n";
		++$errorCount;
		$retval = "";
	}
	else
	{ // if the field on the form has been filled in
		if(is_numeric($year))
		{
			if($year >=1900 && $year <=2014)
			{
				$retval = $year;
			}
			else
			{
			++$errorCount;
			echo "<p>You must enter a year between 1900 and 2014.</p>\n";
			}
		}
		else
		{
			++$errorCount;
			echo "<p>The year must be a number.</p>\n";
		}
		
	} //ends the else for empty
	return($retval);
} //ends the function
function displayForm() {
?>
<form action = "<?php echo $_SERVER['SCRIPT_NAME']; ?>" method = "post"> 
<p>Year of Birth: <input type="text" name="year" /></p>
<p><input type="reset" value="Clear Form" />&nbsp; &nbsp;<input type="submit" name="submit" value="Show Me My Sign" /></p> 
</form>
<?php
}

function StatisticsForYear($year) {
		global $year_count;
	$counter_file = "counts/$year.txt";
	if (file_exists($counter_file)) {
		$year_count = file_get_contents($counter_file);
		file_put_contents($counter_file, ++$year_count);
	} else {
		$year_count = 1;
		file_put_contents($counter_file, $year_count);
	}
return ($year_count);
}?>	
	
</head>
<body>
<?php

$showForm = true;
$errorCount = 0;
//$year=$_POST['year'];

$zodiac="";
$start_year =1900;
if (isset($_POST['submit'])) 
	$year = $_POST['year'];
	$year = validateInput($year,"Birth Year");

	if ($errorCount==0)
		$showForm = false;
	else
		$showForm = true;

if ($showForm == true)
{
		//call the displayForm() function
		displayForm();
}
else
{ //begins the else statement
	//determine the zodiac
	 
	$zodiacArray = array("rat", "ox", "tiger", "rabbit", "dragon", "snake", "horse", "goat", "monkey", "rooster", "dog", "pig");
	
	 switch (($_POST['year'] - $start_year) % 6) {
	case 0:
			$zodiac = $zodiacArray[0];
	break;		
	case 1:
			$zodiac = $zodiacArray[1];
	break;		
	case 2:
			$zodiac = $zodiacArray[2];
	break;		
	case 3:
			$zodiac = $zodiacArray[3];
	break;		
	case 4:
			$zodiac = $zodiacArray[4];
	break;		
	case 5:
			$zodiac = $zodiacArray[5];
	break;		
	case 6:
			$zodiac = $zodiacArray[6];
	break;		
	case 7:
			$zodiac = $zodiacArray[7];
	break;		
	case 8:
			$zodiac = $zodiacArray[8];
	break;		
	case 9:
			$zodiac = $zodiacArray[9];
	break;		
	case 10:
			$zodiac = $zodiacArray[10];
	break;		
	case 11:
			$zodiac = $zodiacArray[11];
	break;
	default:
		echo "<p>The Zodiac for this year has not been determined.</p>\n";
	break;
	} //ends the switch statement

	echo "<p>You were born under the sign of the " . $zodiac . ".</p>\n";
	echo "<p>You are person " . $year_count . "to enter " . $year . "</p>\n";
} //ends the else statement	
	
?>
</body>
</html>

Edited by mstevens, 16 October 2014 - 06:36 PM.

I am using a script I adapted from a tutorial to print the contents of a text box to a txt file. Basically, it's a really simple way of seeing who has logged in. I only have a handful of users.

The problem is, although the text file is being created in the proper folder, it isn't being written to and just remains blank.

<div align="center">
<table width="300" border="2" bordercolor="#FFFFFF" style="-moz-border-radius: 18px;
-webkit-border-radius: 18px;" height="120" cellpadding="0" cellspacing="0">
<tr>
<form name="form1" method="post" action="checklogin.php">
<td>
<table width="100%" border="0" cellpadding="3" cellspacing="1" background="images/loginbg.jpg"  style="-moz-border-radius: 15px;
-webkit-border-radius: 15px;">
<tr align="center">
<td colspan="3"><font color="#FFFFFF"><strong>Family Login </strong></font></td>
</tr>
<tr>
<td width="78"><font color="#000000">Username</font></td>
<td width="6">:</td>
<td width="294"><input name="myusername" type="text" id="myusername">
<?php
$myusername = $_POST['myusername'];

$data = "$myusername\n";


//open the file and choose the mode
$fh = fopen("logs/login.txt", "a");

fwrite($fh, $data);

fclose($fh);

?></td>
</tr>
<tr>
<td><font color="#000000">Password</font></td>
<td>:</td>
<td><input name="mypassword" type="password" id="mypassword"></td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td><input type="submit" name="Submit" value="Login">
</td>
</tr>
</table>
</td>
</form>
</tr>
</table>
</div>

I'm not sure what's going wrong but I'm guessing it's the placing of the php, or at least some of it.
I'd quite like to add the time they logged in as well. Any idea's anyone?