PHP - Help For Writing Code Without Mysql_result

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!

Code: [Select]
<?php  $liked = ("SELETCT SUM(user_id) as totaluser_id FROM liked WHERE `user_id`='26762'");
$liked_results = mysql_query($liked);
$sum_liked = mysql_result($liked, 0); // <--- Error 
echo $sum_liked;

?>


i got this error
Warning: mysql_result(): supplied argument is not a valid MySQL result resource in

can sombady help?

i am wanting to use the following code but it is not working as i thought it would.

instead of hard coding the field name in the script i wish to have it insert using a variable like below.

what is the correct way i should do this.

Code: [Select]
$c= "fieldname";

$a = mysql_result( $results, 0, $c);

mysql_result() : supplied argument is not a valid result in .....

Code: [Select]

function previously_liked ($article_id) {
                 $article_id = (int) $article_id;
return (mysql_result(mysql_query("SELECT COUNT(`like_id`) FROM  `likes` WHERE `like_id` = " .$_SESSION['user_id']. " AND     `article_id` = $article_id"), 0) == 0 ) ? false : true; // this is the line with the error.
}

Hi there
I am having a problem with this warning message being thrown
"Warning: mysql_result() expects parameter 1 to be resource, boolean given in C:\wamp\www\css\rnfunctions.php on line 302"
and again for line 303
Here is the code

function receiveuserchange()
{
if (isset($_POST['level'])){
$query = "SELECT * FROM users";
$result=mysql_query($query) or die("Invalid Query : ".mysql_error());
$num=mysql_numrows($result);
$i=0;

while ($i < $num) {

////////////////here are lines 302 and 303///////////////////////////////
$user=mysql_result($result,$i,"user");
$id=mysql_result($result,$i,"id");
/////////////////////////////////////////////////////////////////////////////

if (isset($_POST["$user"])){
$level = ($_POST["$user"]);
$query = "update profiles set level='$level' where pid='$id'";
$result=mysql_query($query) or die("Invalid Query : ".mysql_error());
ViewUsers('viewusers.php');
}else{
echo "$user not recieved";
}
$i++;
}

Can anyone see what the problem is?

don't know what to do next with this error popping up,,,any help?


<?php
      //set up the variables
      $loguser = $_POST['loguser'];
      $logpass = md5($_POST['logpass']);
      $login = $_get['login'];
      
      //connect with the server and the database
      mysql_connect("localhost","root","") or die("Can't connect with the Server!");
      mysql_select_db("comp3project") or die("Can't connect with the database!");
      
      if($login="yes")   
      {
         $get = mysql_query("SELECT count(cid) FROM customer_data WHERE username = '$loguser'
            and password = '$logpass'");
         $result = mysql_result($get,0);
         
         if($result!=1){print "Login Failed!";}
         else {print "Login Successful!";}
      }
   ?>

I have this code

Code: [Select]
$id = $_GET['esitysid'];
$esitysnimi = mysql_query("SELECT * FROM varasto WHERE id = '".$id."", $yhteys);
print "esitysnimi $esitysnimi";
$esitysnim = mysql_result($esitysnimi, "0", "nimi");
varasto:
Code: [Select]
nimi hinta maara id lippuja
Esitys Nimi 1 10 14 0 5
Esitys Nimi 2 120 5 1 0
Esitys Nimi 3 950 5 2 0
and it says

Code: [Select]
Warning: mysql_result() expects parameter 1 to be resource, boolean given
How can I fix it? Thank you for help

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);

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? 

Hi everyone.  Still new to PHP and to Object-Oriented Programming.

My goal for today is to write a few dummy Classes in PHP that do enough so that I can see something on my webpage.

I am wondering if there is someone here who would be willing to help in either the forums or one-on-one.

It may sound like a silly request, but this is all new to me as a Procedural Programmer who has actively programmed in about 10 years!!   

Thanks,


TomTees

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'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.