PHP - Add Value To A Variable Without Forming It Before

My login script stores the user's login name as $_SESSION[ 'name'] on login. For some unapparent reason, i'm getting errors stating that $user  and $priv are  undefined variables, though I've attempted to define $user as being equal to $_SESSION['name'], using $user to look up the the user's privilege level (stored as the su column ) in the SQL table, and then where the result of the sql query is $priv which is then evaluated in an if statement. I can't seem to figure out why this might not be working.

The code I'm using:

<?php
session_start();
function verify() {
	//verify that the user is logged in via the login page. Session_start has already been called. 
	if (!isset($_SESSION['loggedin'])) {
	header('Location: /index.html'); 
	exit;
	}
	//if user is logged in, we then lookup necessary privleges. $_SESSION['name'] was written with the login name upon login. Privleges   // are written in db as a single-digit integer of of 0 for users, 1 for administrators, and 2 for special users. 
        $user === $_SESSION['name'];
    //Connect to Databse
	$link = mysqli_connect("127.0.0.1", "database user", "password", "database");
	if (!$link) {
  	  echo "Error: Unable to connect to MySQL." . PHP_EOL;
    	echo "Debugging errno: " . mysqli_connect_errno() . PHP_EOL;
    	echo "Debugging error: " . mysqli_connect_error() . PHP_EOL;
    	exit;
        }
        //SQL Statement to lookup privlege information. 
       if ($result = mysqli_query($link, "SELECT su FROM accounts WHERE username = $user", MYSQLI_STORE_RESULT)) {
         //LOOP TO CYCLE THROUGH SQL RESULTS AND STORE Privlege information as vairable $priv.
            while ($row = $result->fetch_assoc()) {
           $priv === $row["su"];
            }
        }
        // close SQL connection.
        mysqli_close($link);

 // Verify privleges and take action. Only a privlege of "1" is allowed to view this page. A privlege of "2" indicates special 
//accounts used in other scripts that have certain indermediate additional functions, but are not trusted administrators. 

        if ($priv !== 1) {
        echo $_SESSION['name'];
        echo "you have privlege level of $priv";
        echo "<br>";
        echo 'Your account does not have the privleges necessary to view this page';
        exit;
        }
        
}
verify();
?>

 

I have a script that adds points together based upon the placing. This is the actual script:

Code: [Select]
<? $points = 0;

if($place === '1st') {$points = $points + 50;}
elseif($place === '2nd') {$points = $points + 45;}
elseif($place === '3rd') {$points = $points + 40;}
elseif($place === '4th') {$points = $points + 35;}
elseif($place === '5th') {$points = $points + 30;}
elseif($place === '6th') {$points = $points + 25;}
elseif($place === '7th') {$points = $points + 20;}
elseif($place === '8th') {$points = $points + 10;}
elseif($place === '9th') {$points = $points + 10;}
elseif($place === '10th') {$points = $points + 10;}
elseif($place === 'CH') {$points = $points + 50;}
elseif($place === 'RCH') {$points = $points + 40;}
elseif($place === 'TT') {$points = $points + 30;}
elseif($place === 'T5') {$points = $points + 30;}
elseif($place === 'Champion') {$points = $points + 50;}
elseif($place === 'Reserve Champion') {$points = $points + 40;}

echo "Total HF Points: $points";
?>What it *should* do (my friend's script works the same way and it works) it starts at points = 0, than if there is a first place, it adds 50, and so forth until it reaches the end.

It is included into a file, in this area:

Code: [Select]
<div class="tabbertab">
<h2>Records</h2>
<? $query92 = "SELECT * FROM THISTABLE WHERE VARIABLE='$id' OR VARIABLE = '$name' ORDER BY ABS(VARIABLE), VARIABLE";
$result92 = mysql_query($query92) or die (mysql_error());

echo "<table class='record'>
<tr><th>Show</th> <th>Class</th> <th>Place</th></tr>
";

while($row92 = mysql_fetch_array($result92)) {
$class = $row92['class'];
$place = $row92['place'];
$entries = $row92['entries'];
$race = $row92['show'];
$purse = number_format($row92['purse'],2);

echo "<tr><td>$race</td> <td>$class</td> <td>$place</td></tr>";
}
?>
<tr><td colspan='3'><div align='right'><? include('includes/points.php'); ?></div></td></tr>
</table>
</div>

This is the code that is relevant. When ended here, it echoes the last place that appears in the results (such as a 5th place echoing 30 points). When I move it to be included in the while loop, it shows Total Points: 50 Total Points: 25 Total Points: 10 (depending on the results displayed on that page). What am I doing wrong?

I have a form that creates rows of data input textboxes depending on a user input number of things.  I have a naming convention for all these textboxes that basically just keeps incrementing a number suffix for each row.  All this is working fine.

My problem is I need to get the data inserted into this table of textboxes into an array.

Here's my code where I attempt to to this (it does not work):

Code: [Select]
$temp = $_SESSION['Num_Part'];
$count = 1;
while ($count <= $temp){
  $temp2[$count] = "'Participant_P".$count."'";
  //echo $temp2[$count]."<br/>";
  $temp3[$count]=$_POST[$temp2[$count]];  //here's the problem
  $temp4[$count] = "'Result_P".$count."'";
  $temp5[$count]=$_POST[$temp4[$count]];  //here's the problem
  //echo $temp4[$count]."<br/>";
  $count++;
}

The problem is that the $_POST does not work with the variable in the argument position - even though the argument is formatted with single quotes.

Can a variable be used in a POST argument and if so what is the correct syntax?

If not, is there some other simple solution to harvest the data into an array.  I understand I can harvest by explicitly accessing each key in the post assoc array.  But this could be dozens of rows of input fields.

Thanks in advance for your help here.  I couldn't find anything online re this topic.

Hello everyone,

I can get Test 2 to successfully operate the if statement using a variable variable.

But when I try the same method using a session variable (Test 1) the if statement is not executed. Please could you tell me why the if statement in Test 1 is not being executed?

Code: [Select]

<?php
# TEST 1
$_SESSION[test_variable] = "abcd";
$session_variable_name = "_SESSION[test_variable]";
if ($$session_variable_name == "abcd")
{
echo "<br>line 373, abcd<br>";
}

# TEST 2
$test_variable = "efgh";
$test_variable_name = "test_variable";
if ($$test_variable_name == "efgh")
{
echo "<br>line 379, efgh<br>";
}

?>
Many thanks,

Stu

Probably something simple but I have searched high and low and can't figure this one out.

I have a variable that is of the datetime format.
I have another variable that is of the time format.
I need to add them together.

Example:
$var1 = 2012-02-24 06:38:22
$var2 = 02:00:00

$var3 = $var1 + $var2 = 2012-02-24 08:38:22

Thanks for the help!

hi all,

I have an language pack for example:
languages/en.php:
Code: [Select]
$en['mail']['letter closing'] = "regards,\n your friend!";
and in my config:
Code: [Select]
$language = "en";
$include_language = @include("languages/".$language.".php");
if(!($include_language))
{
    $try_default_language = @include("languages/nl.php");
    if(!($try_default_language))
    {
        echo "kan de taalpakket niet vinden<br>";
        echo "Could not find the language pack.<br>";
        echo "example on error: ".$test." shows nothing";
        exit;
    }
}
In my function I want to include the language pack for example i have $language = 'en' so I want to include $en['general']['letter closing']

I will do this:
Code: [Select]
global $language,${$language}['general'];
But that gives an error unexpected '[' blah blah.

How can i call the variable variable array in the valid php way?

I just moved my code from Appserv to EasyPHP and it gave me this error, it was working fine on Appserv...what's with easyPHP ??

I have a function that get's a quick single item from a query:

function gimme($sql) { 
global $mysqli; 
global $mytable; 
global $sid; 
$query = "SELECT ".$sql." FROM ".$mytable." WHERE sid = ".$sid; $result = $mysqli->query($query); 
$value = $result->fetch_array(MYSQLI_NUM); 
$$sql = is_array($value) ? $value[0] : ""; 
return $$sql; // this is what I've tried so far 
$result->close(); 
}

It works great as:

echo(gimme("name"));

Then I realized that I could use that as a variable ('$name' in this case) elsewhere. However, I can't figure out how get that new, variable variable 'outside' of the function.  As such,

echo($name);

isn't working outside the function.  Is there a way to return a variable variable?  In other words, is there a way to make a function that creates a variable variable that will available outside of the function?

 

Thanks

 

Quote

i need to store a variable from database like if i have "copies" in one of my column in my database then i have to store a particular value for copies store it to $copies
here i want that i can store value of copies into $copies




$update_book="update book set copies=copies-1 where bookid='$bookid'";
        $result=mysql_query($update_book,$linkID1);
            if($result)
            {
            print "<html><body background=\"header.jpg\">
            <p>book successfully subtracted from database</p></body></html>";
            }
            else {
            print "<html><body background=\"header.jpg\">
            <p>problem occured</p></body></html>";
            }

    }

I'm trying to loop through a database and assign the results to a variable that is created with the loop with the variables name based on which instance is currently running in the loop.  I'll go ahead and include the code I thought might work, but I obviously have no idea what I'm doing in this case

$i = 0;
while ($i < $num_results) {
$ip = ($i++);
$result = ("$res_" . "$ip");
$loop_date = ("$res_" . "$ip" . "_date");
$get_result = "SELECT * FROM table WHERE id = '$result'";
$que_result = mysql_query($get_display);
while ($sub_res = mysql_fetch_array($que_result, MYSQL_BOTH)) {
$loop_date = $sub_res['date'];
}
$i++;
}
echo $res_1_date;

Can anyone please clue me in on how I can get this to work please.  It works with $result but it isn't working with the last $res_1_date... As I'm writing this I see where the problem is but just not quite sure how to make it work...

Is if($variable) the same as if(!empty($variable)), just different syntax?

Code: [Select]
$word = 'numbers';
$numbers= array('1', '2', '3', '4');

echo $$word[0];
I expected the output to be '1'. It ended up being nothing... 

Why does this not work? Is it not possible to have a variable variable array? And if not, is there a workaround?

Cheers,
Joe

Hi,
I have following code :
Code: [Select]
<?php
class CategoryWork {
    
    public $form = "";
    public $error = "";
    public $add_form = "<br /><br /><p><strong>Add New Category</strong></p><br /><form id=\"form1\" name=\"form1\" method=\"post\" action=\"category.php?action=add\">
  <table width=\"550\" height=\"170\" border=\"0\">
    <tr>
      <td width=\"153\">Name :</td>
      <td colspan=\"2\"><label for=\"cat_name\"></label>
      <input name=\"cat_name\" type=\"text\" id=\"cat_name\" size=\"50\" maxlength=\"50\" /></td>
    </tr>
    <tr>
      <td>Slug :</td>
      <td colspan=\"2\"><label for=\"cat_slug\"></label>
      <input name=\"cat_slug\" type=\"text\" id=\"cat_slug\" size=\"50\" maxlength=\"50\" /></td>
    </tr>
    <tr>
      <td>Description</td>
      <td colspan=\"2\"><label for=\"cat_desc\"></label>
      <textarea name=\"cat_desc\" id=\"cat_desc\" cols=\"48\" rows=\"10\"></textarea></td>
    </tr>
    <tr>
      <td>&nbsp;</td>
      <td width=\"97\"><input type=\"submit\" name=\"button\" id=\"button\" value=\"Submit\" /></td>
      <td width=\"286\"><input type=\"reset\" name=\"button2\" id=\"button2\" value=\"Reset\" /></td>
    </tr>
     <tr>
      <td>&nbsp;</td>
      <td colspan=\"2\">error</td>
    </tr>
  </table>
</form>";
}
?>

At the line of <td colspan=\"2\">error</td>
I want to insert $error, hence I made the line as
<td colspan=\"2\">".$error."</td> and also tried <td colspan=\"2\">".$this->error."</td>

After executing with this, php says syntax error at the line of public $add_form

I am going to declare an error at the $error

When I do <td colspan=\"2\">$error</td>
it says : Parse error:  syntax error, unexpected '"' at line of public $add_form

When I do  <td colspan=\"2\">".$error."</td>
it says : Parse error:  syntax error, unexpected '.', expecting ',' or ';' at the above line i.e. of <td>

I am damn confused, I googled but nothing found any helpful material.
Anybody please help me..

Thanks

So I'm fairly new to php and just need some quick help. Here is my code:

Code: [Select]
function plaintext_category(){
$cat_plain = strip_tags( get_the_term_list($post->ID, 'portfolio_category', '', ', ', '' ) );
$cat_plain = strtolower($cat_plain);
$cat_plain = str_replace(' ','-',$cat_plain);
echo $cat_plain;

}

$pattern = '/title=\"(.*?)\"/';
$replace = 'title="echo $cat_plain"'; /* This is where i need help, how do i echo this? */
$categories = preg_replace($pattern,$replace,$categories);
echo $categories;
So I'm trying to pass the echo value of $cat_plain but when I put echo in front of it, i get an error. If I don't put echo, I just get a blank result.

Please, need some help on this. Thanks guys!

Hi, so I have an easy problem that for some reason I couldn't find an answer to anywhere.

I have a bunch of variables like this:

$pic1fileName $pic2fileName $pic3fileName $pic4fileName $pic5fileName...you get the idea
So I have another variable I'm pulling from a database that specifies which number to show, so I need a variable something like this:

$pic($pic_number)fileName

I just don't know what the proper syntax is.  Anyone?  Thank you freaking much for any help; this is a lame problem.