PHP - Function That Generates A Variable Variable?

I am using two diffrent databases in my php system, MySQL and MSSQL. my issue is i need a list of results from a MySQL query to be part of a Where clause in the MSSQL query. I cant put the the MSSQL query in the while section of the MySQL query as i need to sort by the results of the MSSQL query. if i do the MySQL query and then use  the info from that in the MSSQL query I get only one result. so i decided to make it a function. how ever i need to know how to  make the function into a variable so i can input it in the MSSQL query. Hope this all makes senes. heres what i have that is not working .

function somefunction($vaiable) {
$gers = mysql_query(sprintf("SELECT emp_id, lastname FROM info WHERE active='1' AND oid='$oid' AND pidnumber='4'")) or die(mysql_error());
while ($gerow = mysql_fetch_array($gers)) {
echo "ci.attorney='$lastnamey' OR ";

 }
}

$attlist = getattorney($vaiable);
echo $attlist;

$rscw=odbc_exec($conn, sprintf("SELECT ci.attorney, SUM(ci.totalweight) AS 'citw' FROM info ci, clients cl WHERE ci.ID = cl.ID AND (cl.county = '$xca' OR cl.county ='$xcb' OR cl.county ='$xcc' OR cl.county ='$xcd') AND ($attlist ci.attorney='xyz') GROUP BY ci.attorney ORDER BY citw DESC")) or die (odbc_errormsg());

while (odbc_fetch_row($rscw))
{

echo odbc_result($rscw,"attorney");
echo odbc_result($rscw,"citw");
}


is there a way to make a function be defined as a variable?

Any help is appreciated

Hi,

Im trying to pass a variable ($in_this_instance) into, and then back out of a function.  The variable goes into the function no problem, and is echo'ed out fine.  However the echo after the close of the function, does not give anything out. 

Code: [Select]
$in_this_instance = 'boo';

function in_this_instance($data) {
global $in_this_instance;

echo $in_this_instance;

$in_this_instance = 'hoo';


$rep_val = '[front banner]';

    $test = strpos($data, $rep_val);
    if ($test === false) {
return $data;
    } else {
    return $data;;
    }
}

add_filter('the_content', 'in_this_instance');
echo $in_this_instance;

Could any one give me a pointer please?  Many Thanks!

I have just started using functions, and have been following a few tutorials. I think im misunderstanding returns.

lets say i have the variables called $math1 an $math2

and i want to make $total = $math1 + $math2;

i then use return $total;

I then thought i could echo $total and it would work outside of the function but its giving me an error of an underfines variable.

Could anyone explain what im doing wrong please? thanks

Code: [Select]
class replayer extends users {
--SNIP--
public $islive = false;
--SNIP--
     function chat($id, $nick, $chat) {
          --SNIP--
          <-----------
     }
}
There's my code, I want to take the variable $islive from the above class and use it in the chat() function.

How would I do it? I've tried making it a global inside of the chat() function.

This seems really straight forward but I can't figure out why it's happening.  As a part of an account registration script, I'm randomly generating a password using a function called, generatePassword($length,$strength);
I hash the password (md5) before putting it in the database but I also email the non-hashed version to the user.  My problem is that each time I call the variable, it generates a new password so the one in the database does not match the one received by the user, $user_password_plain = generatePassword($length,$strength);
$user_password = md5($user_password_plain);
Is there a different way to store the variable so that I can access it later in the script and it will be the same?  I tried using a session variable but that didn't seem to work either.  Thanks!

Help I am trying to pass a php variable into a javasript function. Here is my code.

<a href="<?php echo $row['track']; ?>" download="<?php echo $row['track']; ?>" onclick="myhit(<?php echo $row['track']; ?>)">Download</a>

<script>
function myhit(top){
alert(top);

}
</script>

the javascript function is never called

Hi,

I'm having an image upload function I did create for few weeks ago - it has an variable called $newname, which contains the path and file.

I'm then using the imageupload() function in another function called EditFrontPage(), which is used to 'update' some content.

If I update the image, it runs the imageupload function, which is great, it resize and optimize the image, and it moves it to the folder I specified.

What I then want, is within' my EditFrontPage() function, is to echo out the $newname variable from the imageUpload function.

is there a way to do this? in a smart way?

If my code is needed, you can take a look at it he http://codepad.org/poKNSU5H

Thank you a lot guys!

Hello,

I have been trying to replace a variable's value inside a function in case it matches a certain criteria but I can't seem to get it right.

Let me give an example:

Code: [Select]
<?php

function replace_value($obj)
{
if($obj==1)
{
$obj=2;
}
}

$number = 1;
replace_value($number);

echo $number;
So how do I go about making $number = 2, because it still has a value of 1 even after I pas it through the function replace_value.

Hello

I have a question (that i think i know the answer to).

What would be more efficient:

1. Create a global variable and run a php function and store result in there and call back the variable when needed

2. just echo the return value of the function when needed

Now im guessing that calling the function once and storing that in a variable rather than keep calling the function is more efficient

I dont know why I'm having so many problems with this. I need to get a variable that defined in an included file inside of a function I've got this

Inside file.php I've got the $variable1 and $variable2 defined (they're dynamic/different to each user)

include('file.php');

function foo() {
GLOBAL $variable1, $variable2;

echo $variable1;
echo $variable2;
}

foo();

Yet it outputs nothing?

If I Include file.php right inside of the function it works however. Any idea?

Hi All,

I have a function that i want to pass a variable into so that i can do some SQL on it.

The information that i want to pass into it is a data-id on a button that is used to trigger the function.

The button that will be clicked is this

<div class='modaltrigger btn btn-primary' data-id='$itemId' data-toggle='modal'>Manage</div>

This button is being created by another function.

I would like to know either how i pass the variable from one function to another or how i pass it from the data-id to the function in php.

Thanks in advance.

Hi guys,

I have a trouble with my php snippet, when I insert the var function in the url bar something is like:

http://www.mysite.com/delete.php?favorites&id=0

or

http://www.mysite.com/delete.php?whateveritis&id=0

It doesn't get pass the favorites function to delete the id. It is the same things that it goes for each different function.

Here's the current code:

<?php
Code: [Select]
session_start();
    define('DB_HOST', 'localhost');
    define('DB_USER', 'mydbuser');
    define('DB_PASSWORD', 'mydbpass');
    define('DB_DATABASE', 'mydbtablename');

    $errmsg_arr = array();
    $errflag = false;

    $link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD);
    if(!$link) {
  die('Failed to connect to server: ' . mysql_error());
    }

    $db = mysql_select_db(DB_DATABASE);
    if(!$db) {

die("Unable to select database");
    }

   function clean($var){

return mysql_real_escape_string(strip_tags($var));
    }
    $favorites = clean($_GET['favorites']);
    $id = clean($_GET['id']);

if($favorites && $id == ''){
   // both are empty
   $errmsg_arr[] = 'favorites id are missing.';
   $errflag = true;
}

    if($errflag) {
  $_SESSION['ERRMSG_ARR'] = $errmsg_arr;
  echo implode('<br />',$errmsg_arr);
   }
   else {
$insert = array();
if(isset($_GET['id'])) {
    $insert[] = 'id = \'' . clean($_GET['id']) .'\'';
}
if(isset($_GET['favorites'])) {
    $insert[] = 'favorites = \'' . clean($_GET['favorites']) . '\'';
}

if($favorites && $id) {
   mysql_query("DELETE FROM favorites WHERE id='$id'");
   $deleted = mysql_affected_rows();
   if($deleted > 0)
   {
  echo "favorites channels is deleted";
   }
   else
   {
  echo("favorites is already deleted");
   }
 }
}
?>
If you do know how to get pass the favorites function, then please say so as i need your help.

Any advice would be much appreicated.

I have just re-installed Xampp and suddenly my sites are now displaying lots of:

Notice: Use of undefined constant name - assumed 'name' in ...
Notice: Use of undefined constant price - assumed 'price' in ...

this is an example of the line its refering too:

$defineProducts[1001] = array(name=>'This is a product', price=>123);

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 am currently using a javascript to redirect the page using a select tag and the onchange property. Each page calls a different function.

I want to be able to pass the <select> value to a variable that I can use within a php condition statement on the same page. This will generate the required output based on the <select> value. This way I dont require a seperate page for each query.

So onchange.... $select_value == this.value?

Then within my page:
Code: [Select]
if ($select_value == 'alpha') {
    get_query_alpha();
}
if ($select_value == 'bravo') {
    get_query_bravo();
}

This is my current set up...
Code: [Select]
<p>Order by:
<select name="query" id="query" onchange="gotourl(this.value)">
<option value="query1.php" >Sort by A</option>
<option value="query2.php" >Sort by B</option>
<option value="query3.php" >Sort by C</option>
<option value="query4.php" >Sort by D</option>
</select></p>

<table> <?php get_query_1();?> </table>

and js...
Code: [Select]
function gotourl(url){
window.location= url;
}
function selectsubject(){
alert("Please select a subject!");
}

Is there a simple way to do this? Thanks.