PHP - Giving Variable To Function (very Basic)

Hi, Im learning php, but Im a bit confused on this basic problem.

I have the following code:

<?php
include "tax.php";
$price = 95;
echo "Price before tax: $price <br>";
echo "Price after tax: ";
echo add_tax ($price);
?>


Which uses the function below

<?php
function add_tax ($amount) {
     $total = $amount * 1.09;
     return $total;
}
?>

When I execute the code I can see that it does work, but I do not understand why this part works!

echo add_tax ($price);   

There is no multiplier symbol here, so is the function performed on the ($price) variable simply because of where it is located on this line of code?   Thanks

Hi,

I have just started using PHP again after a long break and am very rusty. I am having a few problems with some basic things:

I have defined a function to connect to my MySQL database and it doesn't work. My $hostname, $username, $password variables are stored in a separate php file which I am including first and the standard mysql_connect function works ok, but I wanted to put it inside another user defined function called "condb" to make things quicker later.

my function looks like this:

function condb(){ 
mysql_connect($hostname, $username, $password) OR DIE ('Unable to connect to database! Please try again later.');
}

and I am trying to run it like this:

condb();

Unfortunately it is not working. Probably a basic error but any help appreciated.
Thanks

Hello all, I have some piece of code that is nested like this


$variable = 'This is a global argument';
function parentFunction($variable) {
function childFunction() {
echo 'Argument of the parent function is '.$GLOBALS['variable'];
}
childFunction();
}
parentFunction(5);


What I want to know is - Is there a way to access a variable from the parent function without passing arguments to the child function? (Something like how classes have parent::?). I don't want to use $GLOBALS because it might cause some variable collision, and I didn't want to pass arguments because incase I decide to change the arguments in the parent function I'd have to do it in the child function aswell. From my searching around in the Internet it seems like this is not possible, but if theres a slight chance that there might be something out there, i'm willing to give it a shot .

Thanks in advance

I want to define a function instead of repeating query in all my php pages. I call a function by passing an $id value and from that function i have to get all the info related to that id, like name, description and uom.    I am trying to do this, but i dont know how to get these values seperately.   here is my function  

function items($item_id)
{
$details = array(); 
$result = mysql_query("select item_id, name, uom, description from items where item_id=".$item_id."") or die (mysql_error());
while($row = mysql_fetch_array($result))
{
$details[] = array((stripslashes($row['name'])), (stripslashes($row['uom'])), (stripslashes($row['description'])));
}
return $details;
}

and i call my function like this

$info = items($id);

Can somebody guide me in this    

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

 

This was generated through phpFormGenerator, but I'm getting the following warning message when a file is not uploaded. I didn't set it as a required field.

"Warning: copy() [function.copy]: Filename cannot be empty ..."

If it possible to skip this check since it's not supposed to be required to upload a file? I'm simply trying to make a contact form that emails basic information with the added option of attaching a file/picture.

Code: [Select]
<?php
include("global.inc.php");
$errors=0;
$error="The following errors occured while processing your form input.<ul>";
pt_register('POST','FullName');
pt_register('POST','Email');
pt_register('POST','Phone');
pt_register('POST','QuestionsorComments');
$QuestionsorComments=preg_replace("/(\015\012)|(\015)|(\012)/","&nbsp;<br />", $QuestionsorComments);$AttachPictureforQuote=$HTTP_POST_FILES['AttachPictureforQuote'];
pt_register('POST','Howdidyoufindus');
if($FullName=="" || $Email=="" ){
$errors=1;
$error.="<li>You did not enter one or more of the required fields. Please go back and try again.";
}
if($HTTP_POST_FILES['AttachPictureforQuote']['tmp_name']==""){ }
 else if(!is_uploaded_file($HTTP_POST_FILES['AttachPictureforQuote']['tmp_name'])){
$error.="<li>The file, ".$HTTP_POST_FILES['AttachPictureforQuote']['name'].", was not uploaded!";
$errors=1;
}
if(!eregi("^[a-z0-9]+([_\\.-][a-z0-9]+)*" ."@"."([a-z0-9]+([\.-][a-z0-9]+)*)+"."\\.[a-z]{2,}"."$",$Email)){
$error.="<li>Invalid email address entered";
$errors=1;
}
if($errors==1) echo $error;
else{
$image_part = date("h_i_s")."_".$HTTP_POST_FILES['AttachPictureforQuote']['name'];
$image_list[4] = $image_part;
copy($HTTP_POST_FILES['AttachPictureforQuote']['tmp_name'], "files/".$image_part);
$where_form_is="http".($HTTP_SERVER_VARS["HTTPS"]=="on"?"s":"")."://".$SERVER_NAME.strrev(strstr(strrev($PHP_SELF),"/"));
$message="Full Name: ".$FullName."
Email: ".$Email."
Phone: ".$Phone."
Questions or Comments: ".$QuestionsorComments."
Attach Picture for Quote: ".$where_form_is."files/".$image_list[4]."
How did you find us: ".$Howdidyoufindus."
";
$message = stripslashes($message);
mail("webmaster@gmail.com","Form Submitted at your website",$message,"From: phpFormGenerator");
?>


<!-- This is the content of the Thank you page, be careful while changing it -->

<h2>Thank you!</h2>

<table width=50%>
<tr><td>Full Name: </td><td> <?php echo $FullName; ?> </td></tr>
<tr><td>Email: </td><td> <?php echo $Email; ?> </td></tr>
<tr><td>Phone: </td><td> <?php echo $Phone; ?> </td></tr>
<tr><td>Questions or Comments: </td><td> <?php echo $QuestionsorComments; ?> </td></tr>
<tr><td>Attach Picture for Quote: </td><td> <?php echo $AttachPictureforQuote; ?> </td></tr>
<tr><td>How did you find us: </td><td> <?php echo $Howdidyoufindus; ?> </td></tr>
</table>
<!-- Do not change anything below this line -->

<?php 
}
?>
Thanks for your time and any help you can provide!

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

i want to use a variable that is created inside of a function in another function
some thing like this

function show_category ($cats, $index, $level)
{
global $categories_use_categories_top_only;
$categories_use_categories_top_only='no';
.
.
$catRow = '<tr class="scPL-' . $level . '">' . "\n" . '<th colspan="' . PL_COLCOUNT . '">' . $cats[$index]['categories_name'].'<br><i>'. $categories_use_categories_top_only . '</i><br><i>'. $cat_top . '</i></th>' . "\n" . '</tr>' . "\n";
    return $catRow;
}
function show_product ($cats, $cat_index, $prod_index)
{
global $categories_use_categories_top_only;
.
.
  if ($categories_use_categories_top_only=='no' )  { 
     include(DIR_WS_MODULES . 'options.php');
}
.
.


i want to use $categories_use_categories_top_only which its value determined in the first function in the second function as i showed. the first function acts correctly
how can i do this?

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!

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,

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!

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

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.

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

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.