PHP - Variable Doesn't Show Up In Url

Hello all,

I can't get the data store in the $ART_ID variable to pass into the database.  The original $Artisan variable is set up like this: 1. Artisan Name.  So the explode is taking just the number. If I put an echo after the explode and the $ART_ID variable it outputs the correct information but it doesn't store in the database as that data.  The query is in correct order too.
Thanks in advance. 



$CType_Type = $_REQUEST["CTYPE_Type"];
$Artisan = $_REQUEST['Artisan'];
$Quantity = $_POST['Quantity'];
$HAnswer1 = $_POST["HAnswer1"];
$HAnswer2 = $_POST["HAnswer2"];
$HAnswer3 = $_POST["HAnswer3"];
$HAnswer4 = $_POST["HAnswer4"];

$break = explode(".", $Artisan);
$ART_ID = $break[0];
   
if(!$Quantity)
{
   die('Quantity field is empty. Please enter the quantity of handicrafts made.');
}
else
{
         
   $ctypeQuery = mysql_query("SELECT CTYPE_ID FROM CraftType WHERE CTYPE_Type = '".$CType_Type."'");
         
   while($row = mysql_fetch_array($ctypeQuery))
   {
      $CTYPE_ID = $row["CTYPE_ID"];
               
      $sql = ("INSERT INTO Handicraft VALUES (`HANDI_ID`, '".$Quantity."', 'NULL', '".$CTYPE_ID."', '".$ART_ID."', 'NULL', '1')");
            
      if(!mysql_query($sql))
      {
         die('Error inserting Handicraft Type into table: ' . mysql_error());
      }
      else
                { -----data in this section doesn't affect the rest of the code---- }   
       }
}

How would I write this in php?:

Code: [Select]
If $url does not begin with 'http://'
$url = ''
I want to assign nothing to the variable if it doesn't begin with http://

Thanks! 

When I echo out my variable in the php file it works fine but when I put the variable in a table cell it doesn't echo out. 
 

Hello,

I am writing my first php script (of course) and it works except it only shows 4 variables.
Example I have it to send me:
mail($myemail, $fullname, $email, $message, $phone);
But it will only send me the first four no matter what order they are in. It's driving me crazy. I can't think or find any reason why it would do this.

If I change it to
mail($myemail, $fullname, $email, $phone, $message);
It will send me everything but $message
if I change it to
mail($myemail, $fullname, $message, $email, $phone );
It will send me everything but $phone

Does anyone know why or what I need to change? I apologize if this sounds like a dumb question. I am very very new to php.

Thank you for any guidance you can give. Here is the code:
HTML:

<form action="Contact.php" method="post" id="contact" name="contact" style="margin:0px";>
<h2>Contact Form:</h2>
<p>Fields marked (<span style="color:#F00;">*</span>) are required.</p>

<p>
<label for="text_fullname">Full Name<span style="color:#F00;">*</span>:</label>
<input name="fullname" type="text" class="input" id="fullname" tabindex="1" />

<label for="text_email">Email<span style="color:#F00;">*</span>:</label>
<input name="email" type="text" class="input" id="email" tabindex="2" />

<label for="text_phone">Phone:</label>
<input name="phone" type="text" class="input" id="phone" tabindex="3" />

<label for="text_comments">Message/Comment:</label>
<textarea name="message" cols="23" rows="3" class="input" id="message" tabindex="4"></textarea>
<br>

<div style="padding-left:200px;">
<button type="submit" id="send" name="send">Send!</button> &nbsp;&nbsp; <button type="reset">Reset</button>
</div>

</form>
---------------------------------------
PHP:
<?php
/* Set e-mail recipient */
$myemail = "tkowebworks@gmail.com";

$fullname = $_POST['fullname'] ;
$email = $_POST['email'] ;
$phone = $_POST['phone'] ;
$message = $_POST['message'] ;

mail($myemail, $fullname, $email, $message, $phone);
header( "Location: http://www.ppm-oc.com/thankyou.html" );
?>
----------------------------------

Thank you

hi im having trouble with my login script i keep getting the errors

Deprecated: Function session_register() is deprecated in C:\wamp\www\login.php on line 53

Deprecated: Function session_register() is deprecated in C:\wamp\www\login.php on line 57

Deprecated: Function session_register() is deprecated in C:\wamp\www\login.php on line 61

i just wanted to knw wot they mean the page styll runs ok, is there any way i can make it work

Code: [Select]
$sql = mysql_query("SELECT * FROM myMembers WHERE email='$email' AND password='$pass' AND email_activated='1'");
$login_check = mysql_num_rows($sql);

if($login_check > 0){

    while ($row = mysql_fetch_array($sql)){

        $id = $row["id"];   
        session_register('id');
        $_SESSION['id'] = $id;
       
    $firstname = $row["firstname"];   
        session_register('firstname');
        $_SESSION['firstname'] = $firstname;
       
    $email = $row["email"];   
        session_register('email');
        $_SESSION['email'] = $email;
         
        mysql_query("UPDATE myMembers SET last_log_date=now() WHERE id='$id'");
   
        $my_msg="Hello $firstname, you have succesfully logged in";
        //print "return_msg=$my_msg&id=$id&firstname=$firstname";
       
    } // close while


} else{

$my_msg="Error Your Password Or Email Did Not Match";
 // print "return_msg=$my_msg";
  exit();
}

}
also im having trouble with my my_msg variable it does not seem to show up i keep gettin error

if ($count==1){
				
						
				header("Location:store.php");				
				
								}

very simple I have issolated it and it doesn't redirect maybe u can see where my mistake is

I want to show a cookie of a referral's username on a sign up page. The link is like this, www.mysite.com/signup?ref=johnsmith.

The cookie doesn't show if I go to that url page. But it does show up once I reload the page.  So I'm wondering if it's possible to show the cookie the first time around, instead of reloading the page?

Here is my code.

// This is in the header
$url_ref_name = (!empty($_GET['ref']) ? $_GET['ref'] : null);

if(!empty($url_ref_name)) {

  $number_of_days = 365;
  $date_of_expiry = time() + 60 * 60 * 24 * $number_of_days;
  setcookie( "ref", $url_ref_name, $date_of_expiry,"/");

} else if(empty($url_ref_name)) {
  if(isset($_COOKIE['ref'])) {
    $user_cookie = $_COOKIE['ref'];
  }
} else {}

// This is for the sign up form
if(isset($_COOKIE['ref'])) {
  $user_cookie = $_COOKIE['ref'];
  ?>
  <fieldset>
    <label>Referred By</label>
    <div id="ref-one"><span><?php if(!empty($user_cookie)){echo $user_cookie;} ?></span></div>
    <input type="hidden" name="ref" value="<?php if(!empty($user_cookie)){echo $user_cookie;} ?>" maxlength="20" placeholder="Referrer's username" readonly onfocus="this.removeAttribute('readonly');" />
  </fieldset>
  <?php
}

 

I am displaying rows from a database onto a page using:

while($row=mysql_fetch_array($query)){

echo $row['name'];

}

I need to figure out how to limit the rows shown to the page to 100. And if there are 100 rows on the page, a link will be displayed at the bottom, that says "Next 100". Then this will display the next 100 rows.
Can you give an example how to do this please?

Thanks

Hi,

I would like to do the following but not sure how.

If the you/user is on index.php of http://www.domain.com/ show one page
If not show another

How would I do this?

Thanks

Hi ,

I know my code sucks but i'm learning fast!! I'm trying to show a form if the qty value in a database == 10 or a different form if the value ==20.  I tried but failed.  Any help really appreciated.

Code: [Select]

<?php require_once('Connections/book.php'); ?>
<?php
$colname_cardpayment = "-1";
if (isset($_GET['orderid'])) {
  $colname_cardpayment = (get_magic_quotes_gpc()) ? $_GET['orderid'] : addslashes($_GET['orderid']);
}
mysql_select_db($database_book, $book);
$query_cardpayment = sprintf("SELECT * FROM cards WHERE orderid = '%s' ORDER BY qty ASC", $colname_cardpayment);
$cardpayment = mysql_query($query_cardpayment, $book) or die(mysql_error());
$row_cardpayment = mysql_fetch_assoc($cardpayment);
$totalRows_cardpayment = mysql_num_rows($cardpayment);

 // Database connect
$con = mysql_connect("mysql1.myhost.ie","admin_book","root123");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("book_test", $con);


//Parse Values from Coupon.php Form
$orderid = mysql_real_escape_string(trim($_POST['orderid']));
$name  = mysql_real_escape_string(trim($_POST['name']));
$surname  = mysql_real_escape_string(trim($_POST['surname']));
$add1 = mysql_real_escape_string(trim($_POST['add1']));
$add2 = mysql_real_escape_string(trim($_POST['add2']));
$town = mysql_real_escape_string(trim($_POST['town']));
$county = mysql_real_escape_string(trim($_POST['county']));
$postcode = mysql_real_escape_string(trim($_POST['postcode']));
$phone = mysql_real_escape_string(trim($_POST['phone']));
$email = mysql_real_escape_string(trim($_POST['email']));
$letterstyle = mysql_real_escape_string(trim($_POST['letterstyle']));


$sql="INSERT INTO custdetails (orderid, name, surname, add1, add2, town, county, postcode, phone, email, letterstyle)
VALUES
('$orderid','$name','$surname','$add1','$add2','$town','$county','$postcode','phone','$email','$letterstyle')";

if (!mysql_query($sql))
  {
  die('Error: ' . mysql_error());
  }
?>

<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8" />
<title>Digital Scribe Books</title>
<link href="style.css" rel="stylesheet" type="text/css" />

<script type="text/javascript">
function MM_preloadImages() { //v3.0
  var d=document; if(d.images){ if(!d.MM_p) d.MM_p=new Array();
    var i,j=d.MM_p.length,a=MM_preloadImages.arguments; for(i=0; i<a.length; i++)
    if (a[i].indexOf("#")!=0){ d.MM_p[j]=new Image; d.MM_p[j++].src=a[i];}}
}
function MM_swapImgRestore() { //v3.0
  var i,x,a=document.MM_sr; for(i=0;a&&i<a.length&&(x=a[i])&&x.oSrc;i++) x.src=x.oSrc;
}
function MM_findObj(n, d) { //v4.01
  var p,i,x;  if(!d) d=document; if((p=n.indexOf("?"))>0&&parent.frames.length) {
    d=parent.frames[n.substring(p+1)].document; n=n.substring(0,p);}
  if(!(x=d[n])&&d.all) x=d.all[n]; for (i=0;!x&&i<d.forms.length;i++) x=d.forms[i][n];
  for(i=0;!x&&d.layers&&i<d.layers.length;i++) x=MM_findObj(n,d.layers[i].document);
  if(!x && d.getElementById) x=d.getElementById(n); return x;
}

function MM_swapImage() { //v3.0
  var i,j=0,x,a=MM_swapImage.arguments; document.MM_sr=new Array; for(i=0;i<(a.length-2);i+=3)
   if ((x=MM_findObj(a[i]))!=null){document.MM_sr[j++]=x; if(!x.oSrc) x.oSrc=x.src; x.src=a[i+2];}
}
</script>


</head>

<body onload="MM_preloadImages('images/buttons/home_over.png','images/buttons/books_over.png','images/buttons/cards_over.png','images/buttons/letters_over.png')">


<div id="snow">
<div id="wrapper">


<div id="header">

<div id="logo"><img src="images/digital_scripe.png" width="218" height="91" /></div>


<div id="menu"><a href="index.php" onmouseout="MM_swapImgRestore()" onmouseover="MM_swapImage('Home','','images/buttons/home_over.png',1)"><img src="images/buttons/home_act.png" name="Home" width="131" height="132" border="0" id="Home" /></a><a href="books.php" onmouseout="MM_swapImgRestore()" onmouseover="MM_swapImage('Books','','images/buttons/books_over.png',1)"><img src="images/buttons/books_act.png" name="Books" width="131" height="132" border="0" id="Books" /></a><a href="cards.php" onmouseout="MM_swapImgRestore()" onmouseover="MM_swapImage('Cards','','images/buttons/cards_over.png',1)"><img src="images/buttons/cards_act.png" name="Cards" width="131" height="132" border="0" id="Cards" /></a><a href="letters.php" onmouseout="MM_swapImgRestore()" onmouseover="MM_swapImage('Letters','','images/buttons/letters_over.png',1)"><img src="images/buttons/letters_act.png" name="Letters" width="131" height="132" border="0" id="Letters" /></a></div>
 </div>
 
 
 
 
<div id="content">



<?php 
echo 'Order ID is :  '. $orderid . '.<br />';

if ($row2['qty'] == 10)
    echo "<div>

<form action="https://www.paypal.com/cgi-bin/webscr" method="post">
<input type="hidden" name="cmd" value="_xclick">
<input type="hidden" name="business" value="accounts@agraphics.ie">
<input type="hidden" name="lc" value="IE">
<input type="hidden" name="item_name" value="10 Christmas Cards">
<input type="hidden" name="item_number" value="<? echo $orderid; ?>">
<input type="hidden" name="amount" value="12.99">
<input type="hidden" name="currency_code" value="EUR">
<input type="hidden" name="button_subtype" value="services">
<input type="hidden" name="shipping" value="2.99">
<input type="hidden" name="return" value="http://www.digitalscribe/thanks.php">
<input type="hidden" name="bn" value="PP-BuyNowBF:btn_buynowCC_LG.gif:NonHosted">
<input type="image" src="https://www.paypalobjects.com/en_US/i/btn/btn_buynowCC_LG.gif" border="0" name="submit" alt="PayPal - The safer, easier way to pay online!">
<img alt="" border="0" src="https://www.paypalobjects.com/en_US/i/scr/pixel.gif" width="1" height="1">
</form>
</div>";

if ($row2['qty'] == 20)
    echo "<div>

<form action="https://www.paypal.com/cgi-bin/webscr" method="post">
<input type="hidden" name="cmd" value="_xclick">
<input type="hidden" name="business" value="accounts@agraphics.ie">
<input type="hidden" name="lc" value="IE">
<input type="hidden" name="item_name" value="20 Christmas Cards">
<input type="hidden" name="item_number" value="<? echo $orderid; ?>">
<input type="hidden" name="amount" value="21.99">
<input type="hidden" name="currency_code" value="EUR">
<input type="hidden" name="button_subtype" value="services">
<input type="hidden" name="shipping" value="2.99">
<input type="hidden" name="return" value="http://www.digitalscribe/thanks.php">
<input type="hidden" name="bn" value="PP-BuyNowBF:btn_buynowCC_LG.gif:NonHosted">
<input type="image" src="https://www.paypalobjects.com/en_US/i/btn/btn_buynowCC_LG.gif" border="0" name="submit" alt="PayPal - The safer, easier way to pay online!">
<img alt="" border="0" src="https://www.paypalobjects.com/en_US/i/scr/pixel.gif" width="1" height="1">
</form>
</div>";


?>

</div>


<div id="footer" class="clear"><div id="sign"><div id="sign_text">Personalised<br />
Books</div>
</div></div>

</div></div>






</body>
</html>
<?php
mysql_free_result($cardpayment);
?>

I am trying to allow the user to update a variable he chooses by radio buttons, which they will then input text into a box, and submit, to change some attributes. I really need some help here.  It works just fine until I add the second layer of variables on top of it, and I can't find the answer to this question anywhere.

<?PHP require('connect.php'); ?>

<form action ='' method='post'> <select name="id">

<?php
$extract = mysql_query("SELECT * FROM cars");
   
   while($row=mysql_fetch_assoc($extract)){
   $id = $row['id'];
   $make= $row['make'];
   $model= $row['model'];
   $year= $row['year'];
   $color= $row['color'];

echo "<option value=$id>$color $year $make $model</option>
";}?>

</select>

Which attribute would you like to change?<br />
<input type="radio" name="getchanged" value="make"/>Make<br />
<input type="radio" name="getchanged" value="model"/>Model<br />
<input type="radio" name="getchanged" value="year" />Year<br />
<input type="radio" name="getchanged" value="color"   />Color<br /><br />

   <br /><input type='text' value='' name='tochange'>
   <input type='submit' value='Change' name='submit'>

</form>

//This is where I need help...

<?PHP
if(isset($_POST['submit'])&&($_POST['tochange'])){
mysql_query("
   UPDATE cars
   SET '$_POST[getchanged]'='$_POST[tochange]'
   where id = '$_POST[id]'
         ");}?>

Hello all. I am very new to PHP, and I am not sure where to look or what I'm looking for in my current assignment.

My task is to take in two numbers between 0-100. Once I take in that number, it should state beside it "The __ was accepted."
The program should not accept any numbers greater than 100 or any characters.

Once I do this, I must take a second number and do a similar thing.

Finally, I must have a statement show up at the bottom stating which number is greater.

Essentially, I need help in determining what I should use to place parameters, and how I can keep the program from echo ing any statement until input has been taken and tested for parameters.

Any help you can provide will be greatly appreciated!

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();
?>

 

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