PHP - Passing An Ampersand In A Url

Code: [Select]
<?php 
$parentesco .=' 

<ul><li><a href="children.php?category_id='.$idc.'&name= '. $name2 . '">'. $name. ' </a></li></ul>'; ?>
in the url string $idc is passing but $name2 is not. $name2 is pickup from a query, I echo that value everywhere in the code and it echo its value but is not passing so how.

Help

I have some function or method.  Is there a better design patter to implement this?

function myFunction($a=null,$b=null,$c=null,$d=null,$e=null,$f=null,$g=null,$h=null)
{
   //Do a bunch of stuff
}

Maybe the second function?

function myNewFunction($data=array())
{
   $data=array_merge(array('a'=>null,'b'=>null,'c'=>null,'d'=>null,'e'=>null,'f'=>null,'g'=>null,'h'=>null),$array);
   //Do a bunch of stuff
}

Please provide decision making factors why you would use one approach over the other.

Ok quick run down... I need to pass a variable from one page[form] to another while skipping a page that is in the middle.

The middle page is paypal... after they pay, they are being redirected to a thank you for registering with a email also being sent.
But I need their email from the first page[form].

so what i need to figure out or help is:

1. First page is a form, they enter name, last name, address, email and so on
2. They are redirected to paypal to pay for a service.
3. Are directed back to my site for a email to be sent to them, but I need their email from the first page.

I have the following code that outputs a url. It searches my MySQL database for the name of a given item within a theme:


$relative = 'http://www.domain.com/wp-content/';
$url1= $relative . 'themes/arjuna-x/itemspec.php?type=';
$url2='&item=';

<a class="entry-thumbnails-link" href="<?php echo $url1.'Themes'.$url2.$rowThemes['Theme']; ?>">

The output is as follows:
http://www.domain.com/wp-content/themes/arjuna-x/itemspec.php?type=Themes&item=(insert variable here)

All is good up to this point.

My problem however, is that some items (in this case: theme names) have apostrophes in them. For example: April Fool's.

This results in breaking my php code.

How can I get php to call these variables with apostrophes so that they don't break my php?

Hi guys   I would like to get whole content of html file after I submitted a password. The problem is that the my code doesn't access the value for the file. I tried to create a session but i doesn't work. What other option do I have to get the content of the html file?

session_start();
  	$selected_file = $_POST['radio1'];
  	// get the filename of the file
    $fileinfo = pathinfo($selected_file);
    $filename = $fileinfo['dirname'] . DIRECTORY_SEPARATOR . $fileinfo['filename'];
    $password = 'code';
    $lines = file("$filename.html");
    $_SESSION['selectedfile'] = $selected_file;
    $_SESSION['file'] =  $filename;
    $_SESSION['Scipt'] =  ("$filename.html");
    $_SESSION['Scipttext'] =  $lines;
  	$_SESSION['file2'] = $fileinfo;
	

if (isset($_POST['submitradio']))
{ 
 	 echo '<div class="imageselected">';
	 echo '<img src="'.$_SESSION['selectedfile'].'" /></br>'.PHP_EOL;
	 echo '</div>';
    // check to see if a html file named the same also exists
    if(file_exists($_SESSION['Scipt']))
    {
    	
    	echo "<form action='test_result.php' method='post'>";
    	echo '<div class="Password">';
    	echo 'Type in password to view full Script';
    	
		echo "<label><div class=\"Input\"><input type='password' name='passIT' value='passit'/></div>";
		echo "<input type='submit' name='submitPasswordIT' value='Submit Password'/></div>";
		echo '</div>';
		echo "$filename.html shares the same name as $selected_file";
		

		for($x = 1;$x<=15;$x++) {
		header( "Content-Type: file/html" );
		$lines = ($_SESSION['Scipttext']);
		$new = strip_tags($lines); 
		echo $lines[rand(0, count($lines)-1)]."<br>";
        }
    	// end of forloop
 
		}
  		// end of check
		  // start Sorrytext
		 
    		else
    		{
    		 echo '<div class="NoScript">';
        		echo "Nothing available at the moment.";
        		echo '</div>';
    		}
    		
		 // end Sorrytext
}
// End of submitradio
if($_POST['submitPasswordIT']){
if ($_POST['passIT']== $password ){
echo "You entered correct password";
readfile($_SESSION['Scipt']);
}
else{
echo "You entered wrong password";
}
}
echo '</form>';

I would be grateful for help.

I cannot figure this out and I know I'm just looking past the obvious...

I am creating a simple "you are now leaving this website" page, Everything is setup, and all urls pointing to http://mysite.com?exit=http://externalsite.com, It should redirect to my exit.php which will redirect to the external site after 5 seconds.

The problem is, the "http://" of the external site is causing a 403 forbidden, and even without it, anything with multiple &s isn't passing through.

I cannot figure out how to get it to pass the full url to the exit.php...

I'm creating this URL to pass variables to use from one Page to another:   http://eg10basketbal...Borgman&team=16     My WP site is seeing the first variable but not the second.  After some searching I've found this suggestion, but it's not working for me:

function add_query_vars_filter( $vars ) {

	$vars[] = "nameFull";
	$vars[] = "team";

	return $vars;
}
add_filter( 'query_vars', 'add_query_vars_filter' );

I tried putting this in the wp-includes/functions.php file of the, but ended up with a fatal error.  I have since added it to my theme's functions.php file. There isn't an error, but it doesn't yield recognition of my second variable.     

//i've been looking at this same code for two whole days now and ive changed it lots of times but it can only get worse
//i can't pass the value of the variable $today which has a date datatype
//this is the part where i get the value of $today
<?php
$query="SELECT DISTINCT today FROM arrastre order by today desc";
$result=mysql_query($query);
while($row=mysql_fetch_object($result)){
$today=$row->today;
echo "<option value=$today>$today</option>";
}
?>
//here's is the part where i use $today to get $tcl
//except for a major problem that when the while loop returns more than one $tcl, i get more than one button, but that's not the problem because that is how it's supposed to work. The problem is that when there are more than one buttons, only the last button would work, the only button that displays values
<?php
if(isset($_POST['godate'])){
$today=$_POST['today'];
$query="SELECT DISTINCT tcl FROM arrastre WHERE today='$today' ORDER BY tcl ASC";
$result=mysql_query($query);
while($row=mysql_fetch_array($result)){
$tcl=$row['tcl'];
$today=$row['today'];
?>
<input type="hidden" name="tcl" value="<?php echo $tcl;?>">
<input type=submit name="tcl" value="<?php echo $tcl; ?>">
<?php } }
?>

//thsi is another problem
if(isset($_POST['tcl'])){
$tcl=$_POST['tcl'];
$today=$_POST['today'];
$query = "select * from arrastre where tcl=$tcl and today='$today' order by tcl asc";
$result = mysql_query($query);
$totalcollections = 0;
while($row=mysql_fetch_array($result)){
$orno=$row['orno'];
$billnmbr=$row['billnmbr'];
$payor=$row['payor'];
$arrastre=$row['arrastre'];
$wharfage=$row['wharfage'];
$total=$row['total'];
$today=$row['today'];
//*i deleted the display part to make this shorter a little
}}
//the $today variable is empty in this post and i have no idea how to successfully pass this value
so i could use it in my $query
//thanks for your time

$seat_s is a $var that can have multiple values depending on a form in a previous page.

this code works fine as is, but i need to pass the array into a 3rd page. i have no idea how to do this from the code below.

ive tried passing the array straight into a _post then _get on the next page but that only holds the 1st value and not all of them.

suggestions please.... thanks

<?php

$seat_s = $_POST['seats'];

foreach($_POST['seats'] AS $seat) {

$rowId = substr($seat, 0, 1);

$columnId = substr($seat, 1);

echo $rowId . $columnId . ", ";

}

?>

Can anyone explain to me why this system isnt working? I have no clue why it shouldnt be.

The pin number is passed from a form to this script, which then determines if a users pin is correct and if so, forwards them onto the relevant page. Otherwise, it returns them to the login.


<?php 
session_start();

$pin=$_POST['pin'];

if ($pin='7221') {
$_SESSION['pin']='7221';
$_SESSION['name']='Keith D Whelpton';
header('Location: phonesystem.php');
}
elseif ($pin='3585') {
$_SESSION['pin']='3585';
$_SESSION['name']='Luke S Barnes';
header('Location: phonesystem.php');
}
else {
header('Location: phonelogin.php');
}
?>

The code below is a form the gets information from a database and post it into the appropriate field. I am having a problem with the while loop. All the variables before the while loop work just great and echo out how they should. But everything after the while loop does not work. I know it has something to do with the while loop because when I take the while loop out everything works as it should. Can anyone please help me with this?

  <? $id = $_POST["id"];
  $therapy = $_POST["therapy"];
  db_connect();
  $query=mysql_query("select * from student where student_id = $id") or die ("invalid query");
  $row=mysql_fetch_array($query);

print "<table width='330' border=0 align='center'>
  <form method='POST' action='table.php'>
    <td>*First Name:</td>
    <td>*Last Name:</td>
    <tr>";
?>
      <td width="157"><input size="23" name="first_name" value="<?php echo "$row[first_name]"; ?>" /></td>
      <td width="138"><input type="text" size="23" name="surname" value="<?php echo "$row[surname]"; ?>" /></td>
    </tr>
    <tr>
    <td>Level:</td>
     <td>Therapy:</td>
      
    </tr>
    <tr>
      <td>
      <select name="program_level">
        <option <?php if ($row[program_level]=='1') echo "selected=yes";  ?> value="1">O&A</option>
        <option <?php if ($row[program_level]=='2') echo "selected=yes";  ?> value="2">Student</option>
        <option <?php if ($row[program_level]=='3') echo "selected=yes";  ?> value="3">Supervisor</option>
        <option <?php if ($row[program_level]=='4') echo "selected=yes";  ?> value="4">Manager</option>
        <option <?php if ($row[program_level]=='5') echo "selected=yes";  ?> value="5">Director</option>
        <option <?php if ($row[program_level]=='6') echo "selected=yes";  ?> value="6">Graduate</option>
        </select>
      </td>
            <td>


<!-- Auto drop down for therapist names-->  
         
<? $query2=mysql_query("select * from therapy_info;"); ?>

<select name='therapy'>

<? while ($row=mysql_fetch_assoc($query2)){ ?>

<option value="<? echo "$row[therapy_name]"; ?>"> <? echo "$row[therapy_name]"; } ?> </option>

</select>
 
 <!-- End Auto drop down for therapist names-->



</td> 
   </tr>
    <tr>
      <td>Active:<?php echo "$row[active]"; ?></td>
    </tr>
    <tr>
      <td><select name='active' value='$row[active]'>
        <option <?php if ($row[active]=='1') echo "selected=yes";  ?> value='1'>Active</option>
        <option <?php if ($row[active]=='0') echo "selected=yes";  ?> value='0'>Inactive</option>
        </select>
        </td>
    </tr>
    <?php 
    print "<tr>
      <td><input type='submit' value='Change' name='button' /></td>
      <input type='hidden' name='id' value='$id' />";
  ?>
      <input type='hidden' name='$hardcode_session' />
      <?
 
      
    print "</tr>
  </form>
</table>";

here is my code:
Code: [Select]
$query2 = "SELECT * FROM members_copy WHERE RSUSER = ".$_SESSION["rsUser"]."";
$result2 = mysql_query($query2);
$UserID = $result2['USERID'];
echo $UserID;


When I echo out UserID it doesn't show a number?!

Please help?!

how can i pass more then one value via href 

with this kind of method

echo "<a href=\"submit_docs.php?prop_id=".$prop_id."\">Click </a>";


tnx

This is driving me MAD!

The Variable gets passed through the URL but when I try and echo the result I gte NOTHING.

First Page.

Code: [Select]
<?php include'config.php';


$result = mysql_query("SELECT * FROM carbontrust");

while($entry = mysql_fetch_array($result))


Print "$entry[contact] - $entry[company] :: <a href='edit.php?id=$entry[id]'>Edit</a><br>";

exit;

?>


Second Page. (Where I want the Variable passed to)
Code: [Select]
<?php include'config.php';


$id = $_GET['id'];
echo "ID: $id";


}


?>

Its probably just a stupid error but iv looked through a load of tutorials and this should be correct. :/

Trying to set up an error message when someone tries to upload a file without the approved .ext.
I have it working so it won't up load but I am trying to get an error to print out.

I was thinking that I could do something like this

if the ext are wrong set
$errorMsg1 == 1;
and then the page will refresh and I would pass that variable to echo out

if ($errorMsg1 == 1){
echo "Invalid"; }
else {

but it isn't passing can anyone help me with this? Tyring to keep it simple

this is the code to select an image

else{
 
  $result = mysql_query("SELECT * FROM photos WHERE userID LIKE '$clientID'");
  while ($r=mysql_fetch_array($result))
  {
     $photo_1=$r['photo_1'];
     $photo_2=$r['photo_2'];
     $photo_3=$r['photo_3'];
     $photo_4=$r['photo_4'];
     $photo_5=$r['photo_5'];

   echo " 
    <form enctype='multipart/form-data' action='' method='POST'>
      <input type='hidden' name='MAX_FILE_SIZE' value='500000' />
      <div id='imageTop'>Image &#38;#35;1</div>
        ";
        if ($errorMsg1 == 1){
       echo "Invalid"; }
       echo "

       <div id='imageBottom'>
         <span class='image'>";
           if (empty($photo_1)) { 
            echo " <img src='uploads/noPhoto.gif' width='75' height='75' class='zip'> ";
              }
            else {
             echo "<a href='uploads/$photo_1' ><img src='uploads/$photo_1' width='75' height='75' class='zip'></a> ";
              }
            echo " 
         </span>
         <span class='action'>
            &nbsp;<input type='file' name='photo_1' class='zip'><br><br>
            <input type='checkbox' name='delete_1'>Select to Delete image
          </span>
       </div>

}


this is the code of what to do with that image

  if (isset($_POST['delete_1']))  {
    $query = "UPDATE photos SET photo_1='' WHERE userID='$clientID'";
    $result = mysql_query($query) or die(mysql_error());
    echo " <div id='aboutUpdate'><img src='img/loader.gif'> Information is updating</div> ";
    echo "<meta http-equiv=refresh content=\"0; URL=photos.php\">";
  }
  else if ($one != NULL) {
    $extension = strrchr($_FILES['photo_1']['name'],'.');
    $extension = strtolower($extension);

    if($extension != '.jpg' && $extension != '.gif' && $extension != '.png' && $extension != '.bmp' ){
      $errorMsg1 == 1;
      echo "<meta http-equiv=refresh content=\"0; URL=photos.php\">";
    }
    else {
    $photoNumber="_1";
    $finalName="$clientID$photoNumber";
    $save_path = "uploads/";
    $target_path = $save_path . basename( $_FILES['photo_1']['name']); 
    $NewPhotoName = $finalName; 
    $withExt = $NewPhotoName . $extension;
    $filename = $save_path . $NewPhotoName . $extension; 
    if(move_uploaded_file($_FILES['photo_1']['tmp_name'], $filename)) {
      $query = "UPDATE photos SET photo_1='$withExt' WHERE userID='$clientID'";
      $result = mysql_query($query) or die(mysql_error());
        echo " <div id='aboutUpdate'><img src='img/loader.gif'> Information is updating</div> ";
        echo "<meta http-equiv=refresh content=\"0; URL=photos.php\">";
      }
    }
    else  {