I guess very simple - declaring variables from a user input

[jug]

<div id="lyrLogin" style="position:absolute; left:578px; top:60px; width:166px; height:45px; z-index:20; visibility: visible;">
  <form action="<?php echo $Username; ?>" method="post" name="frmLogIn" id="frmLogIn">
    <p align="left">
      <input name="txtLogInUserName" type="text" id="txtLogInUserName">
    </p>
  <P>
      <input name="imageField" type="image" src="login_button_up.gif" width="105" height="36" border="0" value="Submit" onClick="<?php $Username = $_POST['txtLogInUserName']; ?>" >
    </p>
  </form>
</div>

<img src="final_layout.jpg">

Alright i have a layer which includes an image and a text box - all i want to do is when they click the image whatever they have typed into the text box will be echo-ed out.

I am a total newbie and will allow anyone to laugh at the code but just cant think why this isnt working - makes sense to me

any response will be much appreciated

cheers -- jug

[wildteen88]

Your intentions are completly wrong. PHP runs on the server and generates the output back to the browser. You canno do this:
[code]input name="imageField" type="image"  ... onClick="<?php $Username = $_POST['txtLogInUserName']; ?>" >[/code]
As befoe the HTML is even shown PHP would have created the username var and assign it it the value of whats in $_POST['txtLogInUserName'] variable.

What you'll want to do is something like this:
[code=php:0]<?php

//check that the form has been submitted:
if(isset($_POST['imageField_x']))
{
    // form has been submitted!
 
    // now we set up the username variable:
    $Username = $_POST['txtLogInUserName']

  // Welcome the user:
  echo 'Welcome <i>' . $Username . '</i>, How are you today?<br /><br />';
}
else
{
  // Foprm hasnt been submitted so we'll tell them to:
  echo "Please fill out the form below, click submit when done<br /><br />";
}
?>
<div id="lyrLogin" style="position:absolute; left:578px; top:60px; width:166px; height:45px; z-index:20; visibility: visible;">
  <form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post" name="frmLogIn" id="frmLogIn">
    <p align="left">
      <input name="txtLogInUserName" type="text" id="txtLogInUserName" />
    </p>
    <p>
      <input name="imageField" type="image" src="login_button_up.gif" width="105" height="36" border="0" value="Submit" />
    </p>
  </form>
</div>[/code]


When you first go to the page, [i]Please fill out the form below, click submit when done[/i] will be displayed above the form. But when you fill in the form and press sumit, you'll get a welcome message with the username you put in the txtLogInUserName text field.

[jug]

<?php

if(isset($_POST['imageField_x']))
{
    $Username = $_POST['txtLogInUserName']
    echo 'Welcome <i>' . $Username . '</i>, How are you today?<br /><br />';
}
else
{
echo "Please fill out the form below, click submit when done<br /><br />";
}
?>
 
<div id="lyrLogin" style="position:absolute; left:578px; top:60px; width:166px; height:45px; z-index:20; visibility: visible;">
  <form action="<?php echo $_SERVER['PHP_SELF']; ?>" " method="post" name="frmLogIn" id="frmLogIn">
    <p align="left">
      <input name="txtLogInUserName" type="text" id="txtLogInUserName">
    </p>
  <P>
      <input name="imageField" type="image" src="login_button_up.gif" width="105" height="36" border="0" value="Submit" >
    </p>
  </form>
</div>

<img src="final_layout.jpg">




This is the new combined code - but still i get an error:

Parse error: parse error in c:\program files\easyphp1-8\www\test.php on line 7

= which is = echo 'Welcome <i>' . $Username . '</i>, How are you today?

Obviously i dont know whats up as its practically your code - please enlighten me of the problem

many thanks in advance - jug

ps also what does <i> </i> and ="<?php echo $_SERVER['PHP_SELF']; ?>" just for future reference

[wildteen88]

Oops missed out the semi-colon after this:
Change:
$Username = $_POST['txtLogInUserName']
to this:
$Username = $_POST['txtLogInUserName'];