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  #1  
Old June 13th, 2002, 12:18 AM
maryer maryer is offline
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JS pup-up with php

Hello all!
I need printable version of form. Now when i submit form datas go to the mysql database and on the screen comes "thanks for submitting......". On this screen i need to have "Printables version of submitted form" link which has to open JS pup-up window with latest filled datas. Now how can i transfer datas in php to this new window. i tryed with include("form.php") but it didnt work.
I would be very appreciate if you help me for this case.

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  #2  
Old June 13th, 2002, 04:47 AM
Ben Rowe
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sorry if i have to re-explain your situation, your english is kinda hard to read, basically, you upload some data into a database, then on the next page, you want to give the user the option to print out the data they just submitted??

its quite easy,

when you submit the data, after you run the insert query run something like this

$id=mysql_insert_id();

then what you need to do is print out something like this

printf ("<a href='print.php?id=%s' target='_blank'>Printable version</a>\n", $id);

then in the print page query your database where id=$id

simple

hope that helps

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Old June 13th, 2002, 04:52 AM
mytch mytch is offline
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ok,
well what you need to do is use the JavaScript window.open() function to open the new window on the page where the form has been submitted, something like this:

<script language="JavaScript">

function openIt()
{
window.open("form.php?<?php echo $qs; ?>");
}

</script>

<body onLoad="openIt()">

As you can see, i've included a PHP variable called $qs. You should make $qs like this:

<?php

$qs = "";

foreach($_POST as $key=>$val)
$qs .= "$key=$val&";

?>

This will open something like

form.php?name=Mitchell&age=19&sex=male&

Then in form.php, just set the values of the form fields to the $_GET array, like this:

<input type="text' name="MyName" value="<?php echo$_GET["name"]; ?>">

etc...

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