how to display variables within using echo command within html code

this is part of my code

question?

why the variable $name and $address does not display on the html page
instead of real values I do get variable names
if ($modify == 'print') {
echo '


<table width="100%" border="0" cellpadding="0" cellspacing="0">
<tr>
<td valign="top">Name:</td>
<td><font size="1"> <strong>
<input name="name2" type="text" value="$name" size="100">
<br>
<font size="2">note: name field cannot be
changed</font></strong></font></td>
</tr>
<tr>
<td>Address:</td>
<td><input name="address2" type="text" id="address" value="$address"
size="100"></td>
</tr>
</table>


';
};
//Close connection with MySQL
MySQL_close()
?>
</td>
</tr>
</table>

Hi,

Bartosz Wegrzyn wrote:
> this is part of my code
>
> question?
>
> why the variable $name and $address does not display on the html page
> instead of real values I do get variable names


> <td><input name="address2" type="text" id="address" value="<?php echo $address ?>"

you may also write, with the same result:
value="<?php echo '$address' ?>"
or
value="<?php echo "$address" ?>"

remember html cannot display php values if you do not explicit the
markup <?php ?> for php processor.

php is a server side scripting language, html is interpreted client side
by browsers...

Ciao Gianfranzo

--
--
ICQ#: 21258924
Nick Gianfranzo
Home page http://www.gianfranzo.it
Gianfranzo è contro la guerra
--

I found out that I could use require() or include() functions
What do u think
"Gianmarco Vanni" <gianmarcovanni@tin.it> wrote in message
news:3F1D1804.3000904@tin.it...
> Hi,
>
> Bartosz Wegrzyn wrote:
> > this is part of my code
> >
> > question?
> >
> > why the variable $name and $address does not display on the html page
> > instead of real values I do get variable names
>
>
> > <td><input name="address2" type="text" id="address" value="<?php
echo $address ?>"
>
> you may also write, with the same result:
> value="<?php echo '$address' ?>"
> or
> value="<?php echo "$address" ?>"
>
> remember html cannot display php values if you do not explicit the
> markup <?php ?> for php processor.
>
> php is a server side scripting language, html is interpreted client side
> by browsers...
>
> Ciao Gianfranzo
>
> --
> --
> ICQ#: 21258924
> Nick Gianfranzo
> Home page http://www.gianfranzo.it
> Gianfranzo è contro la guerra
> --
>

this is my new code
the problem is that varialebles are not past to print.php

what am I doing wrong ???

<table width="100%" border="0" cellpadding="0" cellspacing="0">
<tr>
<td><em><font size="1" face="Verdana, Arial, Helvetica,
sans-serif"><strong>LEXON
- Mysql Database</strong></font></em> </td>
</tr>
<tr>
<td align="left"><p><strong>Modify - Delete - Print (Customer &amp;
Workorder)<br>
<br>
</strong></p></td>
</tr>
<tr>
<td align="left">
<?php
$host = "localhost";
$login_name = "xxx";
$password = "xxx";
//Connecting to MYSQL
MySQL_connect("$host","$login_name","$password");

//Select the database we want to use
MySQL_select_db("lexon") or die("Could not select database");

//Assign contents of form to variables
//Customer Data
$name = $_POST['name'];
$address = $_POST['address'];
$city = $_POST['city'];
$state = $_POST['state'];
$zipcode = $_POST['zipcode'];
$phone = $_POST['phone'];
$ss = $_POST['ss'];
$cc = $_POST['cc'];
$exp = $_POST['exp'];
$notes = $_POST['notes'];
//Wordorder Data
$date = $_POST['date'];
$time = $_POST['time'];
$service = $_POST['service'];
$due = $_POST['due'];
$special = $_POST['special'];
$extrawork = $_POST['extrawork'];
$adddue= $_POST['adddue'];
$company= $_POST['company'];
$installer= $_POST['installer'];
$total = $due + $adddue;

$modify = $_POST['modify'] ;
//echo ("$modify");

//function to update customer and workorder
if ($modify == 'modify') {
$updatecustomer = "UPDATE addressindex SET address='$address', city='$city',
state='$state', zipcode='$zipcode', phone='$phone', ss='$ss', cc='$cc',
exp='$exp', notes='$notes' WHERE name='$name'";
$updateworkorder = "UPDATE workorder set date='$date', time='$time',
service='$service', due='$due', special='$special', extrawork='$extrawork',
adddue='$adddue', company='$company', installer='$installer', total='$total'
where name='$name'";
$result = mysql_query($updatecustomer);
$result2 = mysql_query($updateworkorder);

//Code to check if statement executed properly and display message
if ($result) {
echo("Customer data updated sucessfully<br>");
} else {
echo("An error has occured<br>");
}

if ($result2) {
echo("Work Order data updated sucessfully<br>");
} else {
echo("An error has occured<br>");
}
};

//function to delete customer and workorder

if ($modify == 'delete customer') {
$deletecustomer = "DELETE from addressindex WHERE name='$name'";
$result3 = mysql_query($deletecustomer);

//Code to check if statement executed properly and display message
if ($result3) {
echo("Customer data deleted sucessfully<br>");
} else {
echo("An error has occured<br>");
}

};

if ($modify == 'delete workorder') {
$deleteworkorder = "DELETE from workorder where name='$name'";
$result4 = mysql_query($deleteworkorder);

//Code to check if statement executed properly and display message
if ($result4) {
echo("Work Order data deleted sucessfully<br>");
} else {
echo("An error has occured<br>");
}
};

if ($modify == 'print') {
echo ("name is $name<BR>");
echo ("Preparing workorder - Please wait...");

echo '

<META HTTP-EQUIV="Refresh"
CONTENT="4;URL=http://lexon.ws/login/print.php">



';
};
//Close connection with MySQL
MySQL_close()
?>
</td>
</tr>
</table>

"Bartosz Wegrzyn" <blwegrzyn@lexon.ws> wrote in message
news:B4cTa.24155$BM.7571069@newssrv26.news.prodigy .com...
> I found out that I could use require() or include() functions
> What do u think
> "Gianmarco Vanni" <gianmarcovanni@tin.it> wrote in message
> news:3F1D1804.3000904@tin.it...
> > Hi,
> >
> > Bartosz Wegrzyn wrote:
> > > this is part of my code
> > >
> > > question?
> > >
> > > why the variable $name and $address does not display on the html page
> > > instead of real values I do get variable names
> >
> >
> > > <td><input name="address2" type="text" id="address" value="<?php
> echo $address ?>"
> >
> > you may also write, with the same result:
> > value="<?php echo '$address' ?>"
> > or
> > value="<?php echo "$address" ?>"
> >
> > remember html cannot display php values if you do not explicit the
> > markup <?php ?> for php processor.
> >
> > php is a server side scripting language, html is interpreted client side
> > by browsers...
> >
> > Ciao Gianfranzo
> >
> > --
> > --
> > ICQ#: 21258924
> > Nick Gianfranzo
> > Home page http://www.gianfranzo.it
> > Gianfranzo è contro la guerra
> > --
> >
>
>

"Bartosz Wegrzyn" <blwegrzyn@lexon.ws> wrote in message
news:UjcTa.24161$BM.7573218@newssrv26.news.prodigy .com...
> this is my new code
> the problem is that varialebles are not past to print.php
>
> what am I doing wrong ???
>
<Deleted Lines>
> //Assign contents of form to variables
> //Customer Data
> $name = $_POST['name'];
<Deleted Lines>

> if ($modify == 'print') {
> echo ("name is $name<BR>");
<Deleted Lines>

Hi,

We are taking it on trust that the html form has a field called 'name' and
that it is assigned a value by the user to be posted to the PHP script,
That being so it should display - Remember that PHP is case sensitive for
variable names

so having name = 'Fred' in html and
$Fred =$_POS['fred'] in PHP isn't going to work. Fred !== fred

I noticed earlier in the thread you were struggling with variables within
strings

echo " Name = '$fred' <br />"; will display Name = 'value of fred'
whereas

echo ' Name = "$fred" <br />'; will display Name = "$fred"

The outer most quotes have to be double quotes for it to work.

Returning to your posted variable problem, just add a debug line to print
$name as soon as you have it - to make sure it really has arrived.
The other way you can lose variables is inside functions, your code refers
to what appears to be inline code as a function. if it really is code taken
from a function ( i.e. not inline as shown ) the variable scope of the
function will not include variables declared outside of the function unless
declared as global. in other words there will be two variables called $name,
one existing inside the function (un initialised) and one outside the
function loaded with the posted data and otherwise unused.

Cheers

Ron