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[SOLVED] convert ASP coding to PHP


Nazirul

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Hi...

 

i want to declare XML in PHP just like ASP code below :

 

<!--#include file="ADOVBS.INC"-->
<%
Set ConnObj = Server.CreateObject("ADODB.Connection")
Set RstObj = Server.CreateObject("ADODB.Recordset")

ConnObj.Open "spa"

sql="select *from CUSTOMER"

set RstObj=ConnObj.Execute(sql,RecordsAffected) %>

<?xml version="1.0" encoding="windows-1252"?>
<?xml-stylesheet type="text/xsl" href="ViewAllSpaCust.xsl"?>

<Senarai_Pelanggan>

<% While not rstobj.eof %>

<senarai> </senarai><senarai><Nama><%=rstobj("CUSTNAME")%></Nama> 

<NoIc><%=rstobj("CUSTIC")%></NoIc> <Gender><%=rstobj("CUSTGENDER")%></Gender> 

<Address><%=rstobj("CUSTADDRESS")%> </Address>

<Race><%=rstobj("CUSTRACE")%></Race> 

<Dob><%=rstobj("CUSTDOB")%></Dob> 

<MobileNo><%=rstobj("CUSTMOBILENO")%></MobileNo> 

<HomeNo><%=rstobj("CUSTHOMENO")%></HomeNo>

<MemberDate><%=rstobj("CUSTMEMBERDATE")%></MemberDate> 

</senarai> 

<%rstobj.MoveNext%>

<%wend%>

</Senarai_Pelanggan>

 

what is the similar syntax for the XML embedded in PHP just like the ASP.. i've tried :

 

<?php

//php code goes here

?>

<?xml version="1.0" encoding="ISO-8859-1"?>
..

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/2002/REC-xhtml1-20020801/DTD/xhtml1-strict.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>...

 

but no luck.. it give error :

 

Parse error: syntax error, unexpected T_STRING in C:\wamp\www\collegeXML\ViewStudent.php on line 28

 

hmm.. :D

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well whats the php code inside your php tags? Cant tell u what the error is without anycode to look at...

 

and you more likley would have something along these lines...

<?php

//php code goes here to query DB

?>

<?xml version="1.0" encoding="windows-1252"?>
<?xml-stylesheet type="text/xsl" href="ViewAllSpaCust.xsl"?>

<Senarai_Pelanggan>

<?php
//your php goes here....just echo/print any xml eg...

echo '<tag>'. $phpVariableHoldingValue .'</tag>';
?>

</Senarai_Pelanggan>

 

or you can also look into php5's XMLWriter class...

well... the error was in line 28 .. that refer to the <?xml version="1.0" encoding="windows-1252"?>

 

im not sure if it either php version problem...im using php 4.4.8 right now..

 

i've tried this

 

<studNumber><? $Student_Number ?></studNumber>

 

but no luck...

ah...yes....i've come accross that before my work around for that was just to echo the xml declaration...

 

<?php
echo '<?xml version="1.0" encoding="windows-1252"?>';
?>

 

Its the only way I was able to do it, php interpreter might confuse the <?xml for a short php tag.....

lol.. use ----> php 5

 

<?php
echo '<?xml version="1.0" encoding="ISO-8859-1"?>';
echo '<?xml-stylesheet type="text/xsl" href="ViewStudent2.xsl"?>';
echo '<student>';
echo '<room_number>'. $room_number .'</room_number>';
echo '<student_number>'. $student_number .'</student_number>';
echo '<student_name>'. $student_name .'</student_name>';
echo '<student_course>'. $student_course .'</student_course>';
echo '<semester>'. $semester .'</semester>';
echo '<address>'. $address .'</address>';
echo '<under>'. $Undergraduate_From .'</under>';
echo '</student>';
?>

 

that works... but the XSL cant be used ..

echo '<?xml-stylesheet type="text/xsl" href="ViewStudent2.xsl"?>';

 

 

how about sending some headers specifying that its an XML document...then maybe your document be rendered correctly...

<?php
header('Content-type:text/xml');
echo '<?xml version="1.0" encoding="ISO-8859-1"?>'; //after echo and before <? theres a ' 
echo '<?xml-stylesheet type="text/xsl" href="ViewStudent2.xsl"?>';
echo '<student>';
echo '<room_number>'. $room_number .'</room_number>';
echo '<student_number>'. $student_number .'</student_number>';
echo '<student_name>'. $student_name .'</student_name>';
echo '<student_course>'. $student_course .'</student_course>';
echo '<semester>'. $semester .'</semester>';
echo '<address>'. $address .'</address>';
echo '<under>'. $Undergraduate_From .'</under>';
echo '</student>';
?>

 

not work..

 

i think the reason why it works might be.. the php ignore the '<room_number' n etc... and only display the $value

 

my friend email me the working code and it is :

 

<?php
ob_start();
session_start();
?>
<?xml version="1.0"?>
<?xml-stylesheet type="text/xsl" href="detail.xsl"?>
<!DOCTYPE recipe SYSTEM "detail.dtd">
<recipe>
<?php
$id = $_GET['id'];
include("library/config.php");
include("library/opendb.php");
$sql = "SELECT * FROM recipe WHERE id = '$id' AND act = '0'";
$result = mysql_query($sql);
$row = mysql_fetch_array($result);
$cat_id = $row['category'];
$sql = "SELECT cat_name FROM category WHERE id = '$cat_id'";
$result2 = mysql_query($sql);
$r = mysql_fetch_array($result2);
?>
<data>
<recId><?php echo $id; ?></recId>
<recTitle><?php echo $row['title']; ?></recTitle>
<ingredient><?php echo $row['ingredient']; ?></ingredient>
<instruction><?php echo $row['instruction']; ?></instruction>
<suggestion><?php echo $row['suggestion']; ?></suggestion>
<comment><?php echo $row['comment']; ?></comment>
<source><?php echo $row['source']; ?></source>
<sender><?php echo $row['name']; ?></sender>
<email><?php echo $row['email']; ?></email>
<date><?php echo $row['date']; ?></date>
<catId><?php echo $cat_id; ?></catId>
<catName><?php echo $r['cat_name']; ?></catName>
</data>
<formFunction>
<approve><a href='approve.php' class='content'>Approve</a></approve>
</formFunction>
<?php
include("library/closedb.php");
?></recipe>

 

great thanks : DarkSuperHero   ;D

you are very helpful... :D

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