PHP4 script to PHP5 HELP!

mediasix · November 20, 2007

i currently have this script in place, but i need to make it compatible with php5.

please can you help me convert it?!?

thanks in advance

mediasix

<?php

if ($HTTP_POST_VARS['submit']) {

mysql_connect("xxxxxx","xxxxxx","xxxxxx");

mysql_select_db("xxxxxx");

$title=$HTTP_POST_VARS['title'];

$text=$HTTP_POST_VARS['text'];

$query ="INSERT INTO weblog (title,content)";

$query.=" VALUES ('$title','$content')";

$result=mysql_query($query);

if ($result) echo "<b>Successfully Posted!</b>";

else echo "<b>ERROR: unable to post.</b>";

}

?>

revraz · November 20, 2007

Change

$HTTP_POST_VARS

to

$_POST

mediasix · November 20, 2007

thanks that worked a treat, do you think you could help on this piece of code too?

thanks again

mediasix

<?

mysql_connect("xxxxxx","xxxxxx","xxxxxx");

mysql_select_db("xxxxxx");

$query="SELECT * FROM weblog WHERE title='$id'";

$result=mysql_query($query);

$num=mysql_numrows($result);

mysql_close();

$i=0;

while ($i < $num) {

$date=mysql_result($result,$i,"date");

$title=mysql_result($result,$i,"title");

$content=mysql_result($result,$i,"content");

?>

revraz · November 20, 2007

I don't see any PHP problems with that code myself. Maybe a missing $i++

mediasix · November 20, 2007

OK, probably would have been better giving you the full code, here it is.

when i navigate to the page providing the correct variable the page is blank.

Thank you very much

mediasix

<?

mysql_connect("xxxxxx","xxxxxx","xxxxxx");

mysql_select_db("xxxxxx");

$query="SELECT * FROM weblog WHERE title='$id'";

$result=mysql_query($query);

$num=mysql_numrows($result);

mysql_close();

$i=0;

while ($i < $num) {

$date=mysql_result($result,$i,"date");

$title=mysql_result($result,$i,"title");

$content=mysql_result($result,$i,"content");

?>

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">

<head>

<title>mediasix blog - <? echo "$title"?></title>

</head>

<body>

</div>

</body>

</html>

<?

$i++;

}

?>

PFMaBiSmAd · November 20, 2007

Change all the short open tags <? to full open tags <?php

mediasix · November 20, 2007

i have done that, but i still get a blank page

when i turn on php4.4.7 it works fine, but as soon as i turn on php5.2.3 is breaks!

im baffled ???

please please help!

(oh and it can be found at www.mediasix.co.uk click on the test blog entry on the right panel)

revraz · November 20, 2007

Your entire display relies on $i < $num

echo $num to check it's value

This line may be returning unexpected results

$query="SELECT * FROM weblog WHERE title='$id'";

PFMaBiSmAd · November 20, 2007

Blank pages are usually caused by fatal php parse errors. Check your web server log for errors.

mediasix · November 21, 2007

OK, is it possable to remove the $i < $num and it still function?

whe i do i get the errors:

Warning: mysql_result() [function.mysql-result]: Unable to jump to row 0 on MySQL result index 2 in /home/.putz/mutual_mediasix/mediasix.co.uk/weblog/weblogentry.php on line 9

Warning: mysql_result() [function.mysql-result]: Unable to jump to row 0 on MySQL result index 2 in /home/.putz/mutual_mediasix/mediasix.co.uk/weblog/weblogentry.php on line 10

Warning: mysql_result() [function.mysql-result]: Unable to jump to row 0 on MySQL result index 2 in /home/.putz/mutual_mediasix/mediasix.co.uk/weblog/weblogentry.php on line 11

being a n00b sucks, please help ???

thanks mediasix

~n[EO]n~ · November 21, 2007

Try this

<?php
mysql_connect("localhost","root",'');
mysql_select_db("******");
$query="SELECT * FROM weblog"; 
$result=mysql_query($query);
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>mediasix blog - <? echo "$title"?></title>
<link href="../css/mediasix.css" rel="stylesheet" type="text/css" />
<link href="../css/mediasix-print.css" rel="stylesheet" type="text/css" media="print" />
<script type="text/javascript" src="../scripts/equalcolumns.js"></script>
</head>

<body>
<div id="header"><img src="../images/mediasix-web-logo.gif" alt="logo" width="600" height="100" /></div>
<div id="holder">
<div id="contentholder">
<div class="nav" align="center" id="nav"><?php include("../includes/menu.php");?></div>
<?php 
$i=0;
while ($row = mysql_fetch_array($result)) {
$date=mysql_result($result,$i,"date");
$title=mysql_result($result,$i,"title");
$content=mysql_result($result,$i,"content"); 
?>
<div id="content">
<h1><? echo "$title"?></h1><h2><? echo "$date"?></h2>
<p><? echo "$content"?></P></div>
<?
$i++;
}
?>
<div id="othercontent"><?php include ("../includes/othercontent.php");?></div>
<div id="footer"><?php include("../includes/footer.php");?></div>
</div>
</div>
</body>
</html>

Tested on

PHP version :

5.2.4

sharifbaba · November 21, 2007

Change

$HTTP_POST_VARS

to

$_POST

mediasix · November 21, 2007

OK i had to change the layout of the the code slightly, as i need the title of the html page to be dynamicaly created too.

the code now works on php5, but not exactly how i want it.

it now displays all the entries in form 'weblog' but i need it to take a specific vairiable from the url using GET, in my last piece of code i had $query="SELECT * FROM weblog WHERE title='$id'";

but when i put that back into the code below it breaks!

thanks again

mediasix

<?php
mysql_connect("xxxxxx","xxxxxx","xxxxxx");
mysql_select_db("xxxxxx");
$query="SELECT * FROM weblog";
$result=mysql_query($query);
?>
<?php 
$i=0;
while ($row = mysql_fetch_array($result)) {
$date=mysql_result($result,$i,"date");
$title=mysql_result($result,$i,"title");
$content=mysql_result($result,$i,"content"); 
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>mediasix blog - <?php echo "$title"?></title>
<link href="../css/mediasix.css" rel="stylesheet" type="text/css" />
<link href="../css/mediasix-print.css" rel="stylesheet" type="text/css" media="print" />
<script type="text/javascript" src="../scripts/equalcolumns.js"></script>
</head>

<body>
<div id="header"><img src="../images/mediasix-web-logo.gif" alt="logo" width="600" height="100" /></div>
<div id="holder">
<div id="contentholder">
<div class="nav" align="center" id="nav"><?php include("../includes/menu.php");?></div>
<div id="content">
<h1><?php echo "$title"?></h1><h2><? echo "$date"?></h2>
<p><?php echo "$content"?></P></div>
<?
$i++;
}
?>
<div id="othercontent"><?php include ("../includes/othercontent.php");?></div>
<div id="footer"><?php include("../includes/footer.php");?></div>
</div>
</div>
</body>
</html>

~n[EO]n~ · November 21, 2007

Try this, you are using WHILE loop above other html tags, your LOGO and other stuff will repeat below the WHILE loop

<?php
mysql_connect("localhost","root",'');
mysql_select_db("********");
$query="SELECT * FROM weblog WHERE title='$id'";
$result=mysql_query($query);

// you are using WHILE loop here and it ends below, so the code below this line will repeat
// are you expecting like this all your LOGO stuff and other THINGS will repeat 
while ($row = mysql_fetch_array($result)) {
$date=$row['date']; // change here with your table field name 
$title=$row['title'];
$content=$row['content']; // till here 
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>mediasix blog - <?php echo "$title"?></title>
<link href="../css/mediasix.css" rel="stylesheet" type="text/css" />
<link href="../css/mediasix-print.css" rel="stylesheet" type="text/css" media="print" />
<script type="text/javascript" src="../scripts/equalcolumns.js"></script>
</head>

<body>
<div id="header"><img src="../images/mediasix-web-logo.gif" alt="logo" width="600" height="100" /></div>
<div id="holder">
<div id="contentholder">
<div class="nav" align="center" id="nav"><?php include("../includes/menu.php");?></div>
<div id="content">
<h1><?php echo "$title"?></h1><h2><? echo "$date"?></h2>
<p><?php echo "$content"?></P></div>
<?
}
?>
<div id="othercontent"><?php include ("../includes/othercontent.php");?></div>
<div id="footer"><?php include("../includes/footer.php");?></div>
</div>

mediasix · November 21, 2007

i am now rather confused, and i am unsure if i need a loop at all now.

all i want is the page to take the variable 'id' from the url which will look like this:

http://mediasix.co.uk/weblog/weblogentry.php?id=this%20is%20a%20test%20entry

then query the sql database in table 'weblog'

and return the entry that matches the 'id' variable with the 'title' coloumn in the table, i dont need this repeated or anything like that.

all this needs to be before my html so that i can insert the returned values into the header and then into the body.

this worked fine in php4, i need it to work in php5 desperatley

many thanks in advance

mediasix

~n[EO]n~ · November 21, 2007

Have you tried in Version 5.x.x

$query="SELECT * FROM weblog WHERE title='$id'";

Your query will return all the data from the table which matches title=id, if there is 1 you will get 1, if there are many you will get all. So, how many rows you want to display.

And for running in version 5+ i checked this code, i got 5.2.4 and it is running fine.

mediasix · November 21, 2007

i tried that but i get nothing displayed, my page is there with the structure and css but no dynamic data!

here is the code currently being used:

thanks as always

mediasix

**edit**

there should only be on record that can be returned, as all entries will have a different title.

but for arguments sake, only one row

<?php
mysql_connect("xxxxxx","xxxxxx","xxxxxx");
mysql_select_db("xxxxxx");
$query="SELECT * FROM weblog WHERE title='$id'";
$result=mysql_query($query);
$row = mysql_fetch_array($result);
$date=$row['date']; 
$title=$row['title'];
$content=$row['content'];
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>mediasix blog - <?php echo "$title"?></title>
<link href="../css/mediasix.css" rel="stylesheet" type="text/css" />
<link href="../css/mediasix-print.css" rel="stylesheet" type="text/css" media="print" />
<script type="text/javascript" src="../scripts/equalcolumns.js"></script>
</head>

<body>
<div id="header"><img src="../images/mediasix-web-logo.gif" alt="logo" width="600" height="100" /></div>
<div id="holder">
<div id="contentholder">
<div class="nav" align="center" id="nav"><?php include("../includes/menu.php");?></div>
<div id="content">
<h1><?php echo "$title"?></h1><h2><? echo "$date"?></h2>
<p><?php echo "$content"?></P></div>
<div id="othercontent"><?php include ("../includes/othercontent.php");?></div>
<div id="footer"><?php include("../includes/footer.php");?></div>
</div>

~n[EO]n~ · November 21, 2007

Where are you getting this '$id' from, try using 1,2... which is in your table, and keep

error_reporting(E_ALL);

this in your php file at the top

PFMaBiSmAd · November 21, 2007

There was just a recent thread with similar problems - http://www.phpfreaks.com/forums/index.php/topic,168465.0.html Your code is dependent on register globals being on (variables magically appear from post/get/files/session/cookie data). This is not an issue with the PHP version but of php settings. Register globals are apparently off (as the should be) in the PHP5 settings.

What is sad about this is that register globals were turned off by default in php 4.2 in the year 2002. No new code, tutorials, books... should have been written after that point in time that relied on register globals being on. That was 5 years ago. Everyone on the planet should have gotten the message by now to not continue writing code this way. Register globals have been completely removed in PHP6, so everyone will need to fix any code that is dependent on register globals by the time PHP6 is released.

Edit: The posted code has no error checking, error reporting, or error recovery on the mysql function calls and will result in a blank page any time the mysql server cannot be connected to (which happens more than you would think.)

mediasix · November 21, 2007

OK , ive added: error_reporting(E_ALL); to the top of my code and also i changed '$id' to '$1' - same as before, nothing!

the code and link is below -

thanks people

http://mediasix.co.uk/weblog/weblogentry.php?1=this%20is%20a%20test%20entry

<?php
error_reporting(E_ALL);
mysql_connect("xxxxxx","xxxxxx","xxxxxx");
mysql_select_db("xxxxxx");
$query="SELECT * FROM weblog WHERE title='$1'";
$result=mysql_query($query);
$row = mysql_fetch_array($result);
$date=$row['date']; 
$title=$row['title'];
$content=$row['content'];
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>mediasix blog - <?php echo "$title"?></title>
<link href="../css/mediasix.css" rel="stylesheet" type="text/css" />
<link href="../css/mediasix-print.css" rel="stylesheet" type="text/css" media="print" />
<script type="text/javascript" src="../scripts/equalcolumns.js"></script>
</head>

<body>
<div id="header"><img src="../images/mediasix-web-logo.gif" alt="logo" width="600" height="100" /></div>
<div id="holder">
<div id="contentholder">
<div class="nav" align="center" id="nav"><?php include("../includes/menu.php");?></div>
<div id="content">
<h1><?php echo "$title"?></h1><h2><?php echo "$date"?></h2>
<p><?php echo "$content"?></P></div>
<div id="othercontent"><?php include ("../includes/othercontent.php");?></div>
<div id="footer"><?php include("../includes/footer.php");?></div>
</div>

mediasix · November 21, 2007

if anyone could tell me why

error_reporting(E_ALL);

isnt working then i will try and debug myself, but even that wont work!

please help ???

mediasix · November 21, 2007

please anyone!?!?!

revraz · November 21, 2007

Is error reporting on in your php.ini?

mediasix · November 21, 2007

yeah its on, i have tried on my host 'dreamhost' and on my local install,

both times nothing!

i'm really stuck, any more advise welcomed with open arms

thanks

mediasix

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