Help with parsing vars into HTML

Anon

Whew, the whole morning I sat here and worked on my database and PHP. Now it\'s time for testing it and snip there are the problems, as expected

I directly give the code I use here, so I don\'t have to talk around:

This is inserted in a plain HTML text file. Further above I already let the PHP script connect to my database. The problem here is that when I call up this page the echos don\'t seem to work. Here\'s the URL so you can see the result:
http://www.icosmoss.com/projectdetail.php3

Thanks for any corrections and tips

Anon

try
echo $url;
instead of echo( "$url" );

Are you sure the variables exist at all?
Test for that in some wild way.

Anon

Still doesn't work. And the vars were already declared far above this where I also connect to the DB.

I'm really new to all this, and at the moment I can't think of a method to test it.

Anon

Have you tried this?
<?php
echo "Hello World";
?>

If this works, then there is nothing wrong with you echo.

Try this:
<?php
$myvar = "Hello Worlds";
echo $myvar;
?>

Still nothing wrong.

Try that your db varialbes really really exist by
<?php
if ($url != '') { ?>
HEJ HEJ IT EXIST
<?php
} else {
?>
IT DOESN'T EXIST
<?php
}
?>

Anon

Hey thanks. The script reports me: it doesn't exist. 🙁

Although they are declared above. Here's one sample:

<?php
......

// request the text of all the info
mysql_query("SELECT author, url, language FROM table");

// define vars
$author = $row["author"];
$url = $row["url"];
$language = $row["language"];

......
?>

Anon

No wonder that your echo didn't work. You haven't fetched the data.

Should be something like this:

<?php
......

// request the text of all the info
$result = mysql_query("SELECT author, url, language FROM table");
$fetch = mysql_fetch_array($result);

// define vars
$author = $fetch["author"];
$url = $fetch["url"];
$language = $fetch["language"];

......
?>

I think you need to look at some examples or buy a book.

Anon

Hey thanks! No more problems now. Thanks again for your help!

to see the result go to: http://www.icosmoss.com/projectdetail.php3

Now I need to do the second step; I've described it in a new topic in this forum

Anon

More quick help.
If you want to loop through more than one row.
<?php
......

// request the text of all the info
$result = mysql_query("SELECT author, url, language FROM table");
while($fetch = mysql_fetch_array($result))
{

// define vars
$author = $fetch["author"];
$url = $fetch["url"];
$language = $fetch["language"];

......
}
?>

Anon

How do I display the result? You maybe know, I have separated our projects into "completed" and "pending". And I would like to display all the found result into the HTML file just like I have now.

See here for an example: http://www.icosmoss.com/pending.shtml

Anon

One for Pending:
<?php
......

// request the text of all the info
$result = mysql_query("SELECT author, url, language FROM table WHERE status = 'pending'");
while($fetch = mysql_fetch_array($result))
{

// define vars
$author = $fetch["author"];
$url = $fetch["url"];
$language = $fetch["language"];

......
}
?>

And for Complete

<?php
......

// request the text of all the info
$result = mysql_query("SELECT author, url, language FROM table WHERE status = 'complete'");
while($fetch = mysql_fetch_array($result))
{

// define vars
$author = $fetch["author"];
$url = $fetch["url"];
$language = $fetch["language"];

......
}
?>

Anon

Yes, this is what I already have. I mean, how do I display the results in the echo tag; so into viewable HTML?

Anon

What is the error?
Elaborate, explain further, please.

Anon

What is the error?
Elaborate, explain further, please.

By the way, http://www.icosmoss.com/pending.shtml
doesn't work. 404