PHP: problem with index.php


 
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# 1  
Old 07-02-2003
PHP: problem with index.php

iam geting a error with this index script. heres the error

Parse error: parse error in c:\phpdev\www\dev\compulearn\in work\index.php on line 39

Whats wrong??

------------------------

<?php


//display header and left bars
include ('header.php');
include ('left.php');

//connect to the MySql Database


$conn = db_connect();

$sql = "select * from message
where page = '$page'
and published is not null
order by published desc";
$result = mysql-query($sql, $conn);

while ($story = mysql_fetch_array($result))
{
if ($story['picture']) {
$size = getImageSize ($story['picture']);
$width = $size[0];
$height = $size[1];
print "<img src=\"".$story['picture']."\" height=\"$height\" width=\"$width\" align=\"left\" />";

//print test from databasse
print $story['story_text'];
}

//display footer
include ('footer.php');
?>
# 2  
Old 07-02-2003
Are you suppose to have an endwhile at the bottom of that - or a closing } ?
# 3  
Old 07-02-2003
i've solved that, but now I want to read the text from the mySql db and display in that page.
i keep gettin a error :
Warning: mysql_query(): supplied argument is not a valid MySQL-Link resource in c:\phpdev\www\dev\compulearn\in work\main.php on line 18

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in c:\phpdev\www\dev\compulearn\in work\main.php on line 19



heres what ive done. Can sumone solve this


----------------------------
<?php
include ('include_fns.php');
include ('header.php');
include ('left.php');
include_once('db_fns.php');

//connect to the MySql Database

$conn = db_connect();




$sql = "select * from stories
where page = '$pages'
and published is not null
order by published desc";
$result = mysql_query($sql, $conn);
while ($story = mysql_fetch_array($result))
{

$size = getImageSize($story['picture']);
$width = $size[0];
$height = $size[1];
print "<img src=\"".$story['picture']."\" height=\"$height\" width=\"$width\" align=\"left\" />";


print $story['story_text'];
}

include ('footer.php');


?>
# 4  
Old 07-02-2003
Maybe this will show a more specific error?

Replace

$result = mysql_query($sql, $conn);

with

$result = mysql_query($sql, $conn) or die(mysql_error());
# 5  
Old 07-02-2003
now I get this


Warning: mysql_query(): supplied argument is not a valid MySQL-Link resource in c:\phpdev\www\dev\compulearn\in work\main.php on line 18
Access denied for user: 'content@127.0.0.1' (Using password: NO)

ive no user or password on the db.

this is whats in the db_fns.php which is inlcuded once in main.php

<?php
//connect to database and retreive info from tables
function db_connect()
{
$result = @mysql_pconnect('localhost', 'content', '');
if (!$result)
return false;
if (!@mysql_select_db('content'))
return false;

return $result;
}

function get_story_record($story)
{
$conn = db_connect();
$sql = "select * from stories where id = '$story'";
$result = mysql_query($sql, $conn);
return(mysql_fetch_array($result));
}

?>


i then removed the usr and pass stuuf to:

<?php
//connect to database and retreive info from tables
function db_connect()
{
$result = @mysql_pconnect('', '', '');
if (!$result)

and i get this now

Unknown column 'published' in 'where clause'

I have no data in the database.
# 6  
Old 07-02-2003
Quote:
Originally posted by perleo
Warning: mysql_query(): supplied argument is not a valid MySQL-Link resource in c:\phpdev\www\dev\compulearn\in work\main.php on line 18
Access denied for user: 'content@127.0.0.1' (Using password: NO)
Okay, so that wasn't incredibly more helpful. Smilie

I'm not sure what else to suggest - I don't know php so this is beyond me; maybe someone else here knows php...
# 7  
Old 07-02-2003
okay, i have a table named stories. it has 1 picture.
i have the db done fine

how do i just get the text from it and display it and align the pic to the right top side of the text.
i have this but will it work

$conn = db_connect();


$sql = "select * from stories
where page = '$page'
and published is not null
order by published desc";
$result = mysql_query($sql, $conn);

while ($story = mysql_fetch_array($result))
{
print '<h2>'.$story['headline'].'</h2>';
if ($story['picture']) {
$size = getImageSize($story['picture']);
$width = $size[0];
$height = $size[1];
print "<img src=\"".$story['picture']."\" height=\"$height\" width=\"$width\" align=\"left\" />";

}
print $story['story_text'];

}
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