Text and Images Interaction

[xux]

Hi,
    I am trying to implement a page,whereby if images is available it will be displayed but if not it wont display it.My codes are working but with a little bug,if image is not supply it will leave show the image placeholder (the kind that is shown if the image is not available).here are my codes

[code]
<div class="mainContent">
<p><h2 align="center" style="color: #330099;"><u>News and Events</u></h2></p>
<?php
$pageNo = $HTTP_GET_VARS['p'];//the page number
$displayLength = 5;//the number of news to display per page
@$db = mysql_pconnect('localhost', '', '');

if(!$db){
echo "<p style=\"color: #CC0000;\" align=\"center\"><b>ERROR: </b>Could not connect to database. Please try again later</p><p align=\"center\"> Please refresh this page and if the problem persists, contact <a href=\"mailto:[email protected]\">webmaster</a>. Thank you<br /></p>";
exit;
}

mysql_select_db('DB') ;
$query = "SELECT title, picture, content FROM news ORDER BY expiry_date DESC LIMIT ".($pageNo*$displayLength).", ".$displayLength;
$result = mysql_query($query);
$query1 = "select count(*) from news";

if($result){
for($a=0;$a<$displayLength;$a++){
$row = mysql_fetch_array($result);

$title = $row['title'];
$picture_url = $row['picture'];
$content = $row['content'];

//print the news
if($title){
echo "<p style=\"color: rgb(147, 3, 187); font-weight: bold;\"><u>$title</u></p>";
if($picture_url)
echo "<img src=\"$picture_url\" alt=\"Picture\" align=\"right\" border=\"0\"  hspace=\"5\" vspace=\"5\">\n";
echo $content;
}else{
break;
}
}

$query = "SELECT count(*) FROM news";
$result1 = mysql_query($query);
$row = mysql_fetch_array($result1);
$dbSize = $row[0];
$availableNoOfPagesInDB = $dbSize/$displayLength;

$nextPage = $pageNo+1;
$previousPage = $pageNo-1;

echo "<p align=\"right\">";

if($previousPage >= 0){
echo "<a href=\"news.php?p=".($previousPage)."\"><strong>&lt;previous</strong></a>";
}

if($nextPage <= $availableNoOfPagesInDB){
echo "&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<a href=\"news.php?p=".($nextPage)."\"><strong>next&gt;</strong></a>";
}
echo "</p>";
}
?>


[/code]
and it is giving this error message
[code]
Parse error: syntax error, unexpected T_ELSE in /home/sm/public_html/news.php on line 41
[/code]

Thanks

[trq]

Sorry, but the formatting applied to your code makes it near impossible to read. Please take the time to make your post clear.

[xux]

[code]


<?php
$pageNo = $HTTP_GET_VARS['p'];//the page number
$displayLength = 5;//the number of news to display per page
@$db = mysql_pconnect('localhost', '', '');

if(!$db){
echo" ERROR: </b>Could not connect to database. Please try again later";
exit;
}

mysql_select_db('DB') ;
$query = "SELECT title, picture, content FROM news ORDER BY expiry_date DESC LIMIT ".($pageNo*$displayLength).", ".$displayLength;
$result = mysql_query($query);
$query1 = "select count(*) from news";

if($result){
for($a=0;$a<$displayLength;$a++){
$row = mysql_fetch_array($result);

$title = $row['title'];
$picture_url = $row['picture'];
$content = $row['content'];

//print the news
if($title){
echo "<p style=\"color: rgb(147, 3, 187); font-weight: bold;\"><u>$title</u></p>";
if($picture_url)
echo "<img src=\"$picture_url\" alt=\"Picture\" align=\"right\" border=\"0\"  hspace=\"5\" vspace=\"5\">\n";
echo $content;
}else{
break;
}
}

$query = "SELECT count(*) FROM news";
$result1 = mysql_query($query);
$row = mysql_fetch_array($result1);
$dbSize = $row[0];
$availableNoOfPagesInDB = $dbSize/$displayLength;

$nextPage = $pageNo+1;
$previousPage = $pageNo-1;

echo "<p align=\"right\">";


?>

[/code]
Thanks,I have stripped it of the formating

[intrigue]

whats at line 41?
matt

[trq]

[quote]Thanks,I have stripped it of the formating[/quote]

Your code still starts almost off the page. Make it easy for us to help you.

[xux]

[code]
<?php
$pageNo = $HTTP_GET_VARS['p'];//the page number
$displayLength = 5;//the number of news to display per page @$db = mysql_pconnect('localhost', '', '');
if(!$db){
echo" ERROR: </b>Could not connect to database. Please try again later";
exit;
}

mysql_select_db('DB') ;
$query = "SELECT title, picture, content FROM news ORDER BY expiry_date DESC LIMIT ".($pageNo*$displayLength).", ".$displayLength;
$result = mysql_query($query);
$query1 = "select count(*) from news";

if($result){
for($a=0;$a<$displayLength;$a++){
$row = mysql_fetch_array($result);

$title = $row['title'];
$picture_url = $row['picture'];
$content = $row['content'];

//print the news
if($title){
echo "<p style=\"color: rgb(147, 3, 187); font-weight: bold;\"><u>$title</u></p>";
if($picture_url)
echo "<img src=\"$picture_url\" alt=\"Picture\" align=\"right\" border=\"0\"  hspace=\"5\" vspace=\"5\">\n";
echo $content;
}else{
break;
}
}

$query = "SELECT count(*) FROM news";
$result1 = mysql_query($query);
$row = mysql_fetch_array($result1);
$dbSize = $row[0];
$availableNoOfPagesInDB = $dbSize/$displayLength;

$nextPage = $pageNo+1;
$previousPage = $pageNo-1;

echo "<p align=\"right\">";

?>

[/code]
Thanks,I appreciate

[xux]

[code]
<?php
$pageNo = $HTTP_GET_VARS['p'];//the page number
$displayLength = 5;//the number of news to display per page @$db = mysql_pconnect('localhost', '', '');
if(!$db){
echo" ERROR: </b>Could not connect to database. Please try again later";
exit;
}

mysql_select_db('DB') ;
$query = "SELECT title, picture, content FROM news ORDER BY expiry_date DESC LIMIT ".($pageNo*$displayLength).", ".$displayLength;
$result = mysql_query($query);
$query1 = "select count(*) from news";

if($result){
for($a=0;$a<$displayLength;$a++){
$row = mysql_fetch_array($result);

$title = $row['title'];
$picture_url = $row['picture'];
$content = $row['content'];

//print the news
if($title){
echo "<p style=\"color: rgb(147, 3, 187); font-weight: bold;\"><u>$title</u></p>";
if($picture_url)
echo "<img src=\"$picture_url\" alt=\"Picture\" align=\"right\" border=\"0\"  hspace=\"5\" vspace=\"5\">\n";
echo $content;
}else{
break;
}
}

$query = "SELECT count(*) FROM news";
$result1 = mysql_query($query);
$row = mysql_fetch_array($result1);
$dbSize = $row[0];
$availableNoOfPagesInDB = $dbSize/$displayLength;

$nextPage = $pageNo+1;
$previousPage = $pageNo-1;

echo "<p align=\"right\">";

?>


[/code]