PHP - Displaying A Php/mysql Table - Error?

Greetings,   My current code logs into a database, opens a table named randomproverb, randomly selects 1 proverb phrase, and then SHOULD display the proverb in the footer of my web page. As of right now, the best I can do is get it to display "Array", but not the text proverb... this code below actually causes my whole footer to not even show up. Please help!

<?php

include("inc_connect.php"); //Connects to the database, does work properly, already tested

$Proverb = "randomproverb";
$SQLproverb = "SELECT * FROM $Proverb ORDER BY RAND() LIMIT 1";
$QueryResult = @mysql_query($SQLproverb, $DBConnect);
	while (($Row = mysql_fetch_assoc($QueryResult)) !== FALSE) {
	echo "<p style = 'text-align:center'>" . {$Row[proverb]} . "</p>\n";
	}
	$SQLString = "UPDATE randomproverb SET display_count = display_count + 1 WHERE proverb = $QueryResult[]";
	$QueryResult = @mysql_query($SQLstring, $DBConnect);
...
?>

Hey guys,
Having a slight problem with part of the code in my index.php file

Code: [Select]
mysql_select_db('db_name', $con);

$result = mysql_query("SELECT * FROM spy ORDER BY id desc limit 25");
$resulto = mysql_query("SELECT * FROM spy ORDER BY id desc");
$count = mysql_num_rows($resulto);
while($row = mysql_fetch_array($result))
  {
?>
<div class="contentDiv">Someone is looking at <?=$row[title];?> Stats for  "<a href="/<?=$row[type];?>/<?=$row[code];?>/<?=$row[city];?>"><?=$row[code];?> <?=$row[city];?></a>"</div>
<?}?>
</div>
  <div id="login"></div>
          <? include("footer.php"); ?>
</div>
</body>
</html>
I'm getting the following error when viewing the file

Code: [Select]
Warning: mysql_num_rows() expects parameter 1 to be resource, boolean given in ~path to file/index.php on line 70

Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in ~path to file/index.php on line 71

where lines 70 and 71 are

$count = mysql_num_rows($resulto);
while($row = mysql_fetch_array($result))

Any ideas on how to fix this?

I want to display my pictures I stored in Mysql in a 4 column, 2 row table WITH pagination.

Here's the code I use to display the data currently:
Code: [Select]
//your username
$username = "username";
//your password
$password = "password";
//your mySQL server
$host = "host";
//The name of the database your table is in
$database = "database";

//connect, but if there is a problem, display an error message telling why there is a problem
$conn = mysql_connect($host,$username,$password) or die("Error connecting to Database!<br>" . mysql_error());
//Choose the database from your mySQL server, but if there is a problem, display an error telling why
$db = mysql_select_db($database) or die("Cannot select database!<br>" . mysql_error());

// find out how many rows are in the table
$sql = "SELECT COUNT(*) FROM myphotos";
$result = mysql_query($sql, $conn) or trigger_error("SQL", E_USER_ERROR);
$r = mysql_fetch_row($result);
$numrows = $r[0];

// number of rows to show per page
$rowsperpage = 4;
// find out total pages
$totalpages = ceil($numrows / $rowsperpage);

// get the current page or set a default
if (isset($_GET['currentpage']) && is_numeric($_GET['currentpage'])) {
// cast var as int
$currentpage = (int) $_GET['currentpage'];
} else {
// default page num
$currentpage = 1;
} // end if

// if current page is greater than total pages...
if ($currentpage > $totalpages) {
// set current page to last page
$currentpage = $totalpages;
} // end if
// if current page is less than first page...
if ($currentpage < 1) {
// set current page to first page
$currentpage = 1;
} // end if

// the offset of the list, based on current page
$offset = ($currentpage - 1) * $rowsperpage;

// get the info from the db
$sql = "SELECT * FROM myphotos ORDER BY id ASC LIMIT $offset, $rowsperpage";
$result = mysql_query($sql, $conn) or trigger_error("SQL", E_USER_ERROR);

/******  build the pagination links ******/
// range of num links to show
$range = 3;

// if not on page 1, don't show back links
if ($currentpage > 1) {
// show << link to go back to page 1
echo "<span class=\"pagination\"><a href='{$_SERVER['PHP_SELF']}?currentpage=1'>First</a></span> ";
// get previous page num
$prevpage = $currentpage - 1;
// show < link to go back to 1 page
echo " <span class=\"pagination\"><a href='{$_SERVER['PHP_SELF']}?currentpage=$prevpage'>Previous</a></span> ";
} // end if

// loop to show links to range of pages around current page
for ($x = ($currentpage - $range); $x < (($currentpage + $range) + 1); $x++) {
// if it's a valid page number...
if (($x > 0) && ($x <= $totalpages)) {
// if we're on current page...
if ($x == $currentpage) {
 // 'highlight' it but don't make a link
 echo " <span class=\"paginationDown\"><b>$x</b></span> ";
// if not current page...
} else {
 // make it a link
 echo " <span class=\"pagination\"><a href='{$_SERVER['PHP_SELF']}?currentpage=$x'>$x</a></span> ";
} // end else
} // end if
} // end for

// if not on last page, show forward and last page links       
if ($currentpage != $totalpages) {
// get next page
$nextpage = $currentpage + 1;
// echo forward link for next page
echo " <span class=\"pagination\"><a href='{$_SERVER['PHP_SELF']}?currentpage=$nextpage'>Next</a></span> ";
// echo forward link for lastpage
echo " <span class=\"pagination\"><a href='{$_SERVER['PHP_SELF']}?currentpage=$totalpages'>Last</a></span><br> ";
} // end if
/****** end build pagination links ******/

// while there are rows to be fetched...
while ($list = mysql_fetch_assoc($result))
{
extract ($list);
// echo data
$url = $list['url'];;
$title = $list['title'];
$description = $list['description'];

echo("$title<br><a rel=\"example_group\" title=\"$description\" href=\"$url\"><img src=\"$url\" alt=\"\" width=\"\" height=\"\" class=\"gallery_images\" /></a>");
} // end while

/******  build the pagination links ******/
// range of num links to show
$range = 3;

// if not on page 1, don't show back links
if ($currentpage > 1) {
// show << link to go back to page 1
echo "<span class=\"pagination\"><a href='{$_SERVER['PHP_SELF']}?currentpage=1'>First</a></span> ";
// get previous page num
$prevpage = $currentpage - 1;
// show < link to go back to 1 page
echo " <span class=\"pagination\"><a href='{$_SERVER['PHP_SELF']}?currentpage=$prevpage'>Previous</a></span> ";
} // end if

// loop to show links to range of pages around current page
for ($x = ($currentpage - $range); $x < (($currentpage + $range) + 1); $x++) {
// if it's a valid page number...
if (($x > 0) && ($x <= $totalpages)) {
// if we're on current page...
if ($x == $currentpage) {
 // 'highlight' it but don't make a link
 echo " <span class=\"paginationDown\"><b>$x</b></span> ";
// if not current page...
} else {
 // make it a link
 echo " <span class=\"pagination\"><a href='{$_SERVER['PHP_SELF']}?currentpage=$x'>$x</a></span> ";
} // end else
} // end if
} // end for

// if not on last page, show forward and last page links       
if ($currentpage != $totalpages) {
// get next page
$nextpage = $currentpage + 1;
// echo forward link for next page
echo " <span class=\"pagination\"><a href='{$_SERVER['PHP_SELF']}?currentpage=$nextpage'>Next</a></span> ";
// echo forward link for lastpage
echo " <span class=\"pagination\"><a href='{$_SERVER['PHP_SELF']}?currentpage=$totalpages'>Last</a></span><br>";
} // end if
/****** end build pagination links ******/

The above code just displays the pictures vertically. Here's this code live:
http://www.djsmiley.net/gallery/albums/my_photos.php
(you can't view the page in IE)

Now how would I make it so that the first 8 images in the database display in a 4x2 table, etc.?

So I have a jobs database with the following columns: id, jobtext, jobdate, and id. This is how it looks right now: http://prahan.com/jobs/display.html.php

I have another table called author. In the authorid column in need the results of this query, SELECT name FROM author WHERE id = (SELECT authorid FROM job) , to be displayed for each row.

I also want to be able to customize the header title for each column.

Thanks in advance!

Not really sure how to get the images I have stored in MySQL into a html form. I can call-up the text fields from the database but it cannot seem to find the index for the images.

Here is my code:-

<?php
session_start();
mysql_connect("localhost","root","abc") or die ("Error! Cannot connect to database");
mysql_select_db("theimageworks") or die ("Cannot find database");
$query = "SELECT * FROM jobs";
$result = mysql_query($query) or die (mysql_error());
?>


<?php
//display data in html table
echo "<table>";
echo "<tr><td>Username</td><td align='center'>Message</td><td>Product Image</td></tr>";

while($row = mysql_fetch_array($result)) {
    echo "</td><td>";
    echo $row['username'];
    echo "</td><td>";
    echo $row['message'];
    echo "</td></tr>";   
    echo $row['image'];
     }
echo "</table>";
?>

The error message I get is "Notice: Undefined index: image in....."

Thanks in advance!