PHP - Placing An Element At The Bottom Of A Page Centered

This is the code I have now however when you click next to view more results it says please enter a search..., (there is more code above this however I thought that the problem would be within the code provided.)

$currPage = (($s/$limit) + 1);

//break before paging
  echo "<br />";

  // next we need to do the links to other results
  if ($s>=1) { // bypass PREV link if s is 0
  $prevs=($s-$limit);
  print "&nbsp;<a href=\"$PHP_SELF?s=$prevs&q=$var\">&lt;&lt;
  Prev 10</a>&nbsp&nbsp;";
  }

// calculate number of pages needing links
  $pages=intval($numrows/$limit);

// $pages now contains int of pages needed unless there is a remainder from division

  if ($numrows%$limit) {
  // has remainder so add one page
  $pages++;
  }

// check to see if last page
  if (!((($s+$limit)/$limit)==$pages) && $pages!=1) {

  // not last page so give NEXT link
  $news=$s+$limit;

  echo "&nbsp;<a href=\"$PHP_SELF?s=$news&q=$var\">Next 10 &gt;&gt;</a>";
  }

$a = $s + ($limit) ;
  if ($a > $numrows) { $a = $numrows ; }
  $b = $s + 1 ;
  echo "<p>Showing results $b to $a of $numrows</p>";
 
?>

Hi there,

I have a database with table column called EXPAMOUNT

if EXPAMOUNT value was £5.50
it only displays £5.5. when i echo it. How do i get it to display 2 decimal places. So £5.50

My code at moment

echo “Total Expenses &nbsp &pound” . $row[‘SUM(EXPAMOUNT)’]. “

I have two different tables and I am trying to condense them into one table.   The structure of the tables is   cc -- ci -- cc -- cn -- alpha_2   countries -- id -- cn -- alpha_2 -- alpha_3   What I want to do is place the info from the column alpha_2 in the countries table into the alpha_2 column in the cc table. I have tried inner joins, left joins, Insert and Select index. Everything that I have tried gives me an error message. I do not know what I am doing wrong and I was hoping that someone would be able to give me a hand.   I forgot to mention that I want the placement of the alpha_2 column to be based on the cn columns in both tables. So that the information in the alpha_2 column lines up with the right name in the cn column
Edited by mdmartiny, 09 August 2014 - 01:07 PM.

Okay,

I would like to redirect my page within a specific time using java script:

<meta http-equiv="Refresh" content="3;url=<?PHP echo 'progress.php?user=';$session->username?>">

Basically, I would like it to redirect to this: progress.php?user=$session->username.

But it redirects to: progress.php?user=

Please help, thanks ...

Hi,

I have a project, where i want to place a text over the image dynamically, just like if it is sold, i want to make it as SOLD text over the image

Looking for idea to implement this,

Thanks,

Hi,

$_POST['arr_name'];


I want to make the search so that am getting the array values and oputting in IN(val1, val2, val3)


Thanks

Hi There,

I would like to place the last for month names into an variable- for example:

February 2011
January 2011
December 2010
November 2010

But I would like them to all have the same variable ($date) - so when I use it in a mssql_fetch_assoc - it increments with every instance of $date, as there will be one instance of $date for every row returned, getting older with every row.

Hope that is enough info?

Cheers
Matt

Hi Guys,

I'm going to try and watermark my uploaded images with a small logo (placed onto the original image) so they are all watermarked, is there any tutorials anyone can recommend to do this? i can't seem to find any online.

thanks for any help guys

Graham

I believe the solution the my problem should be simple I feel that it's staring me right in the face. I have a Cron Job that sends an email message to users who's bill "due date" falls on the current date. I want to make the email more personalized and say:   Dear John Doe:   You have the following bills due today:   Rent Cable Internet   Please login to pay your bills   Thanks,.   Here's my following PHP code

<?php
header("Content-type: text/plain");
 
// OPEN DATA BASE
define("HOSTNAME","localhost");
define("USERNAME","");
define("PASSWORD","");
define("DATABASE","");
 
mysql_connect(HOSTNAME, USERNAME, PASSWORD) or die("Connetion to database failed!");
mysql_select_db(DATABASE); 
 
joinDateFilter();
 
// BILL QUERY
function joinDateFilter(){
     $query = mysql_query("SELECT bills.billname, bills.duedate, bills.firstname, bills.email, bills.paid FROM bills JOIN users ON bills.userid=users.userid WHERE DATE(bills.duedate) = CURDATE() AND bills.PAID != 'YES'");
     $mail_to = "";
     while ($row = mysql_fetch_array($query)){
      echo $row['firstname']." - ".$row['email']."\n";
  $mail_to = $row['email'].", ";
     }
     if (!empty($mail_to)){
      sendEmail($mail_to);
     }
}
 
// SEND EMAIL
function sendEmail($mail_to) {
$from    = "MyEmail@myemail.com";
$message = "Dear " . $row['firstname'].", <br><br>"
 
."You have the following bills due today.<br><br>"
.$row['billname']. "<br><br>"
."Please login to pay your bills";
 
$headers = 'From: '.$from."\r\n" .
'Reply-To:'.$_POST['email']."\r\n" .
"Content-Type: text/html; charset=iso-8859-1\n".
'X-Mailer: PHP/' . phpversion();
mail($mail_to, "Today is your due date", $message, $headers);
}
?>

Edited by demeritrious, 26 December 2014 - 07:21 PM.

Hi There,

I want to be able to populate 2 variables,

$previousmon and $previousfri

with 2 dates, being the previous Monday and the previous Friday to the week you are on. So for example, if it is Monday 31st Jan, then $previousfri would be 28/01/2011 and $previousmon would be 24/01/2011.

Is this easy enough to do?

Cheers
Matt

Basically i have a folder with 100+ images they are NOT all the same extension, what im wanting to do is use PHP to find all the images and put them all in a database. how would i go about doing this?

thanks

Hello all. I am very new to PHP, and I am not sure where to look or what I'm looking for in my current assignment.

My task is to take in two numbers between 0-100. Once I take in that number, it should state beside it "The __ was accepted."
The program should not accept any numbers greater than 100 or any characters.

Once I do this, I must take a second number and do a similar thing.

Finally, I must have a statement show up at the bottom stating which number is greater.

Essentially, I need help in determining what I should use to place parameters, and how I can keep the program from echo ing any statement until input has been taken and tested for parameters.

Any help you can provide will be greatly appreciated!

Quesion: Show each movie in the database on its own page, and give the user links in a "page 1, Page 2, Page 3" - type navigation system. Hint: Use LIMIT to control which movie is on which page. I have provided 3 files: 1st: configure DB, 2nd: insert data, 3rd: my code for the question. I would appreciate the help. I am a noob by the way.

First set up everything for DB:

<?php
//connect to MySQL
$db = mysql_connect('localhost', 'root', '000') or
    die ('Unable to connect. Check your connection parameters.');

//create the main database if it doesn't already exist
$query = 'CREATE DATABASE IF NOT EXISTS moviesite';
mysql_query($query, $db) or die(mysql_error($db));

//make sure our recently created database is the active one
mysql_select_db('moviesite', $db) or die(mysql_error($db));

//create the movie table
$query = 'CREATE TABLE movie (
        movie_id        INTEGER UNSIGNED  NOT NULL AUTO_INCREMENT,
        movie_name      VARCHAR(255)      NOT NULL,
        movie_type      TINYINT           NOT NULL DEFAULT 0,
        movie_year      SMALLINT UNSIGNED NOT NULL DEFAULT 0,
        movie_leadactor INTEGER UNSIGNED  NOT NULL DEFAULT 0,
        movie_director  INTEGER UNSIGNED  NOT NULL DEFAULT 0,

        PRIMARY KEY (movie_id),
        KEY movie_type (movie_type, movie_year)
    )
    ENGINE=MyISAM';
mysql_query($query, $db) or die (mysql_error($db));

//create the movietype table
$query = 'CREATE TABLE movietype (
        movietype_id    TINYINT UNSIGNED NOT NULL AUTO_INCREMENT,
        movietype_label VARCHAR(100)     NOT NULL,
        PRIMARY KEY (movietype_id)
    )
    ENGINE=MyISAM';
mysql_query($query, $db) or die(mysql_error($db));

//create the people table
$query = 'CREATE TABLE people (
        people_id         INTEGER UNSIGNED    NOT NULL AUTO_INCREMENT,
        people_fullname   VARCHAR(255)        NOT NULL,
        people_isactor    TINYINT(1) UNSIGNED NOT NULL DEFAULT 0,
        people_isdirector TINYINT(1) UNSIGNED NOT NULL DEFAULT 0,

        PRIMARY KEY (people_id)
    )
    ENGINE=MyISAM';
mysql_query($query, $db) or die(mysql_error($db));

echo 'Movie database successfully created!';
?>



********************************************************************
***********************************************************************

second file to load info into DB:



<?php
// connect to MySQL
$db = mysql_connect('localhost', 'root', '000') or
    die ('Unable to connect. Check your connection parameters.');

//make sure you're using the correct database
mysql_select_db('moviesite', $db) or die(mysql_error($db));

// insert data into the movie table
$query = 'INSERT INTO movie
        (movie_id, movie_name, movie_type, movie_year, movie_leadactor,
        movie_director)
    VALUES
        (1, "Bruce Almighty", 5, 2003, 1, 2),
        (2, "Office Space", 5, 1999, 5, 6),
        (3, "Grand Canyon", 2, 1991, 4, 3)';
mysql_query($query, $db) or die(mysql_error($db));

// insert data into the movietype table
$query = 'INSERT INTO movietype
        (movietype_id, movietype_label)
    VALUES
        (1,"Sci Fi"),
        (2, "Drama"),
        (3, "Adventure"),
        (4, "War"),
        (5, "Comedy"),
        (6, "Horror"),
        (7, "Action"),
        (8, "Kids")';
mysql_query($query, $db) or die(mysql_error($db));

// insert data into the people table
$query  = 'INSERT INTO people
        (people_id, people_fullname, people_isactor, people_isdirector)
    VALUES
        (1, "Jim Carrey", 1, 0),
        (2, "Tom Shadyac", 0, 1),
        (3, "Lawrence Kasdan", 0, 1),
        (4, "Kevin Kline", 1, 0),
        (5, "Ron Livingston", 1, 0),
        (6, "Mike Judge", 0, 1)';
mysql_query($query, $db) or die(mysql_error($db));

echo 'Data inserted successfully!';
?>



**************************************************************
****************************************************************

MY CODE FOR THE QUESTION:


<?php
$db = mysql_connect('localhost', 'root', '000') or
die ('Unable to connect. Check your connection parameters.');

mysql_select_db('moviesite', $db) or die(mysql_error($db));

//get our starting point for the query from the URL
if (isset($_GET['offset'])) {
   $offset = $_GET['offset'];
} else {
   $offset = 0;
}

//get the movie
$query  = 'SELECT
   movie_name, movie_year
FROM
   movie
ORDER BY
   movie_name
LIMIT ' . $offset . ' , 1';
$result = mysql_query($query, $db) or die(mysql_error($db));
$row = mysql_fetch_assoc($result);
?>
<html>
<head>
<title><?php echo $row['movie_name']; ?></title>
</head>
<body>
<table border = "1">
<tr>
<th>Movie Name</th>
<th>Year</th>
</tr><tr>
<td><?php echo $row['movie_name']; ?></td>
<td><?php echo $row['movie_year']; ?></td>
</tr>
</table>
<p>
<a href="page.php?offset=0">Page 1</a>,
<a href="page.php?offset=1">Page 2</a>,
<a href="page.php?offset=2">Page 3</a>
</p>
</body>
</html>

I have a problem with one div, I want to fix it at the bottom of the page and when I zoom the page I can see the div without scroll.

I have been staring at this code for hours trying to figure out what is wrong, I guarantee that someone is going to look at it and make me feel like a tool.  The first query is not going through the list, the second is.  It worked till I started updating to PDO. I have tested the SQL, that works, probably a flipping comma or a quotation mark screwing with my brain. Please look and spot the obvious balls up and point it out to me.

as in I get a result that looks like this 

Heading
question
question
question

and it stops.
 

I am expecting
Heading
question
question
question

Heading
question
question
question

Why is this failing? Please show a doddering old fool what I am doing wrong...

$query = $pdo->prepare("SELECT * FROM faq_cats WHERE faqc_site = :site ORDER BY faqc_id ASC LIMIT 0,4");
		$query->bindParam(":site", $site);
		$query->execute();

		while($row = $query->fetch(PDO::FETCH_ASSOC)) {	
				$fid = $row["faqc_id"];
				$faqc = $row["faqc_name"];
					echo "<h2> ".$faqc." </h2>";
						
							$query = $pdo->prepare("SELECT * FROM faqs WHERE faq_cat = :fid ORDER BY faq_id ASC");
							$query->bindParam(":fid", $fid);
							$query->execute();
							while($row1 = $query->fetch(PDO::FETCH_ASSOC)) {	
								$faid=$row1["faq_id"];
								$faqn=$row1["faq_question"];
					echo "<a href=\"#".$faid."\">".$faqn."</a><br />";
				}
		}