Please help me on this project!

[panic36]

Hello, I'm new to PHP but I'm trying to make a download site and I need to basically have ( i think this should be pretty simple )

downlaodname<br>
<a href="downloadlocation"><img src="downloadlocationaddress"></a><p>

Basically that would be the listing... I have been using a piece of code i found here

<?php
  // If the user wants to add a data
  if (isset($adddata)):
?>

<FORM ACTION="<?php echo($PHP_SELF); ?>" METHOD=POST>
<P>Type your data base here :<br>
<TEXTAREA NAME="database" ROWS=10 COLS=40 WRAP>
</TEXTAREA><BR>
<INPUT TYPE=SUBMIT NAME="database VALUE="SUBMIT">
</FORM>

<?php
  else:

// Connect to the database server
$dbcnx = @mysql_connect("localhost",
         "roott", "password");
if (!$dbcnx) {
  echo( "<P>Unable to connect to the " .
        "database server at this time.</P>" );
  exit();
}

// Select the jokes database
if (! @mysql_select_db("database
  echo( "<P>Unable to locate the database
        "database at this time.</P>" );
  exit();
}

// If a joke has been submitted,
// add it to the database.
if ("SUBMIT" == $submitjoke) {
  $sql = "INSERT INTO datas SET " .
         "dataText='$database', " .
         "datadate=CURDATE()";
  if (mysql_query($sql)) {
    echo("<P>Your data has been added.</P>");
  } else {
    echo("<P>Error adding submitted data: " .
         mysql_error() . "</P>");
  }
}

echo("<P> Here are all the data " .
     "in our database: </P>");

// Request the text of all the data
$result = mysql_query(
          "SELECT database FROM data");
if (!$result) {
  echo("<P>Error performing query: " .
       mysql_error() . "</P>");
  exit();
}

// Display the text of each data in a paragraph
while ( $row = mysql_fetch_array($result) ) {
  echo("<P>" . $row["database"] . "</P>");
}

// When clicked, this link will load this page
// with the data submission form displayed.
echo("<P><A HREF='$PHP_SELF?adddata=1'>" .
     "Add a data!</A></P>");

  endif;

?>

Sorry, I'm not trying to spam... I've talked to someone about this, and two of my hardest points right now is i'd need the listings to be alphabetical order by downloadname and I’d only want 20 outputs per page and for multiple pages to run... I'm sorry for all questions, I’m just a bit lost right now... the last piece of information someone told me was

Well, you can either COUNT() the rows with the same SELECT query that would retrieve them minus the LIMIT clause, and pass that number around, or use CALC_SQL_FOUND_ROWS and check it directly. Once you have this number, simply divide by the page size, and voila.

I don't actually understand what he means, if anyone could help me out that would be great.

Thanks you,

Robert P.

[bpat1434]

What he means is that you can use the mySQL function COUNT() to count the number of rows matching your query. So if you have 40 items, and you want to view all the items... the COUNT() would return 40.

Now, to limit how many you show per page, that's as easy as defining how many per page you want to show, and what page you're on. Obviously, what page your on will be a $_GET variable. You then take the page number, subtract by one, and multiply that by 20 to get your starting row.

Page 2 - 1 = 1 20 = start at row 20
Page 1 - 1 = 0 20 = start at row 0

It makes sense if you know SQL. But anyway....

Now, to limti how many you show, it's a matter of adding the LIMIT clause to your query. The mySQL manual can help you in understanding COUNT() and LIMIT().

[panic36]

Thank you for your quick response. I have looked up COUNT/LIMIT from mysql.org in the reference manuals and they do provide a partial understand of the functionality and purpose of them. But when it comes to implementing it, that's where i'm becoming a loss... Also using the limit starting at 1-20, and row 20-40, so on giving you the 20 results per page, but how do i also implement it aranging it in ABC order by downloadname... Sorry for all the hassals

Thanks for your understanding,

Robert P.

[bradgrafelman]

panic36 wrote:

Also using the limit starting at 1-20, and row 20-40, so on giving you the 20 results per page, but how do i also implement it aranging it in ABC order by downloadname

Using 'ORDER BY' like so:

 SELECT downloadname, downloadlocation, downloadaddress FROM tableName ORDER BY downloadname DESC LIMIT 0,20

With an example of pagination using links such as "script.php?page=2" :

$page_number = (is_numeric(@$_GET['page']) ? $_GET['page'] : 1);
$query = 'SELECT downloadname, downloadlocation, downloadaddress FROM tableName ORDER BY downloadname DESC LIMIT ' . (($page_number - 1) * 20) . ', 20';

[panic36]

Thank you so much for your reply

I am now showing 20 results per page, so half of the mission is resolved. I am still having one problem though and that's it's putting them in order the name was added into the database. Would having it be in strict ABC order would it be a SQL command? or a function in PHP...

Thank you for your help,

Robert P.

EDIT

I'm guessing the following would be wrong?

    // Request the text of all the jokes
    $page_number = (is_numeric(@$_GET['page']) ? $_GET['page'] : 1);
	$result = mysql_query(
              "SELECT JokeText, JokeDate FROM jokes ORDER BY JokeText DESC LIMIT ' . (($page_number - 1) * 20) . ', 20';");
    if (!$result) {
      echo("<P>Error performing query: " .
           mysql_error() . "</P>");
      exit();
    }

Because I get a

Error performing query: You have an error in your SQL syntax. Check the manual that corresponds to your MySQL server version for the right syntax to use near '' . ((1 - 1) * 20) . ', 20'' at line 1

[bradgrafelman]

Yes, because you just copied my example instead of integrating it with your script. What you need to do is close the string (ending quote) before you do the $page_number - 1 bit of code. I used a single quote to close the string because I started the string with a single quote in my example. In your code, you've used double quotes.

(The reason I used single quotes is because I always use single quotes in strings that I'm not expecting PHP to parse for variables, i.e. 'this is a string'. The only time I use double quotes is when I do have a variable in the string, such as "this is a $variable". The reason being PHP does not parse single-quoted strings for variables, speeding up the processing of the script. True, this speed is usually negligible, but it's still a good practice if you ask me.)

[panic36]

Thank you so much for your help, I think this is working

    

$page_number = (is_numeric(@$_GET['page']) ? $_GET['page'] : 1);
$result = mysql_query(
              'SELECT JokeText FROM jokes ORDER BY JokeText DESC LIMIT ' . (($page_number - 1) * 20) . ', 20');

and when doing test.php?page=2, page=3 and so on the data is different!

I'm still having one problem though and that’s it seems the order of the data is being displayed when the JokeText was submitted to the database instead of going in ABC Order... Do you know any ways to solve this?

I really appreciate your support, I'm not going to lie, and I am a NEWB lol... I didn't mean to offend you also by copying your code like that; I was kind of lost in what I was doing... I've been reading a lot of books and it seems the only time I actually do learn anything is when talking to other people about the subject

Thanks again while I go through this trial and error period

Robert P.

[bpat1434]

From what you posted, it should be ordering proplery (based on the JokeText field). Can we see some examples?

[panic36]

Yea, http://www.aliandadele.com/mans1.php

I deleted everythign in the database and now it's going in order? Weird... One thing now is that it's backwards... Anyways of reversing it?

[panic36]

I do Apologize for my error it seems to be working now that i flushed the database and changed DESC to ASC... Sorry about that.

With the testing?page=2 is there a way to count them and devide links at the bottom of the page example

1 2 3 4 5 6 7 8 9 10 11 12

Thank you for your support.

Robert P.

[bradgrafelman]

Is this a MyISAM table? If so, get the number of rows by adding a 'COUNT(*)' to your fields selected, like so:

$result = mysql_query( 
              'SELECT JokeText, COUNT(*) as TotalRows FROM jokes ORDER BY ...';

Then, when you're retrieving data, use TotalRows divided by however many rows per page you're using (the above code you've been using has 20 per page) and round up using [man]ceil/man. That's how many pages you need.

[panic36]

Thank you for all your help! I really appreciate everything and things seem to be running ok... However, I'm still having problems with the showing the page number results divide i.e. having at the bottom of the page 1 2 3 4 5 6 7 8 9, which is the number of pages and having those numbers be URLS to the other pages. I was going to just make a next button but it would just keep showing pages even if there are no results... But this would make me have "deep links" ... and search engines don't really like that when you have 20k files divided in 20 items per page viewable... I thank you for your help...

Robert P.

So fare if you want to see what I've done so far (I know it looks shitty but I’m working on it) you can go to

1-800-Fonts

[bradgrafelman]

panic36 wrote:

However, I'm still having problems with the showing the page number results divide i.e. having at the bottom of the page 1 2 3 4 5 6 7 8 9, which is the number of pages and having those numbers be URLS to the other pages.

Well I pretty much gave you the SQL query to do this, and left the actual coding of it up to you. What did you try? Show us the code you've used.

Basically, your problem deals with MySQL pagination, a problem that has been covered countless times on this board. Try searching for keywords such as "pagination" and see what you get. You'll probably find a lot of coding examples.

[panic36]

Yes thank you for the query it was more of me forgetting to echo $pagination -.- lol....

I had a question though, say my page has 4 require(different.php); on it in tables and each contains mysql connection, and queiry, is this wrong? or will it slow the page down?

Another question is possible say I have affiliate 100 affiliate links, and I want 4 display on page I’d....

SELECT affiliatelink from affiliates order by.... limit 0,4

I'm not sure how I’d make the order random... pulling out 4 different affiliate links on each page load.... is this possible?

I thank you for all your help! I'm learning heaps.

Robert P.

EDIT : I tried ORDER BY RAND() but that doesn't do it?

 SELECT field FROM table ORDER BY RAND() LIMIT 0,4

[bpat1434]

User Notes are your friend....

If "SELECT * FROM tab ORDER BY RAND()" doesn't work for you. Try to put a random value between the brackets.

SOmething like:
SELECT field FROM table ORDER BY RAND(NOW()) LIMIT 0,4
SELECT field FROM table ORDER BY RAND(UNIX_TIMESTAMP()) LIMIT 0,4

I had a question though, say my page has 4 require(different.php); on it in tables and each contains mysql connection, and queiry, is this wrong? or will it slow the page down?

You should always just check to see if the current connection is made. I.e.

$conn = mysql_connect('host', 'user', 'pass');
if(!$conn)
  $conn = mysql_connect('host2', 'user', 'pass');

If you do that in each file (check to see if there is a pre-existing connection) you can speed it up a little bit. Now, you may need to select a different DB, but that's easy. Then you've got 1 connection instead of a possible 4 for the same queries....

[panic36]

Thanks for everyones, help, but I was wondering if someone could help me with simple issue.. I don't get what is going through my head.. I've read my PHP tutorials... And I've done this before, but I'm not sure... Info*

What I'm trying to do is have the pagenation recognize the form imput box so when they press on next they will see their customized input be carried onto the next page, so basically, I thought it wuold be something like ?page=2&amp;$fonttext but this isn't working... The input box only shows the information on the first page, not on any other page... Any help would be great thanks so much!

Here is the code if it helps you, i know it's bad :S and I thank everyone for their help and support. Cheers!

<?php

echo('<center><p><br><p>' . $abclinks) . '<FORM action="zfonts.php?page="' . $i . '" method="get">Font Text: <INPUT type="text" name="fonttext"><INPUT type="submit" value="Send"></form><BR></FORM>';

// Connect to the database server
$dbcnx = @mysql_connect("localhost",
         "aliandad", "tutu123~");
if (!$dbcnx) {
  echo( "<P>Unable to connect to the " .
        "database server at this time.</P>" );
  exit();
}

// Select the jokes database
if (! @mysql_select_db("aliandad_fonts") ) {
  echo( "<P>Unable to locate the fonts " .
        "database at this time.</P>" );
  exit();
}

// If a joke has been submitted,
// add it to the database.
if ("SUBMIT" == $submitfont) {
  $sql = "INSERT INTO zfonts SET " .
         "fontname='$fonttext', " .
		 "fontaddress='$fontaddress', " .
		 "fontimage='$fontimage', " .
         "fontdate=CURDATE()";
  if (mysql_query($sql)) {
    echo("<P>Your font has been added.</P>");
  } else {
    echo("<P>Error adding submitted font: " .
         mysql_error() . "</P>");
  }
}

echo("<P> Here are the fonts " .
     "in our Z Fonts database: </P>");

// Request the text of all the jokes
$page_number = (is_numeric(@$_GET['page']) ? $_GET['page'] : 1);
$result = mysql_query(
          'SELECT fontname FROM zfonts ORDER BY fontname ASC LIMIT ' . (($page_number - 1) * 20) . ', 20');
if (!$result) {
  echo("<P>Error performing query: " .
       mysql_error() . "</P>");
  exit();
}

// Display the text of each joke in a paragraph
while ( $row = mysql_fetch_array($result) ) {
  echo("<BR>" . '<a href="http://1-800-fonts.com/winfonts/Z-Fonts/' . $row["fontname"] . '.zip"><img src=font.php?text=' . $fonttext . '&amp;font=/Z-Fonts/' . $row["fontname"] . '&amp;size=36&amp;color=#ff3333></a><BR>');
}

/* Set current, prev and next page */
$page = (!isset($_GET['page']))? 1 : $_GET['page'];
$prev = ($page - 1);
$next = ($page + 1);

/* Max results per page */
$max_results = mysql_num_rows($result);

/* Calculate the offset */
$from = (($page * $max_results) - $max_results);

/* Query the db for total results. You need to edit the sql to fit your needs */
$result = mysql_query("select fontname from zfonts");

$total_results = mysql_num_rows($result);

$total_pages = ceil($total_results / $max_results);

$pagination = '';

/* Create a PREV link if there is one */
if($page > 1)
{
$pagination .= ' <a href="zfonts.php?page='.$prev.'">Previous</a> ';
}

/* Loop through the total pages */
for($i = 1; $i <= $total_pages; $i++)
{
if(($page) == $i)
{
$pagination .= $i;
}
else
{
$pagination .= ' <a href="zfonts.php?page=' . $i . '">' . $i . '</a> ';
}
}

/* Print NEXT link if there is one */
if($page < $total_pages)
{
$pagination .= ' <a href="zfonts.php?page=' .$next . '&amp;' .$fonttext . '">Next</a>" ';
}

/* Now we have our pagination links in a variable($pagination) ready to print to the page. I pu it in a variable because you may want to show them at the top and bottom of the page */

/* Below is how you query the db for ONLY the results for the current page */


while ($i = mysql_fetch_array($result))
{
/* This is where you print your current results to the page */
}

echo('<p><center>' . $pagination);

echo('<p>' . $abclinks . '</center>');

?>