Jump to content

techker

Members
 View Profile  See their activity

  • Posts

    812

  • Joined

  • Last visited

 Content Type 

Posts posted by techker

    paganation help please..

    in PHP Coding Help

    Posted

    hey guys i have this script i found it works in all but it does not link to page 2?it shows like 10 results on 2 pages bu the second page is not there?

     

    <? //---------------------------------------------------------------------------------------
////////////////////////////////////////////
//  PAGINATION FUNCTION  //
//  by: Karl Steltenpohl          //
////////////////////////////////////////////
$con = mysql_connect("localhost","t","k"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("techker_softcore", $con);



function pagination($table, $order, $searchstring, $pre, $pos, $nav, $page, $pages)
{
///////////////////////
//  Get Current Url  //
///////////////////////
$webpage = basename($_SERVER['PHP_SELF']);
global $webpage;

////////////////////////
//  Sorter and Pagination Query Begin  //
/////////////////////////////////////////
//$pre = $_REQUEST['pre'];
//$pos = $_REQUEST['pos'];
//$nav = $_REQUEST['nav'];
//$page = $_REQUEST['page'];
//$pages = $_REQUEST['pages'];


///////////////////////////////////////////
//  Set Initial Pre Pos and Page Limits  //
///////////////////////////////////////////
if($pre == "" and $pos == "" and $page == "")
{
$pre = 0;
$pos = 9;
$page = 1;
}


///////////////////////////////
//  User Navigates Previous  //
///////////////////////////////
if($nav == "prev")
{
$pre = ($pre - 10);
$pos = ($pos - 10);
$page = ($page - 1);
}


///////////////////////////
//  User Navigates Next  //
///////////////////////////
if($nav == "next")
{
$pre = ($pre + 10);
$pos = ($pos + 10);
$page = ($page + 1);
}


/////////////////////////////
//  If page number to low  //
/////////////////////////////
if($page < 1)
{
$pre = 0;
$pos = 9;
$page = 1;
}

//////////////////////////////
//  If page number to high  //
//////////////////////////////
if($page > $pages)
{
$pre = 0;
$pos = 9;
$page = 1;
}


//////////////////////////////////////////
//  Select for total number or results  //
//////////////////////////////////////////
$r = "SELECT DISTINCT * FROM $table $searchstring";
$re = mysql_query($r) or die("error 12547");
$nums = mysql_num_rows($re);


////////////////////////////////////////////
//  Select for current displayed results  //
////////////////////////////////////////////
$request = "SELECT DISTINCT * FROM $table $searchstring ORDER BY $order DESC LIMIT $pre, 10";
$result = mysql_query($request) or die("error 25352");
$num = mysql_num_rows($result);


///////////////////////////////////////
//  Determine total number of pages  //
///////////////////////////////////////
$pages = ceil($nums/10);


/////////////////////////////////
//  Create Navigation Display  //
///////////////////////////////// 
$navigation = "
$nums entries on $pages Page(s)<br>
<a href=\"$webpage?page=$page&nav=prev&pre=$pre&pos=$pos&pages=$pages&view=view\">Previous</a> |
Page $page |
<a href=\"$webpage?page=$page&nav=next&pre=$pre&pos=$pos&pages=$pages&view=view\">Next</a><br>
Results $pre - $pos
";

/////////////////////////////////
//  Create Paginagtion Array   //
/////////////////////////////////
// result is the result of the limited query
$pagination = array($navigation, $result, $num);


/////////////////////////////////
//  Return Paginagtion Array   //
/////////////////////////////////
return $pagination;
}//end function



//--------------------------------------------------------------------------------------------
//Here is an example of how to impliment it
//------------------------------------------------------------------------------------------

//this is the search string to the sql db
$searchstring = "WHERE `type` = 'sketch'";

//this is the table being searched
$table = "softcore";

//this is the table field to order the results by
$order = "id";

// call the function
$pagination = pagination($table, $order, $searchstring, $pre, $pos, $nav, $page, $pages);

//this pulls out the display
$navigation = $pagination[0];

//this pulls out the results
$result = $pagination[1];

//this pulls out the num of results
$num = $pagination[2];

//echo the display onto the site
while ($row = mysql_fetch_array($result)) 
	{

	 echo '<th scope="row"><a href="/admin/art/'. $row['pic'] .'" target="_blank" >
<img src="/admin/imgsize.php?w=60&h=50&img=art/'. $row['pic'] .'" border="0" alt="" />
</a>
<th />'; 

	}
echo"<br>$navigation</br>";

//now all you need to do is loop the result based on the num
//each page will display 10 results.?>

     

    link:http://softcoredesign.com/test2.php?page=1&nav=next&pre=0&pos=9&pages=2&view=view

    event calendar

    in MySQL Help

    Posted

    hey guys i have an event calendar script that insert in a databse the event like so:

     

      id  uid  m  d  y  start_time  end_time  title  text 

     

    now i want to do some stats on my maine page.

     

    like upcomming event,# of evevent in the week and month..

     

    so my questions i got the day

     

    SELECT `d` 
FROM pec_mssgs
WHERE `d` >= day( curdate( ) ) 
LIMIT 0 , 30

    how can i select the week?

    [SOLVED] list box

    in MySQL Help

    Posted

    hey guys i have this list box that selects from a database.

     

    as is it only selects the first name.i need it to select both first and last name?

     

    $link = mysql_connect("localhost", "techker_techker", "techker");
mysql_select_db("techker_trainerstool");
  
$query = "SELECT client_first FROM trainers_clients";
$result = mysql_query($query);

print "<SELECT name=client>";
while ($line = mysql_fetch_array($result))

{
foreach ($line as $value)

{

print "<OPTION value='$value'";
}
print ">$value</OPTION>";
}


mysql_close($link);
print "</SELECT>";

     

    i have tryed:

     

    adding another query for last name.but showed first name on one line then last on another line..

     

    select client_fisrt,client_last from...(did not work,only showed the first name)

     

    select (client_fisrt,', ',client_last) no success same as the other

     

    i have tryed lots more combos but no succes or error messages?

     

    is there a way to do it?

    variable

    in PHP Coding Help

    Posted

    All you need do is concatinate it into your query. eg;

     

    $sql = "SELECT * FROM foo WHERE username = '" . $check['name'] ."'";

     

    Or imbed it directly within a double quoted string....

     

    $sql = "SELECT * FROM foo WHERE username = '{$check['name']}'";

     

    thxclose but no cigar..  " . $check['name'] ." i forgot the"'"

     

    what is this called so i can can study on this?

     

     

    variable

    in PHP Coding Help

    Posted

    ok..if i can't find the original $check var..then i want to use it.Can i intergrate it then?

     

    $check['name']

     

    in

    another variable that i can make?

     

    and in a query ?

     

    $fileSQL=" SELECT *

    FROM `trainers_clients`

    WHERE `trainer` = $check['name'] ////here

    LIMIT 0 , 30 ";

    $fileLIST=mysql_query($fileSQL);

     

    variable

    in PHP Coding Help

    Posted

    see my client is using login manager.in a page there is the includes.

     

    so all the variable are alredy done.

     

    like this $check =$CheckSecurity->page_check($USERNAME, $PASSWORD,$dbhost,$dbusername,$dbpass,$dbname);

     

    so when i do $check['name'] it gives me the client login name.

     

    now i need to create a variable that will include the client login name.

     

    $documents=Scheck['name']

    variable

    in PHP Coding Help

    Posted

    hey guys in my script i used session and when a client logs in it shows his name

     

    <? echo $check['name']; ?>

     

    i need to make a config page and i whant to do a variable that will eco the username

    i have tryed this but no success

    $trainer=$check['name'];

    a litle math..

    in Applications

    Posted

    hey guys i have a math prob here..

     

    i need to do a body fat calculator..

     

    i have the number

     

    Body Fat Formula For Men

    Factor 1 (Total body weight x 1.082) + 94.42

    Factor 2 Waist measurement x 4.15

    Lean Body Mass Factor 1 - Factor 2

    Body Fat Weight Total bodyweight - Lean Body Mass

    Body Fat Percentage (Body Fat Weight x 100) / total bodyweight

     

    i need to put it in a form like here  http://www.bmi-calculator.net/body-fat-calculator/

     

    i have no clue how to do math with php?

     

    sorry i found it in javascript.

     

    http://javascript.about.com/library/blfat2.htm

     

    thx anyways..

    if

    in PHP Coding Help

    Posted

    ok i got it going i remove the " " cause it was giving me a white space error..

     

    <? 
// Connects to your Database
mysql_connect("localhost", "l_l", "l") or die(mysql_error());
mysql_select_db("techker_l") or die(mysql_error());

$id = mysql_escape_string($_GET['id']);
$q = "SELECT * FROM cardio_select WHERE id = '$id'";
$res = mysql_query($q);
$row = mysql_fetch_assoc($res);


$plan =$row['simple'];
$plan2 =$row['combo'];


if ( $plan == "on" ) {
echo "<meta http-equiv=Refresh content=2;url=client_cardio.php>";;
}
elseif($plan2 == "on"){
echo "<meta http-equiv=Refresh content=2;url=emplacement.php>";;
} 
?>

     

    is this the write way to do it?

    if

    in PHP Coding Help

    Posted

    sorry fergot to post it..lol

     

    ok so i have 2 questions in this:

    <? 
// Connects to your Database
mysql_connect("localhost", "l", "techker") or die(mysql_error());
mysql_select_db("techker_llls") or die(mysql_error());

$id = mysql_escape_string($_GET['id']);
$q = "SELECT * FROM cardio_select WHERE id = '$id'";
$res = mysql_query($q);
$row = mysql_fetch_assoc($res);



$plan = "$row['simple']";      //is this ok?will it get table simple?
$plan2 = "$row['combo']";

if ( $plan == "on" ) {;         //if that table's value is on then it should get page x
}
elseif($plan2 == "on"){;       //if this one is on then it gets  page b
} 
?>

     

    now it's the getting that page the part were i freeze..

     

    i was looking at fetch but i think it is only for arays.

     

    then i looked at another page i did and the echo 

    echo "<meta http-equiv=Refresh content=4;url=emplacement.php>";

     

    but was woundering if 2 would the script get confused?

    if

    in PHP Coding Help

    Posted

    hey guys i have a script that selects client from a databse.now that client is either in a programme simple or combo.

     

    there is a link on that page that gets a grid with exercises on it.

     

    now i need a script between that process thats like if the client is simple fetch page...if client is combo fetch page x..

    css with echo

    in CSS Help

    Posted

    hey guys i want to to a picture gallery for my cuz.now i made an admin section to upload the pics and it inserts the name and description in a database..

     

    now when i use a php echo witha select all pics from..

     

    it shows all the pics put in one straight line..

     

    is there a way to make it show like 4 images per row..

     

    pic  pic  pic  pic

    pic  pic  pic  pic

     

     

    http://www.humanflytattoo.com/tattoo.php  (like this)

    form insert help.

    in PHP Coding Help

    Posted

    first off i ma not sure if you mean radio instead of check boxes but in anycase you have 2 problems

    1. you requesting $_POST['pic']; which is the ID not the name.. the name is "select"

    2. it will only take the last ticked value..

     

    so to fix this.. change

    <input type="checkbox" name="select" id="pic" value='<? $row['name'] ?>'/>

    to

    <input type="checkbox" name="pic[]" id="pic" value='<?php $row['name'] ?>'/>

     

    and then

    change

    $pic=$_POST['pic'];

    to

    $pic=$_POST['pic'][0]; //assuming you only tick 1 box

     

    so what would be the value?i need the named of the image that is ticked.

    whats the [0] added?

    form insert help.

    in PHP Coding Help

    Posted

    hey guys im trying to do a select pic to insert the name in a database.

     

    the insert works but it does not insert the pic name only an id..

     

    i can't seem to figure it out..

     

    <? $dbh = mysql_connect("localhost","user","pass") or die("There was a problem with the database connection.");
    $dbs = mysql_select_db("db", $dbh) or die("There was a problem selecting the categories.");

$type=$_POST['type'];


$sql = "SELECT * 
FROM `gymball` 
WHERE `type` = '$type' ";
$fileLIST=mysql_query($sql);?>

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>GymGraph</title>
</head>

<body>


<p align="center"><u>Sélectionner UNE photo</u></p>
<table width="346" border="0" align="center" cellpadding="0" cellspacing="0">
  <tr><? while($row = mysql_fetch_array($fileLIST)) { ?>
    <td width="173"><?php echo '<a href="/Gymgraph/Gymgraph/gymball/'. $row['name'] .'" target="_blank">
<img src="/Gymgraph/Gymgraph/gymball/'. $row['name'] .'" border="0" alt="" width=115/>
</a> 
<br />'; ?></td>
    <td width="173"><label>
      <input type="checkbox" name="select" id="pic" value='<? $row['name'] ?>'/>
    Selection de photo</label></td>
  </tr>
  <? } ?>
</table>
<p>
  <p>
    <label>
    <div align="center">
      <select name="location" id="location">
        <option value="pic1">1</option>
        <option value="pic2">2</option>
        <option value="pic3">3</option>
        <option value="pic4">4</option>
        <option value="pic5">5</option>
        <option value="pic6">6</option>
        <option value="pic7">7</option>
        <option value="pic8">8</option>
      </select> 
      Emplacement de la photo
    </div>
    </label>
  </p>
  <label>
  <div align="center">
    <input type="submit" name="submit" id="submit" value="Submit" />
  </div>
  </label>
  <div align="center">  Envoyer la sélection
  </div>
</p>
</form>
</body>
</html>

     

    and this is the insert pic

    
<?php
     
    
$pic=$_POST['pic'];
    
$location=$_POST['location'];
     
     
     
    
mysql_connect("localhost", "user", "t") or die(mysql_error()) ;
    
mysql_select_db("techker_gymgraphpics") or die(mysql_error()) ;
     
     
     
    
//Writes the information to the database
    
mysql_query("INSERT INTO $location (name) ".
    
        "VALUES ('$pic')");
     
     
    
//Tells you if its all ok
    
$id= mysql_insert_id();
    
    		echo "<p>This file has the following Database ID: <b>$id</b>";
    
echo "You'll be redirected to Home Page after (4) Seconds";
    
          echo "<meta http-equiv=Refresh content=4;url=gymball_select.php>";
     
    
?>
    

						
    
					
    
					
						
    
							

	
    	
		
    
			
				
				
    
					
				
    
			
			
			
		
    
	
    

						
    
					
				
    		
			
    
		
    
	
    26. 

			

		

	
	

		


	
	

	





			

		


	




						


					

					


					

				

			

			
		

		
		


	

		×
		

			
				

					
				

			

			

			

			
		

	






	

		×
		

			
    
				
  • Create New...
    • 
				
			

		

	



		






















	

		

			
Important Information

			
We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.