pietbez

July 19, 2013

thanks, got it

July 19, 2013

can someone please point out where i m going wrong.

this code is working fine, but i need the results in a variable

$result = mysql_query("SELECT * FROM `gifts` WHERE `id` =$id ");

while($row = mysql_fetch_array($result))

  {
  echo  $row['price'] , $row['units'];
  }

so, when i do this, i only get the first variable???

$result = mysql_query("SELECT * FROM `gifts` WHERE `id` =$id ");

while($row = mysql_fetch_array($result))

$price = $row['price'];
$units = $row['units'];

  {
  echo  $price , $units;
  }

please help.

June 22, 2013

sorry for that.

that was a stupid mistake. cant believe i didnt pick that up.

sometimes when you stare at the code to long you start missing things.

thanks for that.

June 22, 2013

here is the form with the radio buttons

thanks

 <?php
 $id = $_REQUEST['id'];
 $guest = $_REQUEST['guest'];
 
 ?>
<html>
<head>
    <title>RSVP</title>
    <link rel='stylesheet' type='text/css' href='style.css' />
</head>

<body style="body">
    <ul id="nav">
     <li><a href="home.php?id=<?php echo $id; ?>&guest=<?php echo $guest; ?>">invitation</a></li>
     <li><a href="rsvp.php?id=<?php echo $id; ?>&guest=<?php echo $guest; ?>">rsvp</a></li>
     <li><a href="menu.php?id=<?php echo $id; ?>&guest=<?php echo $guest; ?>">menu</a></li>
     <li><a href="sleep.php?id=<?php echo $id; ?>&guest=<?php echo $guest; ?>">sleep</a></li>
     <li><a href="gifts.php?id=<?php echo $id; ?>&guest=<?php echo $guest; ?>">gifts</a></li>
    </ul>
 

    <div id="rsvp-wrap">
        <h2><?php echo $guest; ?></h2>
        <h2>Will you be joining us on our special day?</h2>
     <form method="post" action="rsvp_update.php">
      <table>
        <input type="hidden" name="guest" value="<?php echo $guest; ?>">
	<input type="hidden" name="id" value="<?php echo $id; ?>">
        
        <input type="radio" name="attend" value="Yes"/>- Yes yes yes! cant wait<br><br>
        <input type="radio" name="attend" value="Maybe"/>- Too early to say, gotta check my diary. but pencil me in<br><br>
        <input type="radio" name="attend" value="sorry"/>- Would love to be there, but wont be able to make it. Let me buy you a present to make up for it<br><br>
        <input type="radio" name="attend" value="No"/>- No. not gonna bother. never realy liked you guys that much anyway<br><br><br>
        <input type="Submit" value="Submit" />
      </table>
     </form>
    </div>
  </body>
</html>

June 22, 2013

pulling my hair out here, please help.

why am i getting a "maybe!" result no matter what my radio box selection is.

i know the variable gets passed to this php succesfully, i tested it with "echo $attend"

then i get the right results

but when i use this code, my result is always "maybe"

Please hepl

<?php

  $guest = $_REQUEST['guest'];
  $attend = $_REQUEST['attend'];
  $id = $_REQUEST['id'];
  
  include('include/connection.php');
	
  $query="UPDATE data SET guest = '$guest', attend = '$attend' WHERE id = '$id'";
  mysql_query($query) or die ('Error updating database');
  


  if ($attend == "yes")
  {
  echo "yes!";
  }
  else if ($attend == "no")
  {
  echo "no!";
  }
  else
  {
  echo "maybe!";
  }
  
?>

April 4, 2012

Hi, i know this is probably very basic but i have been banging my head and looking for tuts.

i have built a mysql php dropdown menu.

all displays fine. now, how do i get the menu to actualy take me to a new url?

the new url should be www.mysite.com/"menu selection"

<?

include_once 'includes/db.php';

$result = mysql_query("select * from crimerate WHERE DISTRICT = 'Limpopo'", $con);

if (!$result) {
    die('Invalid query: ' . mysql_error());

}


$options="";

while ($row=mysql_fetch_array($result)) {

    $id=$row["id"];
    $crime=$row["CRIME"];
    $options.="<OPTION VALUE=\"$id\">$crime</option>";
}
?>


<SELECT NAME=crime>
<OPTION VALUE=0>Choose
<?=$options?>
</SELECT>

April 3, 2012

thank you so much,

that is exactly what i was looking for

you saved me alot of pain

April 3, 2012

Hi, can anybody please tell me how i can make "Arson" into a php variable


<script type="text/javascript" src="js/swfobject.js"></script>
<script type="text/javascript">

swfobject.embedSWF(
  "open-flash-chart.swf", "my_chart",
  "550", "400", "9.0.0", "expressInstall.swf",
  {"data-file":"ofc-chart.php?crime=Arson"} );

</script>

April 3, 2009

thanks, its perfect,

slowly i learn

April 3, 2009

no, thanks, its working fine, it was just in responce to your question about

the first link in the menu is active by default

does that have anything to do with ?

April 3, 2009

oops sorry,

http://76.162.122.173/mygreenclean/index.php

here you can see what im atempting to do

April 3, 2009

i know this might sound like a stupid question to all you pros, but to me this is a mountain ???

what would be the sensible way to code this?

<?php

include('qazwsxedc/config.php');

mysql_connect($server,$username,$password);
@mysql_select_db($database) or die ("Unable to connect to the database");


$button1 = mysql_query("SELECT title FROM menu where id=1");
$but1 = mysql_fetch_array( $button1 );

$button2 = mysql_query("SELECT title FROM menu where id=2");
$but2 = mysql_fetch_array( $button2 );

$button3 = mysql_query("SELECT title FROM menu where id=3");
$but3 = mysql_fetch_array( $button3 );

$button4 = mysql_query("SELECT title FROM menu where id=4");
$but4 = mysql_fetch_array( $button4 );


?>           
<li><a href="#"><span><?php echo $but1['title'];?></span></a></li>
<li><a href="#"><span><?php echo $but2['title'];?></span></a></li>
<li><a href="#"><span><?php echo $but3['title'];?></span></a></li>
<li><a href="#"><span><?php echo $but4['title'];?></span></a></li>

January 25, 2009

i tried this, but no luck.

still getting the same results. any other ideas?

January 22, 2009

yes, thanks for all the help.

my php and flash is now getting along like a house on fire (with a few modifications from my side .)

January 22, 2009

This forum is to "help" people with problems and advance their knowledge, not to do it all for you. The code provided would be simple to modify. Did you even try to modify it for your needs?

You state that you need the data in an array, but your example shows the data in a delimited string. Which is it?

Assuming the second:

im sorry if you feel my post does not belong here, but i am keen to learn php. all my background is flash actionscript.

and yes i did try to modify it for my needs. you have to remember, what looks obvious to an expert can be a mountain for a noob

January 22, 2009

wow! finaly! success!

i have a populated table. thanks to everyone who helped.

but now i have one more problem.

i dont need the output in table form.

i am sending this data to a flash file and the flash needs the data in an aray. this format to be precise.

i know i am a pain in thr arse!

January 21, 2009

OK, to sum up.

i will show various examples of code that people have suggested. but unfortunatly all with similar results.

No1

<?php

include("../inc/db.inc.php"); 


  $query = mysql_query("SELECT * FROM favourite WHERE user_name = 'jojo'");
  while($row = mysql_fetch_row($query))
  {
      $songInfo = mysql_query("SELECT * FROM songs WHERE song_id = '{$row['song_id']}'");
      $songInfo = mysql_fetch_row($songInfo);
      
      echo $songInfo["song_title"] ." / ". $songInfo["artist"] ." / ". $songInfo["year"] ."\n";
  }
  
?>

and this is the output i get:

/ / / / / / / / / /

No2.

<?php

include("../inc/db.inc.php"); 

$query = mysql_query("SELECT favourite.user_name, songs.song_title, songs.artist, songs.year FROM songs,favourite WHERE songs.song_id=favourite.song_id AND favourite.user_name='jojo'

");
  while($row = mysql_fetch_row($query))
  {
  $song_title.=$row["song_title"].'|';
  $artist.=$row["artis"].'|';
  $year.=$row["year"].'|';
}
echo "&song_title=".$song_title."&artist=".$artist."&year=".$year;
?>

and this is the output i get:

&song_title=|||||&artist=|||||&year=|||||

No3.

<?php

include("../inc/db.inc.php"); 

$query = mysql_query("SELECT *
FROM favourite
JOIN songs USING (song_id)
WHERE user_name = '$user_name'");
  while($row = mysql_fetch_row($query))
  {
  $song_title.=$row["song_title"].'|';
  $artist.=$row["artist"].'|';
  $year.=$row["year"].'|';
}
echo "&song_title=".$id."&artist=".$name."&year=".$votes;
?>

and this is the output i get:

&song_title=&artist=&year=

any ideas?

January 20, 2009

Any one know where i can get a small and simple, realtime currency converter. either php or even flash/php.

oh yeah, FREE if pos

January 20, 2009

how can i change the name of the uploaded file by adding a variable to the name.

for example. picture.jpg becomes 1_ picture.jpg

<?php
move_uploaded_file($_FILES['Filedata']['tmp_name'], "./links/".$_FILES['Filedata']['name']);
chmod("./links/".$_FILES['Filedata']['name'], 0777);
?>

January 17, 2009

and here is what i get this time.

SONG ID Username

as if my fields are empty. but they are not.

i think im starting to go mental! i know it must be a small thing

January 17, 2009

$query = mysql_query("SELECT favourite.user_name, songs.song_title, songs.artist, songs.year FROM songs,favourite WHERE songs.song_id=favourite.song_id AND favourite.user_name='jojo'

");
  while($row = mysql_fetch_row($query))
  {
  $Song_title.=$row["Song_title"].'|';
  $artist.=$row["artis"].'|';
  $year.=$row["year"].'|';
}
echo "&Song_title=".$Song_title."&artist=".$artist."&year=".$year;
?>

and this gives me

&Song_title=|||||&artist=|||||&year=|||||

January 17, 2009

<?php
  $query = mysql_query("SELECT * FROM favorites WHERE user_name = 'username'");
  while($row = mysql_fetch_row($query))
  {
      $songInfo = mysql_query("SELECT * FROM songs WHERE song_id = '{$row['song_id']}'");
      $songInfo = mysql_fetch_row($songInfo);
      
      echo $songInfo["song_title"] ." / ". $songInfo["artist"] ." / ". $songInfo["year"] ."\n";
  }
  
?>

this is the output im getting

/ / / / / / / / / / / /

how can that be

January 17, 2009

i need some help with a mysql query

i know this should be quite simple but its to much for my little brain.

one table holds song information like:

table name=songs:

song_id, song_title, artist, year.

one table holds user favourites like:

table name=favourites:

user_name, song_id

so by making the user_name available to php, i want to be able to list that user's favourite songs plus song title artis and year.

so if i give the php a variable of the user name: jojo, the output should be

the wall / pink floyd / 1978

satisfaction / the stones / 1974

hey jude / the beatles / 1976

please can some one help me out here

January 16, 2009

ok, i have figgured something out.

my flash file calls this php in order to populate my flash file with pictures.

it looks like it is this php that messes with the links on main page.

could that be possible?

if i remove the echo from this php file everything works fine, appart ofcource the flash file that does not get populated with pictures

$category=intval($_POST['category']);
$do=mysql_query ("SELECT * FROM items WHERE category = $category ORDER by votes DESC" );
$x=mysql_num_rows($do);
if ($x>0) {
while ($row = mysql_fetch_array($do, MYSQL_ASSOC)) {
  $th.=$row["thumb_pic_path"].'|';} 
}

echo "&total=".$x."&thumbs=".$th;
?>

January 15, 2009

i have added a flash file to my php page and somehow its messing around with the php. some links and buttons on that page stopped workning. has any one ever encountered this?

here is the flash code

<script language="javascript" type="text/javascript">
if (AC_FL_RunContent == 0) {
	alert("This page requires AC_RunActiveContent.js.");
} else {
	AC_FL_RunContent(
		'width', '1000',
		'height', '100',
		'movie', 'cat_thumbs',
            'flashvars','category=<?php echo $item_rec[6];?>', ''
		);
}
</script>

here is an example of a page without the flash. the vote button is working

http://www.officiallythehottest.com/car/ferrari/

here is an example of a page with the flash. the vote button is broken and links are broken

http://www.officiallythehottest.com/car/bugatti-veyron/

any idea what the flash could be doin to my otherwize working php?

Sign In

pietbez

Posts

Joined

Last visited

Content Type

Profiles

Forums

Posts posted by pietbez

store result in variable

store result in variable

if else statement

if else statement

if else statement

drop down menu

passing php variable

passing php variable

[SOLVED] dynamic menu

[SOLVED] dynamic menu

[SOLVED] dynamic menu

[SOLVED] dynamic menu

url link code

[SOLVED] imposible query

[SOLVED] imposible query

[SOLVED] imposible query

[SOLVED] imposible query

currency converter

[SOLVED] image upload

[SOLVED] imposible query

[SOLVED] imposible query

[SOLVED] imposible query

[SOLVED] imposible query

[SOLVED] flash messing with php

[SOLVED] flash messing with php

Browse

Activity

Important Information