Queried Links.......

tuf-web_co_uk

I want to display the id of a tone from my database, and put it into an automated link

so that every time a record is found from a category its id is printed into a link which is submitted to a different page where the tone is displayed according to the id variable.

eg.

<a href=tone_display.php?id=23>Tones Name</a>
<a href=tone_display.php?id=46>Tones Name</a>

Ofcourse the ids wont be in order 1 2 3 4 5 etc as different tones will be placed inside the database at different times

using/editing this code

$sql = "select * from tones where type = '$cat' AND artist LIKE '$letter%'";

// Request Tone

$result = mysql_query($sql);  

if (!$result) {    

echo("<P>Error performing query: " .
mysql_error() . "</P>");    

exit();  }

  // Display Tone Link

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

  echo ("<P><font size=1 face=Verdana color=#000080><b>");
  echo "<a href=tone_display.php?id=";
  echo ("".$row["id"]."");
  echo ("></a></P>");
  }

how can i make it work because at the moment nothing shows up

Thanks In Advance 🙂

Duey

instead of using if(!$result... use


$result = mysql_query($sql) or die(mysql_error());

tuf-web_co_uk

that didn't seem to work 😕

u sure its likes this,

<?php    

$sql = "select * from tones where type = '$cat' AND artist LIKE '$letter%'";

// Request Tone

$result = mysql_query($sql) or die(mysql_error());

  // Display Tone

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

  echo ("<P><font size=1 face=Verdana color=#000080><b>");
  echo "<a href=tone_display.php?id=";
  echo ("".$row[id]."");
  echo ("></a></P>");
  }
include ("includes/footer.php");

?>

dont i need to change anything else in the code

xblue

Hi,

if nothing at all is displayed (not even in the source code), then a) your query is successful, because no error shows up, but b) the while loop is not entered even once, which would mean the query is successful, but returns an empty recordset.

I'd try to echo $sql to see what it looks like.

tuf-web_co_uk

i tried that echo $sql like u said but it just prints

select * from tones where type = '$cat' AND artist LIKE '$letter%'

then i got rid of the link

  echo ("<P><font size=1 face=Verdana color=#000080><b>");
  echo "<a href=tone_display.php?id=";
  echo ("".$row[id]."");
  echo ("></a></P>");

and used this instead to print the id

  echo ("<P><font size=1 face=Verdana color=#000080><b>");
  echo ("".$row[id]."");

and this worked! but why wont it work with my link ?! 😕

please help

tuf-web_co_uk

never mind i got it now, i forgot to put a bit of text in where the link goes

oh well i'm stupid :S

Thanks anyways!! 🙂