how to output query - so many differences

[rayzun]

How is it possible to echo the result after my query??

$query = "select *
FROM list
WHERE DIFFERENCE(text, ‘$models’) >= 1
ORDER BY DIFFERENCE(text, ‘$models’) DESC";

echo "$result<br>"; ???

um
text is a field

Thanks,
Terry

[thorpe]

[man]mysql_result/man

[rayzun]

I had tried something similar but
how come for this particular query I get a warning.

Warning: Wrong parameter count for mysql_result()

$query = "select * FROM list WHERE DIFFERENCE(text, ‘$models’) >= 1
ORDER BY DIFFERENCE(text, ‘$models’) DESC";

$result = mysql_query($query);
$number = mysql_numrows($result);
{
for ($i=0; $i<$number; $i++) {
$text=mysql_result($result,$i,"text");

echo "$text<br>";
}}

Yet this query works fine...

$query = "select * FROM list WHERE SOUNDEX(text) = SOUNDEX('$models')";

$result = mysql_query($query);
$number = mysql_numrows($result);
{
for ($i=0; $i<$number; $i++) {
$text=mysql_result($result,$i,"text");

echo "$text<br>";
}}

[leatherback]

Hey Rayzun,

This is an example of how I typically retrieve data from a table, I then put the output in a variable, and use it in a later phase of the page assembling.
This script creates a drop-down list of citynames from a database.
You can instead of storing the output, also choose to echo the data straight to screen:

//********** Select all cities and create drop-down menu **********

 $plaatsquery = "select * from plaats order by plaatsnaam";
   $plaatsresult = mysql_query($plaatsquery) or die(mysql_error());
   while ($row = mysql_fetch_array($plaatsresult))
     {  // Instead of storing the data, you could use "echo $row[plaatsnr];"
     $plaatsmenu .= "<option $plaatsselect value=\"$row[plaatsnr]\">$row[plaatsnaam]</option>";
     }