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How to display the select result at the same page ?

Posted on 2002-05-18
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Last Modified: 2008-03-10
Hi,

I'm now doing a searching program, and I wish to display the search result at the same page where the searching keyword is on top and the result result is at the bottom of the page. What is the best way of doing it ?? Please provide me some sample code. Thanks..
p/s : Is it possible to do it without using frame ??
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Question by:suelow
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4 Comments
 
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Accepted Solution

by:
lokeshv earned 300 total points
ID: 7019290
of course its possible...

check this ..it will give u idea how to do it..
//my form
<html>
<body>
<form method=post action=<?=$PHPSELF?>>
<input type=text name=search_me><br>
<input type=submit name=submit value=submit>
</form>
</body>
</html>

//now to display the results...
<?
if($submit){
echo "You searched for ".$HTTP_POST_VARS[search_me];
}
?>

Hope this helps..

Lk



0
 

Author Comment

by:suelow
ID: 7021097
Hi lokeshv,

I have tried the following code .. but there is an error
"Parse error: parse error in /usr/local/apache/htdocs/optics/TEST.PHP on line 54"

and I dun know why how to fix the error. Coz I have check the code. It seems there is no syntax error.. pls advice.


Code:
-----------------
<html>
<head>
<title>Testing</title>
</head>
<body bgcolor="#FFFFFF">
<form method="post" action="<?php echo $PHP_SELF?>">
  <p>From Supplier :
    <input type="text" name="fromSupp">
    To Supplier :
    <input type="text" name="toSupp">
  </p>
  <p>From Nd :
    <input type="text" name="fromNd">
    To Nd :
    <input type="text" name="toNd">
  </p>
  <p>From Vd :
    <input type="text" name="fromVd">
    To Vd:
    <input type="text" name="toVd">
    <input type="submit" name="submit" value="Submit">
  </p>
</form>

<?php

if($submit)
{
$db = mysql_connect("localhost", "root", "");
mysql_select_db("Optics", $db);

$result=mysql_query("select * from Glass where supplier >='$fromSupp' and supplier <='$toSupp' and nd >= $fromNd and nd <= $toNd and vd >= $fromVd and vd <= $toVd");
$myrow=mysql_num_rows($result);
if ($myrow=mysql_fetch_array($result)){
   echo "<table width 100% style=font-size:11; font-family:arial><tr BGCOLOR=#8099E6>
<td width=8%><div align=center><b><font color=#ffffff>Glass Type</font></b></div></td>
<td width=10%><div align=center><b><font color=#ffffff>Desc</font></b></div></td>
<td width=8%><div align=center><b><font color=#ffffff>Suppplier</font></b></div></td>
<td width=8%><div align=center><b><font color=#ffffff>ND</font></b></div></td>
<td width=8%><div align=center><b><font color=#ffffff>VD</font></b></div></td></tr>";
        do {
          printf("<TR bgcolor=#f7f7f7><TD><div align=center>%s</div></TD><TD><div align=center>%s</div></TD><TD><div align=center>%s</div></TD><TD><div align=center>%.5f</div></TD><TD><div align=center>%.2f</div></TD></TR>\n",$myrow["glass"],$myrow["desp"],$myrow["supplier"],$myrow["nd"],$myrow["vd"]);          
          } while($myrow = mysql_fetch_array($result));
           echo "</table><br>";
}
 
?>

</body>
</html>

 
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LVL 4

Expert Comment

by:lokeshv
ID: 7021108
one curly braces "}" is missing ..

add that bfore ur clossing PHP tah "?>"


Hope this helps..

Lk
0
 

Author Comment

by:suelow
ID: 7022829
HAHA..I got it... thanks !! :-)
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