how to show this php code submit result at the same page

hello
I found this code somewhere and tried to create online dictionary with it. I managed to insert the code in to my template index.php. But after the user click submit button the result shows in other blank self created page something like index.php?act=sm. I need the results to show on the same page or somewhere with in the same template. I' ve tried every thing I can. Just I couldn't.how can I do that?

I add

Expand|Select|Wrap|Line Numbers

action="<?php echo $PHP_SELF; ?>

but did not work out.

Thank you in advance
Best Regard's

this is the code

Expand|Select|Wrap|Line Numbers

 <?php

echo("<a href=admin.php>ACP</a><br>");
 
if(isset($_GET['act']))

{

$do=$_GET['act'];

switch($do)

{

 case 'form': show_form();break;

 case 'sm' : submited();break;

 default : show_form; break;

}
 
}

else

{

show_form();

}
 
function show_form()

{

?>

<form name=frm method=POST action="index.php?act=sm">
 
Word : <input type=text name="word"> <input type=submit name="sbm" value="Submit">

</form>

<?
 
}

function submited()

{
 
require("dbconn.inc");

//if(isset($_POST['sbm']))

//{

 $word=$_POST['word'];
 
 $sql="select * from words where eng like '".$word."'";

 //echo($sql);

 $result=mysql_query($sql,$link);

 if(@mysql_num_rows($result)!=0)

 {

  while($rows=mysql_fetch_array($result))

  {

   $vie=$rows["vie"];

  }

  echo("Word<b>$word</b> in vietnamese is : <b>$vie</b><br>");

?>

<form name=frm method=POST action="index.php?act=sm">
 
Word : <input type=text name="word"> <input type="submit" name="sbm" value="Submit">

</form>

<?

 }

 else

  echo("Don't know that word !! My database is not update!");

//}

mysql_close($link);
 
}

?>

Mar 27 '11 #1

Subscribe Post Reply

22501

dgreenhouse

250

Expert 100+

Although I didn't test this, this may work:

Expand|Select|Wrap|Line Numbers

 
<!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>Untitled Document</title>

</head>

<body>

<?php

echo("<a href=admin.php>ACP</a><br>");
 
$script = $_SERVER['PHP_SELF'];
 
if(isset($_POST['sbm']) && isset($_POST['word']) ) {

  submited();

} else {

  show_form();

}
 
function show_form() {

?>

  <form name=frm method=POST action="<?php echo $script ?>">

    Word : <input type=text name="word"><br />

    <input type=submit name="sbm" value="Submit">

  </form>

<?php

}
 
function submited() {

  require("dbconn.inc");

  $word=$_POST['word'];

  $sql="select * from words where eng like '".mysql_real_escape_string($word)."'";

  $result=mysql_query($sql,$link); // I assume $link is defined in dbconn.inc
 
  if(@mysql_num_rows($result)!=0) {

    $rows=mysql_fetch_array($result)

    $vie=$rows["vie"];

    echo("Word<b>$word</b> in vietnamese is : <b>$vie</b><br>");

  } else {

    echo("Don't know that word !! My database is not update!");

  }
 
  // The connection will close automatically since

  // this is near the end of the script.

  mysql_close($link);
 
  show_form();

}    

?>

</body>

</html>

Mar 28 '11 #2

dagmawi michael

hello dgreenhouse
Thank you for replying.
It shows parse error in line 36

Expand|Select|Wrap|Line Numbers

 $rows=mysql_fetch_array($result)

    $vie=$rows["vie"];

When I add while loop here it brings me to the same page after clicking submit button but the results are not showing.

best regard's

Mar 28 '11 #3

dgreenhouse

250

Expert 100+

$rows=mysql_fetch_array($result) is not terminated.

Should be:
$rows=mysql_fetch_array($result);

Mar 28 '11 #4

dagmawi michael

Thanks a million,You were very helpful Dgreanhouse.
It works!!

Mar 28 '11 #5

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how to show this php code submit result at the same page

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