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

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  • dagmawi michael
    New Member
    • Mar 2011
    • 5
    #1

    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=s m. 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
    Code:
    action="<?php echo $PHP_SELF; ?>
    but did not work out.

    Thank you in advance
    Best Regard's

    this is the code
    Code:
    <?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);

}
?>
    Last edited by dagmawi michael; Mar 27 '11, 10:32 PM. Reason: Title edited (misspell)
    Tags: None
    • dgreenhouse
      Recognized Expert Contributor
      • May 2008
      • 250
      #2
      Although I didn't test this, this may work:
      Code:
      <!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>

        Comment

        • dagmawi michael
          New Member
          • Mar 2011
          • 5
          #3
          hello dgreenhouse
          Thank you for replying.
          It shows parse error in line 36
          Code:
          $rows=mysql_fetch_array($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

            Comment

            • dgreenhouse
              Recognized Expert Contributor
              • May 2008
              • 250
              #4
              $rows=mysql_fet ch_array($resul t) is not terminated.

              Should be:
              $rows=mysql_fet ch_array($resul t);

                Comment

                • dagmawi michael
                  New Member
                  • Mar 2011
                  • 5
                  #5
                  Thanks a million,You were very helpful Dgreanhouse.
                  It works!!

                    Comment

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