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PHP Echo result


fisher7679
Go to solution Solved by Ch0cu3r,

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Still learning...............So I have now been able to make my dropdown menu pass the results to the php page from the data stored in mysql and show the results with echo on the php page. But I would like those results to be displayed on the same page. How would I start going about this?

Edited by fisher7679
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So the form and database code are in two separate files? For example the the html form is in form.php and it submits the form to add_to_database.php.

 

But what you're wanting it do is display the form and database results on same page? If so make the form in form.php submit to itself (<form action="form.php" method="post">) Then  just include add_to_database.php in form.php where you want the results to appear. Example

<?php include 'add_to_database.php'; ?>

<form action="form.php" method="post">
   // your form fields here
</form>
Edited by Ch0cu3r
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Ok, not sure why but the echo results are not showing on the page. Here is the form code. Also I changed it to .php, added the include and set the action to post to itself like stated above.

<form  action="f0rm.php" method="post">
<p align="center" style="margin-bottom: 0">
<b><font size="3">Test</font></b></p>
<p align="center" style="margin-bottom: 0">
<option value=""></option>
<option value=""></option>
<option value=""></option>
<option value=""></option>
</select>
<br />
<br />
</p>
<p align="center">  <input class="button" type="submit" name="submit" value="Submit">
</p>
</form>
<!--?php include 'data.php'; ?-->

Could it be an issue with the echo?

 

Here us the php code I am using for tetsting

<?php 
mysql_connect("", "", "") or die(mysql_error());  
mysql_select_db("") or die(mysql_error()); 

$term1 = $_POST['term1'];
$term2 = $_POST['term2'];
$term3 = $_POST['term3'];
$term4 = $_POST['term4'];
$term5 = $_POST['term5'];
$term6 = $_POST['term6'];
$term7 = $_POST['term7'];
$term8 = $_POST['term8'];
$term9 = $_POST['term9'];
$term10 = $_POST['term10'];

$sql = ("my sql string here");


$rs_result = mysql_query ($sql);
$num_rows = mysql_num_rows($rs_result);
$total_rows = mysql_num_rows($data2);
$result = ($num_rows/($total_rows));
$percent = round($result*100, 4, PHP_ROUND_HALF_UP);
{ 

} 
echo "$percent%";
?> 

Any further help her would be appreciated as I am stumped. Thanks in advance.

Edited by fisher7679
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