PHP - Sums Without Mysql Side Algorithms

Hello,

I have a datepicker that when a date is selected, the post to a php file will get the sum values for that selected day, month, and year.

I have the "by day" working with this...


<?php  
$choice = (isset($_POST['choice'])) ? date("Y-m-d",strtotime($_POST['choice'])) : date("Y-m-d"); 
$con = mysql_connect("localhost","root","xxxxxxxxxx");  
if (!$con)  {  
die('Could not connect: ' . mysql_error());  
}  
mysql_select_db("inverters", $con);  
$sql = "SELECT sum(power/1000) AS choice FROM feed WHERE date = '".$choice."' group by date"; 
$res = mysql_query($sql) or die('sql='.$sql."\n".mysql_error()); 
$row = mysql_fetch_assoc($res); 
echo $row['choice'].'<br />';  
?>

But by month and year are not clicking...

Help Please

Alan

  I’m trying to write a post-panel where the user can see the preview of his post to the right of the
text area he’s writing into.
  I tried the following into a file called writepost.php :
 

<?php

$text=(isset($_GET['message']))?$_GET['message']:'';
$formatted_text=nl2br(stripslashes(htmlspecialchars($text)));

echo '<form method="post" action="proceedtopost.php?>'.
'<table>'.
'<td style width="50%"><tr>'.
'<fieldset> Write your post here : <br> <textarea cols="50" rows="12" '.
'id="message" name="message">'.$text.'</textarea>'.
'</fieldset> <p> '.
'<input type="submit" name="submit" formaction="writepost.php" value="Preview" /> '.
'<input type="submit" name="submit" value="Sent" /> '.
'</p>'.
'</td>'.
'<td>'.
'<fieldset> Your post will appear as follows :<br><p> '.
$formatted_text.
'</fieldset>'.
'</td></tr>'.
'</table>'.
'</form>';

There are several things wrong with this code :

  1) When the user hits the "Preview" button, I expect writepost.php to be reloaded (this is what happens), and the current content of the textarea to be stored in $_GET['message'] (this is not what happens).

  2) Why does my browser output the preview part under the text area (or in other words outputs the HTML table as a single column of two cells), when I insist in my HTML code for the table to be displayed  as a single row ?
 

I have a web portal which is scripted using client side scripting language i.e HTML. I want to create a login page in .php and want the other pages to be in .hmtl.
Can i do so?
I want to do this because i want the login password to be stored in a database. Also I want the password to get encrypted (by either MD5 javascript or OpenSSL ) when user enters it inside the login form.

I cannot create the entire portal in php because my portal makes use of C code. And php doesn't have an interface with C language.

Can a user get directed to a .html page after a secure login from login.php page ?
My main aim is to secure the access to my web portal using a password.

I tried to authenticate the login using javascript where the password was stored in an array. But i feel any one having moderate knowledge can easily break that password.

Any help would be greatly appreciated!!

is it true that all the server side code runs before the client side code when accessing a webpage?  if that is true, what about PHP code that is embedded inside HTML tags?  wouldn't that result in an error a lot of times?

I have a few tables in php and want them to be side by side instead of under each other like below. I have spent about 2 days trying to figure it out, Anyone got a clue?


This is the code im using

<?php
$con = mysql_connect("localhost","user","password");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("People", $con);

$result = mysql_query("SELECT * FROM wang");

echo "<table border='1'>
<tr>
<th>Wang Total Report</th>

</tr>";

while($row = mysql_fetch_array($result))
  {
  echo "<tr>";
  echo "<td>" . $row['got'] . "</td>";
   echo "</tr>";
  }
echo "</table>";

$result2 = mysql_query("SELECT * FROM miah");


echo "<table border='1'>
<tr>
<th>Miah Total Report</th>

</tr>";

while($row = mysql_fetch_array($result2))
  {
  echo "<tr>";
  echo "<td>" . $row['got'] . "</td>";
   echo "</tr>";
  }
echo "</table>";

$result3 = mysql_query("SELECT * FROM vinc");


echo "<table border='1'>
<tr>
<th>Vinc Total Report</th>

</tr>";

while($row = mysql_fetch_array($result3))
  {
  echo "<tr>";
  echo "<td>" . $row['got'] . "</td>";
   echo "</tr>";
  }
echo "</table>";



$result4 = mysql_query("SELECT * FROM ketarie");


echo "<table border='1'>
<tr>
<th>Ketarie Report</th>

</tr>";

while($row = mysql_fetch_array($result4))
  {
  echo "<tr>";
  echo "<td>" . $row['got'] . "</td>";
   echo "</tr>";
  }
echo "</table>";

mysql_close($con);
?> 

Hi guys,

I am wanting to do the same as Blogger.com (aka Blogspot) and Tumblr. And allow people to set their own domain names for "pages" or blog sites.

So if I was to tell them to set their own CNAME (on domain) to domains.mydomain.com

What would be the PHP part to make this work? Or how would I make this work if its not PHP?

Like Blogger and Tumblr it would not be a redirect (to view) (unless I am wrong about this).

Any and all help is much appreciated.

(PS - Am running Apache/PHP and MySQL on dedicated server).

I have a text box and a submit button. I want that when the submit button is pressed, the value on the text box will be checked if it matches the values that are on my server before proceeding, and if they don't match it should give an error report and stop.

All I can make myself is the text box and the submit button, I have tried for hours to find a way to do the rest, but all I can find on the internet are vague explanations and information that a novice like me cannot use. It would be a shame to let me web site that I have worked hard on to go to waste because I cannot do this myself 

Thank you in advance

hello,

since php is run server side, can i launch a program (.exe) on the server from a link?

Hello guys. I was thinking on posting this under the mysql thread. but i think this can be solved by php.. anyway, i want to dump mysql data on the client side whenever a user logs out. how can i do this? thanks in advance.

I am doing server side validations using php and jquery.Its validating properly.But problem is the erros messages are displaying on the top of the page.I want to display side of field.All of the validations store in array.How to display errors to side of the field.

I'm trying to learn how to validate a form using the php server side method. Everything I search is very different and old. Anyone have knowledge of a good tutorial or source that I could use? None of my books have anything about it either!

Thanks!

I'm in the habit of verifying input client side. For example, there is a field to enter a number, and I ensure that it is within a particular range using Javascript.

My backend PHP code does not do such checking... Is this bad? Should I be doing checking using my backend code and send a response back to the client to display a message? Admittedly, I avoid it, since it's a little extra work -- well more extra than just javascripts, IF-THEN-ALERT type statement.

This part of the question is based off of php so I'll put it here of course.

The idea behind this is jquery/ajax is going to get passed an id (eventid) from a dropdown select box inside a form and it's going to run a query to find all the events in my database with that same id. After running the query it's going to list them out and find the highest number of another field (label) and its going to add one to its value and then what I would like for it to is send it back to the form in the jquery/ajax and and place that label inside of the existing input text field in the form.

So I'm asking is if my code is presented and is doing what I"m attempting it to do on the php side and if so if I can take out the print tag of the input text field.


$eventid = $_GET['eventid'];
$result = mysqli_query("SELECT * FROM `events` WHERE `event_id` = '$eventid'");
$list = mysqli_num_rows($result);
$label = $list + 1;
print '<input type="text" name="label" class="text" readonly="readonly" value="' . $list . '" />';

Hi guys,
I'm using this EOD thingi in order to print a table and get it PDF. I am using TCPDF and following their examples to get the PDF output.
I need to print a php variable's value within the EOD.
Can you please help me out.
Below is the piece of code I'm running.

Code: [Select]
$tbl = <<<EOD
<table cellspacing="0" cellpadding="1" border="1">
    <tr>
        <td rowspan="4"><img src="images/m24.jpg"></td>
        <td align="right" width="150">From the desk of</td>
        <td>: $contat</td>
    </tr>
    <tr>
        <td align="right">E-Mail</td>
<td>: $email</td>
    </tr>
    <tr>
       <td align="right">Telephone</td>
   <td>: $telephone</td>
    </tr>
      <tr>
       <td align="right">Mobile</td>
   <td>: $mobile</td>
    </tr>
</table>
EOD;

the $variables are simply ignored. or even if I echo, there is no PHP value.
Can you guys please shed some light on this issue.
Pretty pleeeaaassseee