PHP - How Do I Get Display Answer In An Input Element?

Hi all,

I want to be able to have one file (configuration file i guess) where i set the phone number, address, staff names and positions, etc.  and then have them display at different places on various pages.   I thought they were inc files, but all i can find is the likes of headers and footers.  Any help much appreciated.  I just don't know what it's called, if someone can point me in the right direction, thanks.

MsKazza

<?php

sorry for posting..i already make it

Hi,

I have a calculation issue in my php. Below is my code:

Code: [Select]
$output = "";
$studentId  = false;
$courseId  = false;
$moduleId  = false;
       
while ($row = mysql_fetch_array($result)){   

        $moduletotal += $row['Mark'];
       
        $modulemark = (int)($moduletotal);
         
//Above here is the calculation for add in the Marks of all the sessions

//Below is the code of the display of the query output


if($studentId != $row['StudentUsername'])  {       

$studentId = $row['StudentUsername'];       
$output .= "<p><strong>Student:</strong> {$row['StudentForename']} {$row['StudentSurname']} ({$row['StudentUsername']})";   

   }   
   
    if($courseId != $row['CourseId']) {   

$courseId = $row['CourseId'];       
$output .= "<br><strong>Course:</strong> {$row['CourseId']} - {$row['CourseName']} <br><strong>Year:</strong> {$row['Year']}</p>";   

}   

if($moduleId != $row['ModuleId'])  {       

   $moduleId = $row['ModuleId'];       
   $output .= "<p><br><strong>Module:</strong> {$row['ModuleId']} - {$row['ModuleName']} $modulemark</p>";   

  } 
   
   $output .= "<p><strong>Session:</strong> {$row['SessionId']} {$row['Mark']}</p>";
   
  }


Student: Mayur Patel (u0867587)
Course: INFO101 - Bsc Information Communication Technology 
Year: 3


Module: CHI2550 - Modern Database Applications 72 (72 is the answer to the calculation for the first module but this is incorrect at it should also add the 67 and thus become 139)

Session: AAB 72

Session: AAE 67

Module: CHI2513 - Systems Strategy 200 (200 is the answer to the calculation for this module but this is incorrect it should be only 61. But what it has done is that it has added the 72 and 67 from the first module and added it with the 61 in this module)

Session: AAD 61



SO WHAT MY QUESTION IS THAT HOW CAN I FIND THE TOTAL OF EACH MODULE MARK WHICH THE VARIABLE $modulemark = (int)($moduletotal); IS TRYING TO DO BY ADDING UP ALL THE SESSIONS MARKS WITHIN EACH MODULE.

I HAVE TRIED USING SUM(gr.Mark) in the query and using GROUP BY SessionId, ModuleId, StudentUsername, and CourseId but it ends up not being able to display any records. That is why I want to do the calculation using php rather than SQL.

BELOW IS THE QUERY IF YOU WANT TO SEE IT:

Code: [Select]
$query = "
          SELECT st.CourseId, c.CourseName, st.Year, st.StudentUsername, st.StudentForename, st.StudentSurname,
          s.ModuleId, m.ModuleName, m.Credits, s.SessionId, s.SessionWeight, gr.Mark, gr.Grade
          FROM Course c
          INNER JOIN Student st ON c.CourseId = st.CourseId
          JOIN Grade_Report gr ON st.StudentId = gr.StudentId
          JOIN Session s ON gr.SessionId = s.SessionId
          JOIN Module m ON s.ModuleId = m.ModuleId
          WHERE
          (st.StudentUsername = '".mysql_real_escape_string($studentid)."')
          ORDER BY c.CourseName, st.StudentUsername, m.ModuleName, s.SessionId
          ";

Hi
I made a curl and don't know what to do with the string I got back, I would need the id and the name of each, should I explode, don't know how to do that, here is what I got as answer:

Quote

string(596) "{"data":[{"name":"Esoterik Forum","category":"Community","id":"180883525283576"},{"name":"Hellsehen und Wahrsagen","category":"Company","id":"177967172214131"},{"name":"Spirituelle Lebensberatung","category":"Website","id":"164411270264299"},{"name":"Esoterik","category":"Website","id":"164524456919432"},{"name":"Kartenlegen und Hellsehen","category":"Company","id":"173933135951523"},{"name":"Exchange","category":"Application","id":"190533567633445"},{"name":"Astrologie","category":"Website","id":"164928140213920"},{"name":"Esoterik Forum","category":"Application","id":"184577371575355"}]}"

I just finished a page of code that passes all the debug tests except for the very last line on the page. </html>. The page has php and html, java script and some SQL. (see I'm learning and hadn't placed a post until I exhausted my limited resources.

The Error Codeis:
PHP Parse error: syntax error, unexpected $end in C:\wamp\www\registerform.php on line 292

Here's my concern:

I read this error statement as an error()is missing. Please clarify something for me, You start the php code with <?php and end it with ?>.

 I also understand that the end() will stop all further operation. Correct. When you have a form call itself  for data checking,   and if the data is correct, send it to the appropriate database table, does one need to place the end() at the end of the php portion of the code? And if so, will it not stop any of the code from being read the first time the page is called?

Also since the FORMs action calls for the page to load itself for data verification, how at the end of the checking do I go call another page?
here's the page:

The page is attached:

I have a function that creates a dropdown list in a form.  The User saves their answer to the mysql table.  How do I get their preselected answer back out of the table when they visit the form again?
I know how to pull it from mysql and into a variable, but how to modify the below code to display the correct answer stumps me.

Here is my dropdown function:
Code: [Select]
function dropdown($array, $id) {
  echo '<select name="'.$id.'" id="'.$id.'" class="select"><option value="">Select one...</option>';
  foreach($array as $key => $value) {
  echo '<option value="'.$value.'">'.$value.'</option>';
  }
  echo '</select>';
  }

Thanks for the help!

Hello PHP Freaks forum,

I am a beginner in web development and seem to be understanding it very quickly, but there is one thing that is putting this into some halt.

I have this code right here, what I want to do is have a "riddle site", where if you answer the riddle correctly, it redirects you to another page (in other words to the next riddle).

Object:
If correct, give access to next page.
If incorrect, stay on page and print "Incorrect".

I've tried to start on it, but the way how HTML and PHP don't mix very well, I feel like in the middle of an ocean feeling helpless.

<html>
 <head>
  <title>PHP Test</title>
 </head>
 <body>
 <p>
<font face="Courier New">ZnJ5cyBwYmFnbnZhcnEgaGFxcmVqbmdyZSBvZXJuZ3V2YXQgbmNjbmVuZ2hm</font>
</p>
CLUE: The base is rotated
 <br/></body>
 
<form action="index-5.php" method="post">
  Answer: <input type="text" name="number" /><br />
</select>
<input name="submit" type="submit">
</form>

 <?php
if(isset($_POST['submit'])){

$number = $_POST['number'];
if ($number == "scuba"){
echo "CORRECT";}
}
?>

</html>


If possible, can you give me an explanation how to do it and a sample of it as well?

Thanks,
Gldnbr

I'm trying to search the 'ad' column where 'ad' = 1 or 3

I'm having problems doing this.

Here is what I currently have, but it is not bringing back 'ad' where ='1' (for obvious reasons)
Quote

    $query = "SELECT * FROM users WHERE ad='3' AND state='$state1'";



Here is what I thought would work, but it displays every listing in my database when I do it this way.

Quote

    $query = "SELECT * FROM users WHERE ad='3' OR ad='1' AND state='$state1'";




Thank you in advance for your answers, google is not being so kind to me right now.

I need to make my client's site distinguish among different tabs in the same browser. (See http://www.phpfreaks.com/forums/index.php?topic=357772.0 for background.)

I create a tab ID when the user first visits the site in a particular tab, and I'm passing it from page to page as a parameter. This seems to be the only way to do what I need.

The tab ID is always passed through POST so that the user won't see it. If it were visible, users could make trouble (or more likely get in trouble) by adding or removing the ID themselves.

This is not simple where a page is loaded by a hyperlink. The anchor tag passes parameters via GET, period.

The solution I found is to link to a script named post.php, which takes the real link target and the tab ID as parameters, sends the browser a form that passes the tab ID to the real link target via POST, and auto-submits the form via JavaScript.

This works, but it requires two round trips to the browser to load a page. It's also a pain to code.

Is there a more efficient way to do this... perhaps a way to make the server send the browser a POST response, even though it made a GET request?

If it matters, the server is Apache 2.x.

I am writing a script that will parse my PHP classes and check for things like coupling, visualize my objects and connections, dependencies, check for convention usage, etc.   So, I have a simple file upload.  I'm never saving the files, just get contents and dump the file and work with the string version.   I'm writing it for me, but I figure I might want to open it for others to use in the future, so I may as well write it that way to begin with -- so I need to validate user input.  Problem is, the user input is supposed to be valid PHP code.  I'm thinking that, as long as I'm careful, I shouldn't be executing any code contained in strings, but I'm no security expert and I want a warm fuzzy that my thought on this is correct.  What kinds of things do I need to look out for?  Is it possible to inject when working with strings?   My initial thought is to regex the entire file and replace key portions with known replacements.  So ( and ) would become !* and !^ or $ would become @~ (combinations that -- I think -- don't make sense to php?)  But that may be completely unnecessary processing time if I'm not in any danger, here.  Thanks ahead of time for any help.   PS - as a side question -- what's the best way to verify a file is a php file?  I know of getimagesize for images, but should I just check for <? to verify it's php?  That seems like it would be too easy to fool -- then again, it might not matter much.   -Adam

I hope I can explain what is happening.  I have created two forms in PHP.  The first 'almost' works, i.e. it shows the data.  But I have two problems - 1) the second pulldown menu is always empty and 2) $value from the first pulldown menu ALWAYS equals the last entry thus the last 'if' in the function subdomains ($domains) is always called (but still empty).  The code may explain this better than me:

<!DOCTYPE html>
<html>
<body>
   <!-- processDomains.php is this file - it calls itself (for testing purposes so I can see what is happening) -->
   <form action="processDomains.php" method="post">
   <?php
       // create the domains array (there are actually several entries in the array but I cut it down for testing)
       $domains = array (1 => 'Decommission', 'Migration');
    
       echo "Select Domain:";
       echo "<br>";
    
       // Make the domain pull-down menu - this displays correctly
       echo '<select name="domain">';
       foreach ($domains as $key => $value) {
          echo "<option value=\"$key\">$value</option>\n";
       }
       echo '</select>';

       // input doesn't matter what is 'submitted', always goes to last $value
       echo '<input type="submit" name="submit" value="Submit">';

       // call function subdomains
       subdomains ($value);
    
       function subdomains ($domains) {
           // define values for each array - each array contains available choices for the subdomain pulldown menu
           $migration = array (1 => 'Application Migration', 'Application Patch', 'Application Upgrade');
           $decommission = array (1 => 'Applications', 'Servers', 'Storage');
    
           if ($domains === 'Migration') {
              echo "Select subdomain:";
              echo "<br>";
              // Make the Migration pull-down menu
              echo '<select name="migration">';
              foreach ($migration as $key => $value) {
                  echo "<option value=\"$key\">$value</option>\n";
              }
              echo '</select>';
           } else if ($domains === 'Decommission') {

              /* ===
               * since 'Decommission' is the last entry in the 'Domains' pulldown list, $value ALWAYS equals
               * 'Decommission' and $domains equals $value. So this menu SHOULD work but is always
               * empty.  Thus, two problems - the pulldown menu is always empty and $value isn't based
               * upon user input.
              */
              echo "Select subdomain:";  // this prints so I know I'm in 'Decommission (I eliminated the echo "$domain" to show I'm always coming here)'
              echo "<br>";
              // Make the 'Decommission' pull-down menu
              echo '<select name="decommission">';
              foreach ($decommission as $key => $value) {
                  echo "<option value=\"$key\">$value</option>\n";
              }
              echo '</select>';
              echo '<input type="submit" name="submit" value="Submit">'
           ) // end of 'if-else'
    } // end of function 'subdomain'
     
?>
</form>
</body>
</html>

Let me say thank you in advance and I appreciate the help!  I know I'm doing something (or more than one thing) wrong and I hope someone can tell me what it is.    Best Regards!
Edited by mac_gyver, 19 January 2015 - 09:37 PM.
code tags around posted code please

Below is my output on the browser:

    Student: Kevin Smith (u0867587)
    Course: INFO101 - Bsc Information Communication Technology Course Mark 70 Grade
    Year: 3
   
   
    Module: CHI2550 - Modern Database Applications Module Mark: 41 Mark Percentage: 68 Grade: B
   
    Session: AAB Session Mark: 72 Session Weight Contribution 20%
   
    Session: AAE Session Mark: 67 Session Weight Contribution 40%


    Module: CHI2513 - Systems Strategy Module Mark: 31 Mark Percentage: 62 Grade: B
   
    Session: AAD Session Mark: 61 Session Weight Contribution 50%

Now where it says course mark above it says 70. This is incorrect as it should be 65 (The average between the module marks percentage should be 65 in the example above) but for some stange reason I can get the answer 65. I have a variable called $courseMark and that does the calculation. Now if the $courseMark is echo outside the where loop, then it will equal 65 but if it is put in while loop where I want the variable to be displayed, then it adds up to 70. Why does it do this.
   
    Below is the code:

Code: [Select]
    $sessionMark = 0;
    $sessionWeight = 0;
    $courseMark = 0;
   
    $output = "";
   
    $studentId = false;
    $courseId  = false;
    $moduleId  = false;
   
    while ($row = mysql_fetch_array($result))
    {
   
   
    $sessionMark += round($row['Mark'] / 100 * $row['SessionWeight']);
    $sessionWeight  += ($row['SessionWeight']);
        $courseMark = ($sessionMark /  $sessionWeight * 100);
   
          if($studentId != $row['StudentUsername'])
        {
   
            //Student has changed
            $studentId = $row['StudentUsername'];
   
            $output .= "<p><strong>Student:</strong> {$row['StudentForename']} {$row['StudentSurname']} ({$row['StudentUsername']})\n";
   
        }
             
        if($courseId != $row['CourseId'])
        {
   
            //Course has changed
            $courseId = $row['CourseId'];
   
            $output .= "<br><strong>Course:</strong> {$row['CourseId']} - {$row['CourseName']} <strong>Course Mark</strong>" round($courseMark) "<strong>Grade</strong>  <br><strong>Year:</strong> {$row['Year']}</p>\n";
   
        }
   
        if($moduleId != $row['ModuleId'])
        {
   
            //Module has changed
            if(isset($sessionsAry)) //Don't run function for first record
            {
   
                //Get output for last module and sessions
                $output .= outputModule($moduleId, $moduleName, $sessionsAry);
            }
   
            //Reset sessions data array and Set values for new module
   
            $sessionsAry = array();
            $moduleId    = $row['ModuleId'];
            $moduleName  = $row['ModuleName'];
        }
             
        //Add session data to array for current module
             
        $sessionsAry[] = array('SessionId'=>$row['SessionId'], 'Mark'=>$row['Mark'], 'SessionWeight'=>$row['SessionWeight']);
             
    }    //Get output for last module
   
    $output .= outputModule($moduleId, $moduleName, $sessionsAry);
   
    //Display the output
    echo $output;
I think the problem is that it is outputting the answer of the calculation only for the first session mark. How in the while loop can I do it so it doesn't display it for the first mark only but for all the session marks so that it ends up showing the correct answer 65 and not 72?

I have a calendar select date function for my form that returns the date in the calendar format for USA: 02/16/2012.  I need to have this appear as is for the form and in the db for the 'record_date' column, but I need to format this date in mysql DATE format (2012-02-16) and submit it at the same time with another column name 'new_date' in the database in a hidden input field.  Is there a way to do this possibly with a temporary table or something?

Any ideas would be welcome.

Doug