Jump to content

could anyone check this coding for me pls?


mtvaran

Recommended Posts

could anyone please help me with the code which is i have already displayed data from two table as a drop-down list and a multi select list but now i need to select one from  drop-down list and one or more from multi select list  the insert into another database table.

here is the coding. it shows no error message( blank page after submit) and nothing inserted on the take able

<?php 

$con = mysql_connect("localhost","root",""); 
if (!$con) 
  { 
  die('Could not connect: ' . mysql_error()); 
  } 
error_reporting(E_ALL); 
ini_set('display_errors', 1);  

  mysql_select_db("uni", $con)or trigger_error('MySQL error: ' . mysql_error()); 



$result  = mysql_query("SELECT * FROM course") or trigger_error('MySQL error: ' . mysql_error()); 


echo '<select name ="cid[]" multiple="multiple" size="10">'; 

while($row = mysql_fetch_array($result)) 
{ 
    echo '<option value="' . $row['CourseID'] . '">' . $row['CourseName'] . '</option>'; 
} 

echo '</select>'; 

// ---------------- 

$result  = mysql_query("SELECT * FROM student") or trigger_error('MySQL error: ' . mysql_error()); 


echo '<select name="sid">'; 

while($row = mysql_fetch_array($result)) 
{ 

    echo '<option value="' . $row['StudentID'] . '">' . $row['StudentName'] . '</option>'; 
} 

echo '</select>'; 


mysql_close($con); 

?> 
<form id="form1" name="form1" method="post" action="update_result.php"> 
  <label> 
  <input type="submit" name="Submit" value="Submit" /> 
  </label> 
</form> 
</body> 
</html>

 

 

update_result.php

<?php 

$con = mysql_connect("localhost","root",""); 
if (!$con) 
  { 
  die('Could not connect: ' . mysql_error()); 
  } 
  
  mysql_select_db("uni", $con)or trigger_error('MySQL error: ' . mysql_error()); 
   
  

if (!empty($_POST['sid']) && !empty($_POST['cid'])) 
{ 
    foreach ($_POST['cid'] as $key => $course) 
    { 
     $courses .= $course."-"; 
     } 
    $courses = rtrim($courses,"-"); 
    $student = $_POST['sid']; 
    $sql = "INSERT INTO test (StudentID, CourseID) VALUES('".mysql_real_escape_string($student)."','".mysql_real_escape_string($courses)."')"; 
    $query = mysql_query($sql) or trigger_error('MySQL error: ' . mysql_error()); 

    if (mysql_affected_rows() > 0) 
    { 
        echo mysql_affected_rows() . ' rows added.'; 
    } 
}   
mysql_close($con); 

?> 
</body> 
</html>

 

 

 

Link to comment
Share on other sites

so you turned on error_reporting in php.ini and restarted apache? then, let's check to make sure error_reporting is on. in a new file called phpinfo.php, place this code, then open the page in your browser:

 

<?php
phpinfo(); 
?>

 

what do you see when you execute that page via browser?

Link to comment
Share on other sites

wow... u were correct... excellent now data is inserted. but it says small error .....

 

foreach ($_POST['cid'] as $key => $course)
    {
     $courses .= $course."-"; //............ this line error
     }

 

 

can you find out it for me plsss

Link to comment
Share on other sites

This thread is more than a year old. Please don't revive it unless you have something important to add.

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

Guest
Reply to this topic...

×   Pasted as rich text.   Restore formatting

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.