PHP - Where To Add This Line Of Code For A Successful Link?

I am running an UPDATE query which runs successfully, but my code throws an error saying that the UPDATE did not occur?!   

Specifically, I get a "MEMBER_UPDATE_FAILED_2126"

Here is my code...

// ************************
// Update Last Activity. *
// ************************
if ((isset($_SESSION['loggedIn'])) && ($_SESSION['loggedIn'] == TRUE)){

// Initialize variables.
$loggedIn = 1;
$memberID = 19; // For Test Purposes
// $memberID = (isset($_SESSION['memberID']) ? $_SESSION['memberID'] : '');


// ************************
// Update Member Record. *
// ************************

// Connect to the database.
require_once(WEB_ROOT . 'private/mysqli_connect.php');

// Build query.
$q = "UPDATE member
SET logged_in=?,
last_activity=now()
WHERE id=?
LIMIT 1";

// Prepare statement.
$stmt = mysqli_prepare($dbc, $q);

// Bind variables to query.
mysqli_stmt_bind_param($stmt, 'ii', $loggedIn, $memberID);

// Execute query.
mysqli_stmt_execute($stmt);

// Verify Update.
if (mysqli_stmt_affected_rows($stmt)!==1){
// Update Failed.
$_SESSION['resultsCode'] = 'MEMBER_UPDATE_FAILED_2126';

// Redirect to Display Outcome.
header("Location: " . BASE_URL . "/members/results.php");

// End script.
exit();
}//End of UPDATE MEMBER RECORD

// Close prepared statement.
mysqli_stmt_close($stmt);

// Close the connection.
mysqli_close($dbc);
}//End of UPDATE LAST ACTIVITY


This was working earlier, so what is going on?!

Thanks,


Debbie

So, I have a script that works downloading a file - excerpt below:

Code: [Select]
header("Pragma: public");
header("Expires: 0");
header('Content-type: "application/octet-stream"');
header("Cache-Control: private",false);
header("Cache-Control: must-revalidate, post-check=0, pre-check=0");
$f = "Content-Disposition: attachment; filename=\"".$myfile['downloadname'].".".$myfile['ext']."\"";
header($f);
header("Content-Transfer-Encoding: binary");
header("Content-Length: ".filesize($file));
    ob_clean();
    flush();
readfile("$file");
exit();
I record elsewhere in a mysql database when the script is started.

In addition, I'd like to be able to record whether the download completes (successfully or otherwise). 

As it is, the script can run and the user press cancel and it looks like a download but isn't.

Is this possible?  If so, how?

Your help, as ever, gratefully received.

we have already see the facility in website gmail and facebook that when user log-in successfully then its redirects the page to another page automatically, can someone tell me how to redirect the page when log-in becomes successful ?   User tries to Log-in -> create session variable -> now redirect the page   I can verify user account details and set the session variable at the same time, but how to reload the page so that it finds the session and reload itself and then redirect the page to new page ??   Thanks for sparing your time here !!

Hello,

I'm trying to store the match I get in $result to a variable so that I can easily compare its other values against other values. Here is my code..

//loop through new orders array
foreach ( $orders as $order ) {
	
	//compare job number and line item against database information for a match
	if ( array_search ( $order[ 'job_number' ], array_column ( $result, 'job_number' ) ) !== false && array_search ( $order[ 'line_item' ], array_column ( $result, 'line_item' ) ) !== false ) {
		
		$db_match = key($result);
		
		echo '<br><br>'.$db_match.'<br><br>';
		
		echo 'Job: ' . $order['job_number'] . '/Line Item: ' . $order['line_item'] . ' was found<br>';
	
	//new order which doesn't exist in database
	} else {
		echo 'Job: ' . $order['job_number'] . '/Line Item: ' . $order['line_item'] . ' was not found<br>';
	}
}

When I do $db_match = key($result); I get 0 each time and when I do: 

$db_match = array_search ( $order[ 'job_number' ], array_column ( $result, 'job_number' ) ) !== false && array_search ( $order[ 'line_item' ], array_column ( $result, 'line_item' ) ) !== false;

I get 1 each time...which I assume is the number of matches which meet that criteria instead of the array key I found a match on.

Edited May 8, 2020 by mongoose00318

I have a function to writes data I want to make sure it was successful.

my function:
Code: [Select]
function save_ini_file($filename,$content)
{
$File = "ini_files/". $filename . ".ini";

echo $File;
$Handle = fopen($File, 'w');
$info = $content;
fwrite($Handle, $info);
fclose($Handle);

  return true;
}
the if statement that calls the function.
Code: [Select]
if(!save_ini_file($serial,$data))
{
echo "did not write";
}else{
echo "did write";
}

the file is being created but I am getting the "did not write" echoed

Can you suggest a better way to write this code (I didn't create it):

<body onload="getParameterByName('url')">

<a href="" id="urllink" >Click Here</a>

</body>

so, I don't have to change the <body> tag?

I looked up 'how to get all POST variables' and I ended up putting this bit of code together.  Im not experienced at all with OOP programming, but I think that when '=>' is used its to do with OOP?  Anyway.  I dont get this code.  I understand Foreach $_POST, as $someVar, (BTW im looking for checkboxes from a form), However, If I leave out the '=> $val' I simply get a lot of 'on' values, indicating the checkboxes I checked.

However, WITH the '=> $val' I end up with the checkbox name.  Now i've changed the name of '$val' so i know its not a keyword.  But i dont get it.  Is there a way to access the value of the checkbox? Can someone explain to me whats going on here?  I know that I can get the value of the checkbox by $_POST['$checkRows'] But, I have a feeling It can be done in the foreach line.

foreach($_POST as $checkRows => $hip)
      {
         echo "POSTED: $checkRows<br>\n";         
      }

Please help!

Hi, can any see is the are mistakes in this for me.

Thanks

echo'<img src="skinFiles/'.$skin['thumb_name'].'"class="skinImage" onclick="changeSkin(\'skinFiles/'.$skin["css_name"].'\')" />';

How do I remove the captcha check on this line and keep it valid?

if ($author && $message && ($capatcha == 6 || controlPosts::isAdmin()))

I tried

if ($author && $message && (controlPosts::isAdmin()))

but the script wont post like this?

for timthumb in my wordpress i use this code to
function first_image() {
  global $post, $posts;
  $first_img = '';
  ob_start();
  ob_end_clean();
  $output = preg_match_all('/<img.+src=[\'"]([^\'"]+)[\'"].*>/i', $post->post_content, $matches);
  $first_img = $matches [1] [0];
 
  if(empty($first_img)){ //Defines a default image
    $first_img = "images/default.gif";
  }
  return $first_img;
}

it works really well, then i thought why not to make it show random images if timthumb dont find any image?

so the 1st googling i did came up with this function:
<?php 
function getRandomFromArray($ar) { 
    mt_srand( (double)microtime() * 1000000 ); 
    $num = array_rand($ar); 
    return $ar[$num]; 
} 

function getImagesFromDir($path) { 
    $images = array(); 
    if ( $img_dir = @opendir($path) ) { 
        while ( false !== ($img_file = readdir($img_dir)) ) { 
            // checks for gif, jpg, png 
            if ( preg_match("/(\.gif|\.jpg|\.png)$/", $img_file) ) { 
                $images[] = $img_file; 
            } 
        } 
        closedir($img_dir); 
    } 
    return $images; 
} 

$root = ''; 
// If images not in sub directory of current directory specify root  
//$root = $_SERVER['DOCUMENT_ROOT']; 

$path = 'images/'; 

// Obtain list of images from directory  
$imgList = getImagesFromDir($root . $path); 

$img = getRandomFromArray($imgList); 

?>

and to show random image i Place the following where i wish the random image to appear:

Code: [Select]
<img src="<?php echo $path . $img ?>" alt="" />
my question is how to place this code in my function first_image??
i need to do something like this, but it doesnt make any sense:
if(empty($first_img)){ //Defines a default image
    $first_img = " <img src="<?php echo $path . $img ?>" alt="" />
this was my thought, if i can mix those 2 normally it should work, right?

or maybe there is another way to show random image in the Code: [Select]
[i] if(empty($first_img)){ //Defines a default image[/i]
  [b]  $first_img = ?

ok, this is a much shorter version, but

i have:
Code: [Select]
$val = "1";

then i have a function
Code: [Select]

function  include_function_name{

echo'
<ul><li>
<a href="index.php?pageID="{$val}"">page name</a>
<li/><ul/>';
}


i have tried all sorts of variations but i can get it to make pageID=1



plaease help
thanks
ricky

I have a page that functions like this:   The user begins by loading ajaxtest.php, then using a drop-down menu, selects a user, after which a DIV on ajaxtest.php is populated with the user's selction to getuser.php?q=John%Doe   Here's a visual representation of what is currently happening, along with the part I don't understand.   getuser.php consists largely of a a large HTML table which interacts through PHP with my SQL database.  Several of the fields in this table are editable, and after the user changes one of these fields, he can press an "update" button, which calls an AJAX script in update.js and runs update.php, which contains all my SQL queries to update the database, without the user being redirected to this update.php page. All of this is processing perfectly - the user makes some sort of edit, hits "update", and the database gets updated without any redirection or refresh.   Here's what I can't figure out though.  Once the user makes a change, and hits "update," the database updates, but the end user doesn't see those changes that he made.  I can't just do a simple page refresh, because the action is taking place on getuser.php, but getuser.php is loaded inside of a #DIV on ajaxtest.php.  If I use something like window.location.reload(), this just refreshes ajaxtest.php and puts the user back to the starting point of having to select a user.  This is not what I want.  I don't want to refresh ajaxtest.php, I want to "refresh" ajaxtest.php with getuser.php?q=John%Doe insidethe DIV on ajaxtest.php.   It seems like this is very common - several forums have this capability, where a user has a page loaded, adds a comment, and sees that comment immediately without a full page refresh.  I just don't know how to do it, and I've tried at lengths to search the web for an answer with no luck.   Can someone walk me through how to do this?

Not sure exactly what my problem is here, but i have looked at this at least a hundred times and I just can't figure out what is going on.  Everything works great except for the sending the email part.  It displays the error "There was a problem sending the mail".  The form field checking works fine, the insert into mysql works fine - - but it won't send the email.  I have tried using double quotes and single quotes for the email information ($to, $subject, etc...) I have even eliminated the form data from the email information and it still doesn't send.  I am hopelessly stuck at this point     Any ideas??  Code below:

<html>
<body bgcolor = 'blue'>
<div align = 'center'>
<h1>test FORM</h1>
</div>
<p>
<?php

If ($_POST['submit']) //if the Submit button pressed
{
//collect form data

$fname = $_POST['FName'];
$lname = $_POST['LName'];
$email = $_POST['Email'];
$tel = $_POST['Tel'];
$mess = $_POST['Mess'];


$errorstring = "";

if (!$fname)
$errorstring = $errorstring."First Name<br>";
if (!$lname)
$errorstring = $errorstring."Last Name<br>";
if (!$email)
$errorstring = $errorstring."Email<br>";

if ($errorstring !="")
echo "<div align = 'center'><b>Please fill out the following fields:</b><br><font color = 'red'><b>$errorstring</b></font></div>";
else
{

$fname = mysql_real_escape_string($fname);
$lname = mysql_real_escape_string($lname);
$email = mysql_real_escape_string($email);
$tel = mysql_real_escape_string($tel);
$mess = mysql_real_escape_string($mess);

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

mysql_select_db("mec", $con);

$sql="INSERT INTO formtest (FName, LName, Title, Tel, Email, Mess)
VALUES ('$fname','$lname','$title','$tel','$email','$mess')";

if (!mysql_query($sql,$con))
{
die('Error: ' . mysql_error());
}
$to = 'myemailaddress';
$subject = 'test form';
$message = 'Hello';
$headers = 'From Me';
$sent = mail($to, $subject, $message, $headers);
if($sent)
{print "Email successfully sent";}
else
{print "There was an error sending the mail";}

mysql_close($con);

}


}

?>
<p>
<form action= 'testmecform.php' method= 'POST'>

                  
  <table width = '640' border = '0' align = 'center'>
    <tr> 
      <td align = 'right'><b>First Name</b></td>
      <td><input type = 'text' name = 'FName' value = '<?php echo $fname; ?>' size = '25'></td>
      <td><div align = 'right'><b>Telephone</b></div></td>
      <td><input type = 'text' name = 'Tel'  value = '<?php echo $tel; ?>' size = '25'></td>
    </tr>
    <tr> 
      <td align = 'right'><b>Last Name</b></td>
      <td><input type = 'text' name = 'LName' value = '<?php echo $lname; ?>' size = '25'></td>
  <td>&nbsp;</td>
      <td>&nbsp;</td>
    </tr>
    <tr> 
      <td align = 'right'><b>Email</b></td>
      <td><input type = 'text' name = 'Email'  value = '<?php echo $email; ?>' size = '25'></td>
      <td>&nbsp;</td>
      <td>&nbsp;</td>
    </tr>
    <tr> 
      <th colspan = '4'><b>Please enter any additional information he </b></th>
    </tr>
    <tr> 
      <th colspan = '4'><textarea name = 'Mess' cols = '50' rows = '10'><?php echo $mess; ?></textarea></th>
    </tr>
    <tr> 
      <th colspan = '4'><b>Please make sure all information is correct before submitting</b></th>
    </tr>
    <tr> 
      <th colspan = '4'><input type = 'submit' name = 'submit' value = 'Submit Form'></th>
    </tr>
  </table>
</form>
</body>
</html>

$tab is a variable for my table's name


<?php
       while($row=mysqli_fetch_assoc($result)){
?>

       <?php echo $row['{$tab}_name']?>




   

I've made up some new code but I have an issue with trying to figure out why it won't work. Do you see any issues with it?

$rss = fetch_feed('<?php echo get_post_meta($post->ID, "linktosource", true);?>');

Thanks in advance