PHP - Add Rel Inside Echo

Im having a problem with getting the quotes correct with this. I can either get the variable to work in the href and img src or in the css. Anyone have any ideas on this. Cuase im really stuck.

This allows the $address and $cos to echo in but not the css variables


<td><?php echo "<a href='planet_profile.php?planet=$address'>
<img src='images/star.jpg' id='$cos' 
style='position:absolute;' left:'$b_x px;' top:'$b_y px;'></a>"; ?></td>


and this allows only the css variables to work


 
<td><?php echo '<a href="planet_profile.php?planet="' . $address . '">
<img src="images/star.jpg" id="' . $cos . '" 
style="position:absolute; left:' . $b_x . 'px; top:' . $b_y . 'px;"></a>'; ?></td>

Currently whether data from the queries are found or not, its surrounded inside the div tags...

i want to somehow echo the div tags inside the while loop, but only echo once.

the confusing part for me is that the queries are connecting to 2 different tables for data.. and both need to be found for the div to be echo.
 
if($item['quantity'] >= 1){   
if($item2['type'] == 'weapon'){


code below....


Code: [Select]

<div class="g_content"><h3 style="text-align: left">&nbsp;&nbsp;Weapons</h3><div class="g_text">
<table align='center' cellspacing='10'>
<?php

$current_col = 1;
$max_col = 4;


$query = $db->execute("select * from items where `player_id`=?", array($player->id));

while($item = $query->fetchrow())
{
$query2 = $db->execute("select * from `blueprint_items` where `id`=?",array($item['item_id']));
$item2 = $query2->fetchrow();


//DISPLAY IF QUANTITY IS 1 OR MORE.
if($item['quantity'] >= 1){
if($item2['type'] == 'weapon'){
    //Open new row if first column
    if($current_col==1)
    {
        echo "<tr>\n";
    }

    //Display current record

    echo "<td width='25%'>";
echo "<center><img src=\"{$item2['img']}\" width='80' height='80' style=\"border: 1px solid #CC9900\"></center>";
    echo "<center><a href=\"../description.php?id={$item2['id']}\">{$item2['name']}</a> [x".$english_format_number = number_format($item['quantity'])."]</center>";
    echo "<center>$".$english_format_number = number_format($item2['value'])."</center>";
echo "<center>[<a href='../item.php?sell=".$item['id']."'>Sell</a>] [<a href='../item.php?market=".$item['id']."'>Market</a>] <br>[<a href='../item.php?send=".$item['id']."'>Send</a>] [<a href='../item.php?equip=".$item['id']."'>Equip</a>]</center><br>";
    echo "</td>\n";
 //Close row if last column
    if($current_col==$max_col)
    {
        echo "<tr>\n";
        $current_col = 0;  //<---Changed
    }
    $current_col++;
}
}

}
//Close last row if needed
if ($current_col!=1)
{
    for(; $current_col<=$max_col; $current_col++)
    {
        echo "<td>&nbsp;</td>\n";
    }
}

?>
</table>
</div></div>

Hi

I am trying to use the nl2br function like this

while ($row = mysqli_fetch_array($query)) {
//May need this later to output pictures   
//     $imageURL = 'upload/'.rawurlencode($row["filename"]);
  echo
  
   "  <div class='divTableRow'>
<div class='divTableCell'>{$row['User']} ;</div>
<div class='divTableCell'>nl2br({$row['CommentText']});</div>
</div>


    \n";
}

However the output just looks like the attached picture.

When I check in the sql db I can see the line breaks when doing a select * from Table ;

 

I'm redoing my login script using functions and a basic switch function checking against a $_GET variable (hope you guys know what I'm talking about).

What I want to do is create two functions:

1 that displays the login form
and 1 that processes the information

The form tag would look like this:
<form action=\"<?php $_SERVER['PHP_SELF']?>?action=process\" method=\"post\">
</form>

Here's my switch statement:
Code: [Select]
<?php

//****************************************
//****************Action******************
//****************************************

switch ($_GET["action"]) {
default:
case "index":
if (!$_SESSION["member"]) {
if (!$timeout) {
display_form();
} else {
echo $timeout_error;
}
} else {
echo "You are already logged in.";
}
break;
case "process":
if (!$_SESSION["member"]) {
if (!$timeout) {
process_form();
} else {
echo $timeout_error;
}
} else {
echo "You are already logged in.";
}
break;
}

?>

This runs as soon as "login.php" loads.

It'll automatically run the commands under case "default" and "index". It'll first check to see if the member's logged in. If not, it'll then check to see if "$timeout" is true ($timeout becomes true if the member has attempted to login 5 times and failed). If not, it'll display the login form, by running "display_form()".

Once the form has been filled and submitted, the commands under case "process" will be performed. Again it will first check to see if the member's logged in. If not, it'll check to see if "$timeout" is true. If not, it'll start validating the forms.

For the validation, I've created a variable called "$errors_found", and scripted one if statement checking to see if the email and password exist. If so, the variables "$rm_field_un" and "$rm_field_pw" become true, as well as "$errors_found".

So, back to submission of the form...

If "$errors_found" is true, display the form (when the form is displayed, there will be an if statement within that says "<?php if ($rm_field_un) { echo "Username is wrong."; } ?>", which will be displayed right underneath the username field and label. Same with the password elements). If "$errors_found" is not true, go ahead and register the member.

Now, here's where I need help, because I'm really confused as to how to accomplish this.

The "display_form()" function will contain a single variable called "$login_form". It will contain a value of the HTML constructed login form, and the function will return the variable.

Remember how I stated in the third paragraph up that I will have <php> statements within the form which would display errors if necessary? Well, how do I put those if statements within the HTML, which is contained within the variable?

If that last question confuses you, allow me to present you with an instance:
Code: [Select]
<?php

function display_form() {
$login_form = "
<form name=\"login_form\" method=\"post\" action=\"<?php $_SERVER['PHP_SELF']?>?action=process\">
Username:
<input type=\"text\">
<?php if ($rm_field_un) { echo "Username is wrong"; } ?>
";

return $login_form;
}

?>

See what I mean?

This code confuses me ALOT. I'm not sure if the <php> tags are needed as it is already contained within existing ones, or what. Can somebody please help me out?

Any and all help is much appreciated. =D

hello,  im trying to add a hyperlink that launches in a new window to the following (in the last column)  any ideas?

Code: [Select]
  echo "<tr>";
  echo "<td align='center'>" . $row["ID"] . "</td>";
  echo "<td align='center'>" . $row["Name"] . "</td>";
  echo "<td align='center'>" . $row["jobNO"] . "</td>";
  echo "<td align='center'>" . $epn . "</td>";
  echo "<td align='center'>" . $cname . "</td>";
  echo "<td align='center'>" . $cadd . "</td>";  //want to add a hyperlink here
  echo "</tr>";
  }
echo "</table>";


}

Hi I am new to PHP and this is my first post one here so appologies is this questions seems a bit dumb!

I have an if clause such that if a button is pressed on my web page then i want to reload the page and include a new form on it.

I am having a problem getting the $_SERVER['PHP_SELF'] command to work from iside a echo command. I must not be escaping the code correctly with back slashes:

I currently have the line :

 
echo"<form method=\"POST\" action=\"\<?$_SERVER['PHP_SELF']?\>\">";


However this doesnt seem to work as my page just doesnt display in the browser.

Any advice is much appreciated.

Thanks for taking the time to read.

Trying to figure out how to make it so name is a link to the profile when its echo
anyone know how to do this?

im at a huge stand still


Code: [Select]
<?php 

                                                $sql = "SELECT name FROM users WHERE DATE_SUB(NOW(),INTERVAL 5 MINUTE) <= lastactive ORDER BY id ASC"; 
$query = mysql_query($sql) or die(mysql_error());
$count = mysql_num_rows($query);
$i = 1;
while($row = mysql_fetch_object($query)) {
 $online_name = htmlspecialchars($row->name);


   echo '<a href="Inbox.php">"'[$goauld]'</a>"'; ?>

how do i get the image to show as the code i have is echoing what i have been told is binary, not confirmed.  and not the images.

all images are .jpg

Code: [Select]
$allowed_types = array('png','jpg','jpeg','gif');
$imgdir = '/home/mysite/ftpfolder';
$imgFilesArray = scandir($imgdir);
foreach ($imgFilesArray as $imgkey => $imgvalue)
{
$imgInfo =  pathinfo($imgdir . '/' . $imgvalue);
$imgExtension = $imgInfo['extension'];
if(in_array($imgExtension, $allowed_types))
{
readfile($imgdir . '/' . $imgvalue);
//echo '<img src = "'. $imgdir  .'/'. $imgInfo['basename'].'" alt="" />';
}
}

images are FTP'd to the ftpfolder and can not be placed anywhere else but this folder due to security reasons.

Hi! My question is probably simple but I've been scratching my head for hours now... I'm pretty new to php. I have an online form for orders; when submitted, an email is sent to the shop manager containing the info the client has filled in. So I pass all the info in the code below, but I can't manage to echo the array containing the order's items/qty/item_code. (I'm using Wordpress, Oxygen Builder, Metabox and Code Snippets, if that helps). Here's the code (used as a Code Snippet):

add_action( 'rwmb_frontend_after_process', function( $config, $post_id ) {
    if ( 'command-form' !== $config['id'] ) {
        return;
    }

    $name = rwmb_meta( 'name', '', $post_id );
    $email = rwmb_meta( 'email', '', $post_id );
    $phone = rwmb_meta( 'phone', '', $post_id );
    $orders = rwmb_meta( 'order', '', $post_id ); */ ---> this is the array I wish to show the elements from (array --> [0]:{item:"item", qty:"qty", code:"code"]}, [1]:{}, etc...} */
    $comments = rwmb_meta( 'comments', '', $post_id );
    $reception = rwmb_meta( 'reception', '', $post_id );
    $destination = rwmb_meta( 'destination', '', $post_id );
    $payment = rwmb_meta( 'payment', '', $post_id );

    $body = "Vous avez reçu une nouvelle commande en ligne!

Nom: $name
Courriel: $email
Téléphone: $phone
Commande: */ ---> this is where I want it displayed, but since it is within $body = " ", I can't seem to be able to use foreach( ) or even print_r( )... */
Commentaires: $comments
Réception: $reception
Paiement: $payment";
     

    $headers = ['Content-type: text/html', "Reply-To: $email"];
    wp_mail( 'address@domain.com', 'Nouvelle commande en ligne', $body );
}, 10, 2 );

 

Help with this would be greatly appreciated! Please let me know if any details are missing. Thanks in advance!

Jordan

Edited June 23 by JordanC

<td><label for='images'> <b>File to upload:</b> </label></td>
            <td><input type='file'  name = 'drama_image' '<?php echo $row['drama_image']; ?>'/></
                       </tr>
         
   <?php
            $target_path = "images/";
            $target_path = $target_path . basename( $_FILES['images']['name']);

            if(move_uploaded_file($_FILES['images']['tmp_name'], $target_path)) {
                echo "The file ".  basename( $_FILES['images']['name']).
             " has been uploaded";
            } else{
                echo $row['drama_image'];
            }
         
     ?>

['drama_image'] is the name of the file

I wanna echo it out in the box of file upload so when I save , the default picture will still be there instead of being overwritten as the box does not have any value in it.

OK, have no idea what's going on... I've done this a million times... why wont this output!??
I must have a major brain meltdown and dont know it yet!!!

Code: [Select]
<?php

// this echoes just fine:
echo $_POST['testfield'];


// but this wont echo:
echo 
if (isset($_POST['testfield']))
{
$_POST['testfield'] = $test;
}
echo $test;


/// or even this DOESNT echo either!:
$_POST['testfield'] = $test;
echo $test;

?>

Hi All,

I'm trying to echo the response from an SLA query, the query works and returns the data when I test it on an SQL application.. but when I run it on my webpage it won't echo the result.

Please help?

<?php   
$mysqli = mysqli_connect("removed", "removed", "removed", "removed");  
$sql = "SELECT posts.message FROM posts INNER JOIN threads ON posts.pid=threads.firstpost WHERE threads.firstpost='1'";  
$result = mysqli_query($mysqli, $sql);   
echo {$result['message']}; 
?>

The Script:

$desired_width = 110;

if (isset($_POST['submit'])) {
    $j = 0; //Variable for indexing uploaded image

    for ($i = 0; $i < count($_FILES['file']['name']); $i++) {//loop to get individual element from the array
        $target_path = $_SERVER['DOCUMENT_ROOT'] . "/gallerysite/multiple_image_upload/uploads/"; //Declaring Path for uploaded images

        $validextensions = array("jpeg", "jpg", "png");  //Extensions which are allowed
        $ext = explode('.', basename($_FILES['file']['name'][$i]));//explode file name from dot(.) 
        $file_extension = end($ext); //store extensions in the variable

        $new_image_name = md5(uniqid()) . "." . $ext[count($ext) - 1];
		$target_path = $target_path . $new_image_name;//set the target path with a new name of image
        $j = $j + 1;//increment the number of uploaded images according to the files in array       
      
	  if (($_FILES["file"]["size"][$i] < 100000) //Approx. 100kb files can be uploaded.
                && in_array($file_extension, $validextensions)) {
            if (move_uploaded_file($_FILES['file']['tmp_name'][$i], $target_path)) {//if file moved to uploads folder
                echo $j. ').<span id="noerror">Image uploaded successfully!.</span><br/><br/>';

                $tqs = "INSERT INTO images (`original_image_name`, `image_file`, `date_created`) VALUES ('" . $_FILES['file']['name'][$i] . "', '" . $new_image_name  . "', now())";
                $tqr = mysqli_query($dbc, $tqs);

                // Select the ID numbers of the last inserted images and store them inside an array.
                // Use the implode() function on the array to have a string of the ID numbers separated by commas.
                // Store the ID numbers in the "image_file_id" column of the "thread" table.

                $tqs = "SELECT `id` FROM `images` WHERE `image_file` IN ('$new_image_name')";
                $tqr = mysqli_query($dbc, $tqs) or die(mysqli_error($dbc));

                $fetch_array = array();
                $row = mysqli_fetch_array($tqr);
                $fetch_array[] = $row['id'];


                /*
                 * This prints e.g.:
                    Array ( [0] => 542 )

                    Array ( [0] => 543 )

                    Array ( [0] => 544 )
                 */
                print_r($fetch_array);

                // Goes over to create the thumbnail images.
                $src = $target_path;
                $dest = $_SERVER['DOCUMENT_ROOT'] . "/gallerysite/multiple_image_upload/thumbs/" . $new_image_name;

                make_thumb($src, $dest, $desired_width);


            } else {//if file was not moved.
                echo $j. ').<span id="error">please try again!.</span><br/><br/>';
            }
        } else {//if file size and file type was incorrect.
            echo $j. ').<span id="error">***Invalid file Size or Type***</span><br/><br/>';
        }
    }
}

Hey, sorry that I am posting this darn image upload script again, I have this almost finished and I am not looking to ask more questions when it comes to this script specifically.   With the script above I have that part where the script should store the ID numbers (the auto_increment column of the table) of the image files inside of one array and then the "implode()" function would get used on the array and then the ID numbers would get inserted into the "image_file_id" column of the "thread" table.   As you can see at the above part the script prints the following:

                    Array ( [0] => 542 )

                    Array ( [0] => 543 )

                    Array ( [0] => 544 )

And I am looking to insert into the column of the table the following:

542, 543, 544

I thought of re-writing the whole image upload script since this happens inside the for loop, though I thought maybe I could be having this done with the script as it is right now.   Any suggestions on how to do this?

This topic has been moved to JavaScript Help.

http://www.phpfreaks.com/forums/index.php?topic=316454.0