PHP - How To Echo The Number Of Pictures I Have In A Folder.

How do you do this? Thanks in advance.

try {
echo "<br>";
foreach($dbh->query("SELECT * FROM test_shot WHERE sold=1 ORDER BY year ASC") as $row) {
	
						if($row['picture'] != "" && $row['picture'] != null)
						{
						echo "<div class='image-holder'><img src ='".$row['picture']."' width=300px /><br>";  
            			}
						if($row['year'] != "" && $row['year'] != null)
						{
							echo $row['year'];
						}
						if($row['description'] != "" && $row['description'] != null)
						{
							echo $row['description'];
						}
						if($row['sold'] == 1) {
							echo "<img src='images/sold1.png'><br>";//Add your image code here
							}
							elseif ($row['sold'] == 0) {
							
							echo "</div><br>";
						}
					}

	}	

catch (PDOException $e) {
  print $e->getMessage();
}

?>

 

I'm getting the dreaded " Invalid parameter number: number of bound variables does not match number of tokens" error and I've looked at this for days.  Here is what my table looks like:

|  id          | int(4)       | NO   | PRI | NULL    | auto_increment |
| user_id     | int(4)       | NO   |     | NULL    |                |
| recipient   | varchar(30)  | NO   |     | NULL    |                |
| subject     | varchar(25)  | YES  |     | NULL    |                |
| cc_email    | varchar(30)  | YES  |     | NULL    |                |
| reply       | varchar(20)  | YES  |     | NULL    |                |
| location    | varchar(50)  | YES  |     | NULL    |                |
| stationery  | varchar(40)  | YES  |     | NULL    |                |
| ink_color   | varchar(12)  | YES  |     | NULL    |                |
| fontchosen  | varchar(30)  | YES  |     | NULL    |                |
| message     | varchar(500) | NO   |     | NULL    |                |
| attachment  | varchar(40)  | YES  |     | NULL    |                |
| messageDate | datetime     | YES  |     | NULL    |        

Here are my params:

            $params = array(                 
               ':user_id' => $userid,                 
               ':recipient' => $this->message_vars['recipient'],                 
               ':subject' => $this->message_vars['subject'],                 
               ':cc_email' => $this->message_vars['cc_email'],                 
               ':reply' => $this->message_vars['reply'],                 
               ':location' => $this->message_vars['location'],                 
               ':stationery' => $this->message_vars['stationery'],                 
               ':ink_color' => $this->message_vars['ink_color'],                 
               ':fontchosen' => $this->message_vars['fontchosen'],                 
               ':message' => $messageInput,                 
               ':attachment' => $this->message_vars['attachment'],                 
               ':messageDate' => $date             );

Here is my sql:

$sql = "INSERT INTO messages (user_id,recipient, subject, cc_email, reply, location,stationery, ink_color, fontchosen, message,attachment) VALUES( $userid, :recipient, :subject, :cc_email, :reply, :location, :stationery, :ink_color, :fontchosen, $messageInput, :attachment, $date);";

And lastly, here is how I am calling it:
            $dbh = parent::$dbh;             
           $dbh->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_WARNING);              
           if (empty($dbh))    return false;             
                  $stmt = $dbh->prepare($sql);             
                 $stmt->execute($params) or die(print_r($stmt->errorInfo(), true));              
           if (!$stmt) {    print_r($dbh->errorInfo());              }

I know my userid is valid and and the date is set above (I've echo'd these out to make sure).  Since the id is auto_increment, I do not put that in my sql (though I've tried that too), nor in my params (tried that too).  What am I missing?  I feel certain it is something small, but I have spent days checking commas, semi-colons and spelling.  Can anyone see what I'm doing wrong?

I got this script:

But it give me error, file_get_contents cannot open stream.

I need to add the FTP connection with user/pass paramaters.

then look in set http url, to get the file contents(images) and transfer to ftp server location.

Can Anyone take alook and tell me if I am going down the right path and how to get there. Please

Code: [Select]
function postToHost($host, $port, $path, $postdata = array(), $filedata = array()) {
$data = "";
$boundary = "---------------------".substr(md5(rand(0,32000)),0,10);
$fp = fsockopen($host, $port);

fputs($fp, "POST $path HTTP/1.0\n");
fputs($fp, "Host: $host\n");
fputs($fp, "Content-type: multipart/form-data; boundary=".$boundary."\n");

// Ab dieser Stelle sammeln wir erstmal alle Daten in einem String
// Sammeln der POST Daten
foreach($postdata as $key => $val){
$data .= "--$boundary\n";
$data .= "Content-Disposition: form-data; name=\"".$key."\"\n\n".$val."\n";
}

// Sammeln der FILE Daten
if($filedata) {
$data .= "--$boundary\n";
$data .= "Content-Disposition: form-data; name=\"".$filedata['name']."\"; filename=\"".$filedata['name']."\"\n";
$data .= "Content-Type: ".$filedata['type']."\n";
$data .= "Content-Transfer-Encoding: binary\n\n";
$data .= $filedata['data']."\n";
$data .= "--$boundary--\n";
}

// Senden aller Informationen
fputs($fp, "Content-length: ".strlen($data)."\n\n");
fputs($fp, $data);

// Auslesen der Antwort
while(!feof($fp)) {
$res .= fread($fp, 1);
}
fclose($fp);

return $res;
}

$postdata = array('var1'=>'today', 'var2'=>'yesterday');
$filedata = array(
'type' => 'image/png',
'data' => file_get_contents('http://xxx/tdr-images/images/mapping/dynamic/deals/spot_map')
);

echo '<pre>'.postToHost ("localhost", 80, "/test3.php", $postdata, $filedata).'</pre>';

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;

?>

Hello, I have problem durring binding update query. I can't find what is causing problem.

public function Update(Entry $e)
    {
        try
        {
            $query = "update entry set string = $e->string,delimiter=$e->delimiter where entryid= $e->id";
            $stmt = $this->db->mysqli->prepare($query);
            $stmt->bind_param('ssi',$e->string,$e->delimiter,$e->id);

            $stmt->close();
           

        }
        catch(Exception $ex)
        {
            print 'Error: ' .$ex->getMessage();
        }
    }

When I run function update I'm getting next error:Warning: mysqli_stmt::bind_param(): Number of variables doesn't match number of parameters in prepared statement   Can you help me to solve this problem ?
Edited by danchi, 17 October 2014 - 10:25 AM.

I need to display a number(the number is retrieved from the db) in the form input field such that only the last 4 digits is visbile, the remaining can be masked as * or X or whatever is applicable.

I know the last 4 can be obtained as follows:

    Code: [Select]
$number=substr($number,-4,4);
 
But when i hit the submit button the form validates the input field and checks if the number is a valid number of a specific format.
Therefore when I click on the submit button then I should still be able to unmask the masked numbers or do something similar that would help me validate the whole number.

  Code: [Select]
  <input type="text" name="no" value="<?php if(!empty($number)){ echo $number;} ?>">

I want to copy everything in templates/blue to the folder code/

However:

shell_exec("cp -r 'templates/blue' 'code'");

Creates a folder called blue inside code. I tried cp -r 'templates/blue/*' 'code', but that didn't do anything.

Any ideas?

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']}; 
?>

Hey

I have a string that looks like the following:
Quote

top-php-tutorials-2.html


I have a script that cycles through each page. The 2 in the quote above is the page number.

How can I extract the number between the - and the .html and replace it with another number?

I've tried
Code: [Select]
substr($engine->selectedcaturl, 0,-6).$v.".html"But then I realised this only works for numbers that are 1 digit long

Any input would be appreciated

Hello, like I promised, I'm back with another problem. This one will be harder xD so squeeze your brain   Now I've got a webpage, with a database shown into a table, under that I've got page numbers. So I got my 20 rows for each page.   https://www.dropbox..../stillnotok.jpg   Great/awesome so far   Now I've got a new problem and it's hard for me to explain this too.   In my database I got allot of tables, in my webpage I'm using 2 of them, called 'artikel' and 'images".   Everything from my artikel table is ok. So we don't need to look at that.   But my problem is with the table 'images'.   In that table I got 2 items.   Like you will see in my code, I need the I_ARTCODE and I_FILE   in I_ARTCODE stands: 14 and for the other image 15   in I_FILE stands: 5.jpg and for the other image 6.jpg   With the code I've got now, I display the word 5.jpg and 6.jpg.   But I need to display the picture behind that, the person who gave me this task sayed, I normally can do this with the pad but it's not a real url, I'll give you guys a picture how my database looks like:   https://www.dropbox....uqplxd/path.jpg   Over here the pad is this: 000000-001000   It doesn't seems normal to me and I can't find a sollution for this on the internet.   Can someone plz help me?   Ooh ya, almost forgot, this is my code:

<?php
include('connect-mysql.php');
if (!empty($_GET["page"])) {
    $page  = $_GET["page"];
} else {
    $page=1;
}; 
$start_from = ($page-1) * 20; 
$sqlget = "SELECT *
FROM artikel, images
LIMIT $start_from, 20
";

$sqldata = mysqli_query($dbcon, $sqlget) or die('error getting');

echo "<table>";
echo "<tr><th>A_ARTCODE</th><th>A_NUMMER</th><th>A_OMSCHRN</th><th>A_REFLEV</th><th>A_WINKEL</th><th>I_ARTCODE</th><th>I_FILE</th></tr>";

while($row = mysqli_fetch_array($sqldata)){
	echo "<tr><td align='right'>";
	echo $row['A_ARTCODE'];
	echo "</td><td align='left'>";
	echo $row['A_NUMMER'];
	echo "</td><td align='left'>";
	echo $row['A_OMSCHRN'];
	echo "</td><td align='left'>";
	echo $row['A_REFLEV'];
	echo "</td><td align='right'>";
	echo $row['A_WINKEL'];
	echo "</td><td align='right'>";
	echo $row['I_ARTCODE'];
	echo "</td><td align='right'>";
	echo $row['I_FILE'];
	//echo "<img src='000000-001000".$row['I_ID']."' />";
	echo "</td></tr>";
	}
	
echo "</table>";

$sql = "SELECT COUNT(A_ARTCODE) FROM artikel";
$rs_result = mysqli_query($dbcon, $sql) or die ("mysqli query dies");
$row = mysqli_fetch_row($rs_result) or die ("mysqli fetch row dies");
$total_records = $row[0];

$total_pages = ceil($total_records / 20);

for ($i=1; $i<=$total_pages; $i++) { 
            echo "<a href='index.php?page=".$i."'>".$i."</a> "; 
};
?>

if you have the urls to pictures in a database how would you display the actual pictures on a webpage?

Here 's the code.  Code: [Select]
<?php
require_once("functions.php"); 
?> 
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
<style type="text/css">
td {
border-top-style: solid;
border-right-style: solid;
border-bottom-style: solid;
border-left-style: solid;
border-top-color: #30C;
border-right-color: #30C;
border-bottom-color: #30C;
border-left-color: #30C;
}
</style>
</head>

<body>
<?php
navBar();
echo "<form action=\"\" method=\"post\" name=\"catalog\">";
  
DatabaseConnection();  

  $query = "SELECT * FROM treats"; 
        $result_set = mysql_query($query) or die(mysql_error());
$i = 0;

        echo "<table>";
        while ($row = mysql_fetch_array($result_set))
        {
            echo"<tr><td width=\"400px\">($row['product_pic']).  <br />{$row['product_id']}.<br />{$row['product_title']}.<br  />  {$row['product_Description']}.<br />{$row['product_price']}</td></tr>";
        }
echo "</table>"; 
?>   
</form>
</body>
</html>

Hi, i would like to know what is the best method to store a picture. Currently what happens, is each user has a different avatar, which when uploaded comes in a folder called Avatar, in my website folder and is also stored as a BLOB in mysql. I would like to know is this ok to store all avatars in one folder or some other step should be used and if so could some one guide me to a good tutorial. As I would also like to make an option, where people could upload their pictures, is it safe to store pictures in the same folder for all users or some how create a different folder for each user.

Hi,   I have a crwal engine and every time when I try to take a picture from a specific website, it also take some bullshit from superfish, creating iframe and div class, that I can`t deleted using jquery after my page is load.   I was looking in my code and I found out that the only thing that trigger that is the <img src="<?php h(getPicture($itemRow['site'], if I remove 'site', I don`t see any other superfish code but also don`t see the picture. What can I do to take only the .jpg picture , without any of the other crap?     <?php for ($i = 0; $i < count($itemRow['pics']); $i++) { ?>

I have a log system that allows 10 logs on each side(Left and right). I am trying to make it so that the left side has the 10 most recent logs, then the right as the next 10. Any ideas?