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Data from a table


asai

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Something is still missing here:

 

I have tried this:

function playVideo(url, filename)
{
    $.post( "update.php", {filename: "filename"})
  .done(function( data ) {
    alert( "Data loaded: " + data );
  });
}

The output from the alert is "filename"

Not the actual filename i want.

Hi again,

I have come a little further with my project.

A small change from earlier, but now the update works. Then change is that I haven't a link, but instead uses a button.

One small thing that doesn't work, that is the movie is not run.

I am not sure how and where to put it.

Heres my movies.php file:

<html>
<head>
<link href="css/style.css" rel="stylesheet" type="text/css" media="screen" />
</head>

<?php

include 'combo.php';
include 'config.php';
include 'opendb.php';

$ndate = $_POST['ndate'];

$result = mysql_query("SELECT *
            FROM DayMovie 
            WHERE FileDate LIKE '$ndate%' ORDER BY FileDate DESC") 
or die(mysql_error());  
echo "<table border='0'>";
echo "<tr> <th>Dato</th><th>Visninger</th><th>Handling</th></tr>";
while($row = mysql_fetch_array( $result )) {
	echo "<tr><td>";
	echo date('d.m.Y', strtotime($row['FileDate']));
	echo "</td><td>";
	echo $row['Counter'];
	echo "</td><td>";
    	
?>

   <form name="form" method="POST" action="update.php">
     <input name='filename' type='hidden' value=<?php echo $row['FileName'];?>>
	 <input name='url' type='hidden' value=<?php echo "alldaymovies/{$row['FileName']}"?>>
     <input type="submit"  value="Se film">
   </form>

<?php
}
   echo "</td></tr>";
echo "</table>";

include 'closedb.php';
?>

</html>

From here I post both filename and url to update.php

update.php looks like this:

<html>
<head>
<link href="css/style.css" rel="stylesheet" type="text/css" media="screen" />
</head>

<?php

include 'config.php';
include 'opendb.php';

$filename = $_POST['filename'];
$url      = $_POST['url'];

$result2 = mysql_query("UPDATE DayMovie SET Counter=Counter+1 WHERE FileName='$filename'") 
or die(mysql_error());

include 'closedb.php';

?>

Right now I am not using the url.

Theres 2 things I would like to happen from here:

1. The movie opens from the url in a new window.

2. The update.php returns to movies.php

 

Is this possible and how?

This is now my update.php

<html>
<head>
<link href="css/style.css" rel="stylesheet" type="text/css" media="screen" />
</head>

<?php

include 'config.php';
include 'opendb.php';

$filename = $_POST['filename'];

$result2 = mysql_query("UPDATE DayMovie SET Counter=Counter+1 WHERE FileName='$filename'") 
or die(mysql_error());

include 'closedb.php';

$url = "http://81.166.2.19/alldaymovies/$filename";
window.open($url);

?>

</html>

The data is database is updated, but no window opens with the url. What's missing?

Sorry if I am a little slow...  :-\

From this button is there a way to call a javascript function to open a new window from the url? :

<form name="form" method="POST" action="update.php">
     <input name='filename' type='hidden' value=<?php echo $row['FileName'];?>>
     <input type="submit"  value="Se film">
   </form>

I haven't read back through all of the pages of posts so sorry if this has been coverered - but it seems more sensible to not use forms, and just put your viewing counter processing code on the page of wherever you're hosting the movie itself

 

Just loop out a set of url's instead of submit buttons

<?php

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

$output.= "<a href='alldaymovies.php?filename=$row[filename]' target='_blank'>$row['filename']</a>";
$output.= "<br>";

}

echo $output;

The use of '_blank' forces a new window to be opened in the browser.

 

Then at the top of the actual movie page...

$filename = $_GET['filename'];


$result2 = mysql_query("UPDATE DayMovie SET Counter=Counter+1 WHERE FileName='$filename'") 
or die(mysql_error());

you need to escape $filename as it's open to SQL injection

 

Sorry if I am a little slow...  :-\

From this button is there a way to call a javascript function to open a new window from the url? :

<form name="form" method="POST" action="update.php">
     <input name='filename' type='hidden' value=<?php echo $row['FileName'];?>>
     <input type="submit"  value="Se film">
   </form>

 

You would do it in the function on return of the ajax response as you have been shown in reply #49

http://forums.phpfreaks.com/topic/294097-data-from-a-table/page-3?do=findComment&comment=1504950

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