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Fergal Andrews

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Posts posted by Fergal Andrews

    PHP can only modify divs declared before the modifier

    in PHP Coding Help

    Posted

    hi fmohrmann,

    I think Nightslyr is correct about the load order of the javascript and html.

    You might be better using jquery to do this. You can check the dom has loaded before the javascript is read.

    It is off-topic for a php forum to go into a lot of detail about javascript but here is an example of some jquery that will show and hide divs. You should be able to adapt this to your needs.

     

     

    <!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01//EN"
    "http://www.w3.org/TR/html4/strict.dtd"
    >
<html lang="en">
<head>
    <title>test</title>
    <script src="http://ajax.googleapis.com/ajax/libs/jquery/1.4/jquery.min.js" type="text/javascript"></script>
    <script type="text/javascript">
    $(document).ready(function() {
        
        // by default show the first gallery
        $('div.gallery').hide();
        $('#div1').show();
       
      $('a.button').click(function() {
         // get the numeric id of the link
        var idNum = this.id.substr(6,this.id.length);
        // create the id for the gallery div
        var galleryDivId = 'div' + idNum;
        // add the hide class to all galleries
        $('div.gallery').hide();
        // show the chosen gallery
        $('#' + galleryDivId).show();
        
      })
    })
    </script>
    <style type="text/css">
    div.gallery {
        border: 1px solid #000;
        padding: 30px;
        width: 70px;
    }
    </style>
</head>
<body>
    <div class="gallery" id="div1">
        test 1
    </div>
    <div class="gallery" id="div2">
        test 2
    </div>
    <div class="gallery" id="div3">
        test 3
    </div>
    <div class="gallery" id="div4">
        test 4
    </div>
    <a href="#" class="button" id="button1">gallery 1</a>
    <a href="#" class="button" id="button2">gallery 2</a>
    <a href="#" class="button" id="button3">gallery 3</a>
    <a href="#" class="button" id="button4">gallery 4</a>
</body>
</html>

     

    Hope this helps,

    Fergal

    I can select one keyword, what about multiple?

    in PHP Coding Help

    Posted

    Hi brown2005,

    I think MySQL's FULLTEXT indexing and searching could be what you need. I'm not sure if you have the access to the database to alter tables in order to index them for FULLTEXT searches but if you do then this could help you achieve what you want. FULLTEXT searches allow you to use multiple search words and other advanced options.

    Here are a couple of URL's about FULLTEXT :

    http://www.petefreitag.com/item/477.cfm

    http://onlamp.com/pub/a/onlamp/2003/06/26/fulltext.html

    Hope that helps,

    Fergal

    Can't find the page

    in PHP Coding Help

    Posted

    Hi xuniter,

    The problem is with the form action. You're missing an end quote:

    <form action="dologin.php method="post">

    should be

    <form action="dologin.php" method="post">

     

    Also, headers need to be sent before anything is output to the page. It is best to move your php code containing the header command to the top of the page before <html>, etc. There's no need to use an else command. If the condition is not met then the code will just continue. If it is met then the header will redirect to the new page.

    Cheers,

    Fergal

     

    <?php 

if (isset($_SESSION['user'])) {
header("Location: index.php");
} 
?>
<html>
<head>


<title>Basic CMS | Admin area Login</title>

    Why won't this for loop INSERT?

    in PHP Coding Help

    Posted

    Hi billy_111, 

    Your form inputs have no value so the form is returning an empty array. Try something like: 


<input type="text" name="author[]" id="author1" value="author1" />
<input type="text" name="author[]" id="author2" value="author2" />
<input type="text" name="author[]" id="author3" value="author3" />
<input type="text" name="author[]" id="author4" value="author4" />
...
...
...
<input type="text" name="author[]" id="author10" value="author10" />

     

    Fergal

    string into byte into file

    in PHP Coding Help

    Posted

    Hi csckid,

    You use the $_FILES array to access uploaded files. When a file is uploaded it is written to a temporary directory. You then need to move it to its correct location. You access the file using $_FILES['myfile'] , where 'myfile' is the name you gave the file field in your form:

     

    $tempName = $_FILES['myfile']['tmp_name'];
$uploadedFilename = $_FILES['myfile']['name'];
$directory = '/uploadedFiles';
move_uploaded_file($tempName, $directory . '/' . $uploadedFilename);

     

    Do a var_dump of $_FILES['myfile'] to see all the array variables you can access.

    all the best,

    Fergal

    Restricting Directory Access

    in Apache HTTP Server

    Posted

    Hi mattspriggs28,

    There are a number of ways to prevent reading of files but allow writing to a directory. You can do it with server configuration but directory permissions is usually easier. If your hosting is unix/linux then check this link for information on how to change the permissions. You can often  do this through your ftp software.

     

    http://www.perlfect.com/articles/chmod.shtml

     

    Also, stick a blank index.html file in there so that it will display instead of a list of files if a user goes directly to the directory url.

     

    all the best,

    Fergal

    redirecting with headers never works

    in PHP Coding Help

    Posted

    Hi  viperjts10,

    I ran your code and the redirection worked for me.

     

    The only thing i added was a class to create the session object with methods is_logged_in() and logout() both of which returned true.

     

    I'm not sure why you are having problems but it must be something in the code preceeding the snippets you posted.

     

    The full code I used is below.

     

    All the best,

    Fergal

     

    <?php

class session
{
    function is_logged_in()
    {
        return true;
    }
    
    function logout()
    {
        return true;
    }
}

$session = new session();

if($session->is_logged_in()) {
        /* Kill session variables */
        $session->logout();
        //$_SESSION = array(); // reset session array
        redirect_to('index.php');
        //echo "<meta http-equiv=\"Refresh\" content=\"0;url=index.php\">";
}

function redirect_to($location = NULL) {
    if ($location != NULL) 
    {
            header("Location: {$location}");
            exit;
    }
}

    PdfLib change font colour

    in PHP Coding Help

    Posted

    Hi,

    I'm trying to create a PDF using the PdfLib extension. I am having great difficulty finding out how to change the colour of fonts. The only info I can find on font colours relates to the block plugin for PdfLib and I don't think the client has that on their server.

     

    Can anyone tell me how to do this?

     

    Thanks in advance,

    Fergal

    Help needed with PHP session management class

    in PHP Coding Help

    Posted

    Hi,

    I am writing a session management class for a project I am working on at the moment. The live server has a very restrictive environment, PHP4, Oracle 9i, no cache available, no writing to files on the server and no usage of PHP sessions.

     

    Consequently, I am writing a class to manage sessions as the project involves a big online form over several pages. I've written the class which uses a cookie to store an ID and writes 'session' data to the database.

     

    However, I need to handle users that have disabled cookies. I was thinking the best solution would be to write the id to a hidden form field, but there might be some security concerns about that. Does anyone have a better solution for this problem?

     

    Many thanks,

    Fergal

    Displays nothing?

    in PHP Coding Help

    Posted

    Hi Justin,

    Using a hyperlink to a php page will not submit form data, so when your link goes to info.php there is no form data. The SQL statement therefore will be

     

    SELECT * FROM info WHERE vipcode=''

     

    which will return no results unless you have a record with a null code value. The conditional statement

     

    if (!$row[levelofproject] || !$row[name] || !$row[id]){

     

    therefore fails and nothing is output to the screen.

     

    All the best,

    Fergal

    You need to use a proper method of submitting the form, either a submit input or javascript and put a form action with info.php as the value in the form tag.

    Fergal

    Editing a link - new to php

    in PHP Coding Help

    Posted

    Hi Iluvatar13,

    If you just want to disable the form and  stop the user's journey through the shopping cart, then nothing you do will HTML or Javascript will affect the PHP.

     

    If you don't need anything from the form, then why show it to the user in the first place? Would it not be better to bring them to a different page and show them the required message instead? You could do this with javascript or PHP.

     

    Fergal

    About cookies and sessions

    in PHP Coding Help

    Posted

    Hi corrupshun,

    Cookies and Sessions are useful for different things. The advantage of cookies is that they can persist after the user has closed their browser. Generally, sessions are more reliable because the user can turn off cookies so you can't depend on the data they contain, though very few people actually do that. Some complex server environments where you may have many load balanced servers and proxy servers can make the use of sessions unreliable. I have done quite a bit of work for the BBC and within their complex server environment, session data can get lost. It really comes down to a judgement call as to which suits the job you are doing best. If you are in doubt then both is the safest option.

    Fergal

    Undefined Variable and Not submitting

    in PHP Coding Help

    Posted

    Hi bullbree,

    The Notice you are getting is because your first usage of pfrontimage is within a conditional statement.

    You can fix this by either putting @ before

     

    @$pfrontimage = mysql_real_escape_string($_POST['pfrontimage']);

     

    which will supress the Notice or by declaring the variable earlier in your code before it is used.

     

    $pfrontimage = '';

     

    I've only had a quick look at your code but I can't understand why your are doing a mysql_real_escape_string on pfrontimage if it is a file. Do you have two form fields named pfrontimage?

    If not and I could be wrong, but you seem to be using the same variable as a string and a file in different parts of your code.

    how do you use readfile() to display a picture?

    in PHP Coding Help

    Posted

    Sorry, I should have spotted the schoolboy error in the modifed script I gave you (it's been a long day!). It was echoing the path to the file and not the data. Try this on your local file and then the remote one.

     

    <?php

     

    if(isset($_GET['player']))

    {

        $idpic = mysql_query("SELECT * FROM phpbb_attachments WHERE post_msg_id='$player'");

        $countid = mysql_num_rows($idpic);

        if($countid!=0)

        {

            $row = mysql_fetch_assoc($idpic);

            $file = './forum/files/'.$row['physical_filename'];

            if (file_exists($file))

            {

                $data = readfile($file);

                header('Content-Type: image/jpg');

                echo $data;

            }

        }

    }

     

    ?>

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