Here is something you might wanna try.

/**
* Check for correct password
*
* @param string $password The password in plain text
* @param string $hash The stored password hash
*
* @return bool Returns true if the password is correct, false if not.
*/
function check_hash($password, $hash)
{
    if (strlen($password) > 4096)
    {
        // If the password is too huge, we will simply reject it
        // and not let the server try to hash it.
        return false;
    }

    $itoa64 = './0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz';
    if (strlen($hash) == 34)
    {
        return (_hash_crypt_private($password, $hash, $itoa64) === $hash) ? true : false;
    }

    return (md5($password) === $hash) ? true : false;
}

Once you hash a password, it cannot be undon. Therefore, you cannot cross check a text password with a hashed one. If this was possible, no password woule be safe on the web.

Try
close your first echo with a quote and semi colon

The only thing i can think of is, 1) your hardrive is spinning up, and 2) some programs have to get loaded, so its slowing down the time to get to the desktop.

Well is the person running this page on his own server? Also, did he change "table1" to what ever the table name is, in the SQL statement?

First off you are trying to pass test.php, and you show your add.php, so nothing is going to happen there. Second, in your add.php, you need a connection and an id that gets passed from the ajax function. I would suggest that you get the example from w3shools so you get a basic idea. Its a learning process, but you will get it.

This is a really good website that i learned from.

W3Schools

I believe because you didnt have

script.onload=scriptLoaded;

on your first post.
you could also include the script

<script type="text/javascript" src="https://ajax.googleapis.com/ajax/libs/jquery/1.8.0/jquery.min.js"></script>
<script type="text/javascript">
$(document).ready(function () {
    $('#main').click(function(){ alert('hi'); });
})
</script>

it looks to me that $checked = a variable with an extra ' in it. It says it in the message (private='0'') with the extra aposrophe. Maybe thats why?

yes, since i know php any mysql, i would use that. just display that data on one page have have them echo a link that uses a keyword or id to specifically display the info on that person. Ex

Members.php
<?php
//database connection
// select statement for database

//from w3schools.com with some modding
<?php
$con = mysql_connect("localhost","peter","abc123");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("my_db", $con);

$result = mysql_query("SELECT * FROM Persons");

while($row = mysql_fetch_array($result))
  {
  echo "<a href='members2.php?id=".$row['id']."'>".$row['name']."</a>";
  }

mysql_close($con);
?>

Members 2.php
<?php
$id = $_POST['id'];
$con = mysql_connect("localhost","peter","abc123");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("my_db", $con);

$result = mysql_query("SELECT * FROM Persons WHERE id = $id");

//took out while so it doesnt repeat your output.
$row = mysql_fetch_array($result)
?>
  //display the info of what you want to display
 <table><tr><td><?php $row['name'] ?></td></tr></table>
 <?php

mysql_close($con);
?>

Hope this helps

you cannot use $result twice. try making one $results and the other $result

First off did you set the content as a blob in mysql. Second, go into your http conf file and just like what veedeoo said, change the values to something like that. Third, no need to have a hidden field for the upload size, just have the file upload and a submit button. Fourth, your $query statement needs to change to

mysql_query("INSERT INTO Persons (name, size, type, content)
VALUES ('$fileName', '$fileSize', '$fileType', '$content')");
or
$query = "INSERT INTO Persons (name, size, type, content)
VALUES ('$fileName', '$fileSize', '$fileType', '$content')";

Hope this helps.

you can try using the LIKE operator in your mysql statement.

SELECT fields FROM database WHERE field LIKE 'variable'

ex
SELECT lastName, firstName FROM employees WHERE firstName LIKE 'a%'

So what is your php problem?

Well you are echoing something that either should be wrapped in quotes or need to make it a variable. You have

<td><input type='text' name='<?php echo mark.$y; ?>' id='textfield' /></td>

Possibilities

<td><input type='text' name='<?php echo "mark". $y; ?>' id='textfield' /></td>
<td><input type='text' name='<?php echo "mark $y"; ?>' id='textfield' /></td>

So whats your problem?

Did you upload the videos to a database, or in a folder?

Are you trying to redirect the page with the nested iframe? Basically if you click delete in the iframe, you want it to redirect the main page.

Well i was searching around googling the work cosymantecbfw and i think it relates to ' co symantec bfw' for symantec. Are you running any symantec programs?

Or you could try something like this.

$("table").css({'background-color': '#ffe', 'border-left': '5px solid #ccc'})

Try this. In this case, you already have the errors being posted in the div, so this will scroll to the div with the errors.

<?php
if(isset($_POST["myform"]) && !empty($errors)){?>
    $('html, body').animate({scrollTop: $("#error").offset().top}, 2000);
<?php } ?>

in XAMPP, there is a build in mail system called mercury mail. if you want to use smpt, you need to get that info from yahoo, google.. unless your using outlook, then just use that info.

You could also try this

<script type="text/javscript">
function signup(){
    var username = document.getElementById("text_username").value;
    var password = document.getElementById("text_password").value;
    alert(username + " " + password);
    return false;
}
</script>

<form action="do_signup" method="post" onsubmit="signup()">
Username : <input type="text" id="text_username" /><br />
Password : <input type="password" id="text_password" /><br />
<input type="submit" name="submit_button" value="Signup" />

</form>

Or if your form and inputs had a name:

<script type="text/javascript">
function greeting(){
    alert("Welcome " + document.forms["frm1"]["fname"].value + "!")
}
</script>
</head>
<body>

What is your name?<br />
<form name="frm1" action="submit.htm" onsubmit="greeting()">
<input type="text" name="fname" />
<input type="submit" value="Submit" />
</form>

Hope this helps

You will probably have to use DOM and get the elements id, like in javascript. I had the same problem in javascript, but i figured it out using jQuery and its simple.

Click Here

Dude,
if you want to sort by desc, or asc, then just have an html option where the values are <?php self...?sort=asc

then have your php code

if(!isset('something')) {
mysql code ORDER BY ASC
}

check out www.w3schools.com. really good play to begin learning.

Mark solved please!

Sahil89 commented: :D ya its really bad thing, newbies should learn it. +2

Now, when you log in as Admin, do you "Sucessfuly" log out so the session is destroyed? The only thing i can think of is the session not being destroyed and the member gets the session because its still active. Or when you log in as Member, what is the "Particular Page" that you go to, and does members session work on all other pages? Or do you log in as Member and you click on a nother page after logging in and you get admin role?

Please post your code of what you have for the php so we can look at it. Right now all we can do is just spitwad at the idae

Did you try taking out the return in your onClick? onClick="dynaBlock()"

You need to make your parent div have a position: relative and #answer 1 & 2 need to have position:absolute; z-index: 100

# .cont a { text-decoration:none; font-family: Calibri, Arial, sans-serif; font-size: 14px; color:#000; position:relative}
#answer1 { border: 1px solid black; color: grey; width: 300px; display:none; margin-left: 115px; margin-top: -15px; padding: 10px; font-size: 12px; font-family: Calibri, Arial, sans-serif; height: 135px; position:absolute; z-index: 150; }
#answer2 { border: 1px solid black; color: grey; width: 300px; display:none; margin-left: 115px; margin-top: -15px; padding: 10px; font-size: 12px; font-family: Calibri, Arial, sans-serif; height: 135px; position:absolute; z-index: 150; }
Albert Pinto commented: Thanx.... +0

$search_id = $_POST['searchID'];
$searchById = "SELECT * FROM tc_tool.forms WHERE doc_id LIKE "%$search_id%"';

But i think the problem is with your table name because database and table names cannot contain “/”, “\”, “.”, or characters that are not permitted in file names.
So..

tc_tool**.forms**

is innapropriate, but

tc_tool

is ok to use.