Oh Ok...I understand the problem now! Sorry about that. Well I'm glad you solved your problem, all I ask is that you mark this thread as solved so others don't attempt to solve this already fixed problem. Thanks!
Well the first problem is you didn't specify a type of input. This isn't the main problem, but causes the document to be an invalid (X)HTML page. I will look more into your problem and see what the real cause of it is...
EDIT: I notice that in the first snippet, you have qty containing an array, but in your PHP you are using it as a single variable.
The other thing you should note is making an input an array only works when there is multiple inputs of the same type with that exact name.
Well, I'm assuming your a beginner to PHP. This would be like asking a small child to launch a rocket to the moon and back. Can it be done...yes, but I think you should try immersing yourself in PHP and learning about the language before your try something ambitious. Never-the-less, here's some ideas:
You will need to create an engine that spiders the web and indexes words for each website. (This shouldn't be done in PHP, since there is a max execution time on most free accounts of 30 seconds. How about trying C or Java)
You will need a PHP script to search through your keyword database for a query and display relevant results.
It is way more complicated than this. First off, there is enough information on the internet to surpass millions of large hard drives. It is just not feasible to spider the web unless you have access to large server banks. Then you will also need to evaluate each website for reliability and relevance, an algorithm that Google, Yahoo, and others have been perfecting for many years now. In short, this is no small project. Might I suggest trying to create a small version that just spiders your personal website. It's the same principle, just on a much smaller (and more manageable) scale. Good Luck!
Another suggestion: If you are worried about search engines accessing pages other than the first, you can either set the robots.txt to allow a web spider access to the first page of every article. You might also want to try setting up a simple script that you include on pages other than the first that checks if the referrer is from a search engine and add some Javascript to manually force Google Analytics to track the user.
If you are satisfied with your solution to the problem, I ask that you close the forum and continue to post on the DaniWeb forums!
Well by simple logic, why not assume that one unique view is viewing the first page of an article. If so, why not just count the unique views on the first page of each article.
A note about search engine bots: Most search engines identify themselves with a special HTTP header. If you can write your own unique view counter (Cookies will work; it's how Google does it!), simply set it to ignore search engine spiders. Googlebot uses the Useragent String "Googlebot", so simply ignore the view counter when the $_SERVER['HTTP_USER_AGENT'] is "Googlebot".
Side Note: I would strongly advise against using an IP to track a user. First, if a user is on a large network running through a central router, all the computers on that network would have the same IP address. Large businesses and colleges usually have several IP addresses, but not a unique IP for each computer. This means that if 100 unique people on one network view an article once in a 24 hour period, it will only register as 1 unique view, instead of 100. Beyond this, IP addresses are easily masked. Users can hide behind a proxy or use a client such as Tomato to change IP addresses.
Try this..you need some quotes (You also need a form action..which can be easily fixed by PHP_SELF which will insert the name of the current PHP page):
You can check for a search by the employee's name. Then you can check for a space. If there is a space you can split the two names with explode and then perform a query that matches first and last name. If there isn't any spaces only perform a query for the first or last name (Depending on what you want).
<?php
if(isset($_POST['searchType'])) //Check first, get values later
{
$method=$_POST['searchType'];
$search=$_POST['query'];
switch ($method)
{
case 'EmployeeFName': //Switch is not like if...you don't need an expression, just a value will do
if(empty($_POST['query'])) exit("Please Fill in a name of an employee you want to search. Click <a href=search.php>here</a> to go back.</p>");
else
{
echo"<i>Your search for {$search} had the following results:</i>";
//Here's the magical splitting of the names
$arr = explode(" ", $search);
$search_first = $arr[0];
$search_last = $arr[1];
$query = mysql_query("SELECT * FROM Employee WHERE EmpFName = '".$search_first."' AND EmpLName = '".$search_last."'");
$result = mysql_query($query);
if(mysql_num_rows($result)==0) echo exit("<font color='red'>No Record Found</font>");
echo"<table class='full'>";
echo"<th></th><th>ID</th><th>Employee Name</th><th>Title</th><th>Team</th><th>Department</th></tr>";
while ($row = mysql_fetch_array($result)) //Ohh! You can only fetch from a result RETURNED by mysql_query, not the actual query
{
//No need to setup individual variables...just use the $row array!
echo"<tr><td class='view'><form action='employee.html' method='post'><div><input type='submit' value='View' class='button' /></div></td><td name='userID' class='id'>".$row["EmployeeID"]."</td></form><td name='userName' class='record'>".$row["EmployeeFName"];." ".$row["EmployeeLName"]." </td><td name='userTitle' class='record'>".$row["EmployeeTitle"]."</td><td name='userTeam' class='record'>".$row["EmployeeTeam"]."</td><td name='userDept' class='record'>".$row["EmployeeDepart"]."</td></tr>";
}
}
break;
//And on, and on, etc.
}
}
function mysqlConnectAndSelect()
{
checkLogin();
sqlConnect();
mysql_select_db("titans", $con);
}
?>
There are also lots more problems beyond this one. I'd be glad to fix them all for …
Alright then, have you replaced all your files (for this project) with one file called paycheck.php and added only my code to it. Line 10 is the end of the first if block. I tried it on my XAMPP and it worked perfectly and didn't have any errors. (And the reason why the form is being filled out again is because a value is missing...it simply fills in the values you gave it and leaves the other required ones blank). On a side note: Don't worry about all the questions. When you are new to a language it's hard at first, but you'll get it eventually and maybe answer people's questions on DaniWeb!
EDIT: One little fix that I just remembered. It shouldn't fix your error, but it should fix and output problem:
<html>
<head>
<title>Paycheck Calculator</title>
</head>
<body>
<?php
//If the form has been submitted
if($_GET['submit']=="Submit"&&isset($_GET["hoursWorked"])&&isset($_GET["wagesRate"]))
{
//Return the paycheck and give user the option to calculate again
echo "Your Paycheck is: <b>".(($_GET["hoursWorked"] * $_GET["wagesRate"]) + (($_GET["hoursWorked"] - 40) * $_GET["wagesRate"] * 1.5))."</b><br/>\n<a href='paycheck.php'>Calculate Again</a>";
}
else
{
//Else return the form with the values that need to be filled
echo "<form action='paycheck.php' method='get'>\n <label for='hoursWorked'>Hours Worked: </label><input type='text' name='hoursWorked' value='".$_GET["hoursWorked"]."' /><br/>\n <label for='wagesRate'>Wage: </label><input type='text' name='wagesRate' value='".$_GET["wagesRate"]."' /><br/>\n <input type='submit' name='submit' value='Submit' />\n</form>";
}
?>
</body>
</html>
I would also suggest being more specific about getting variable from forms. Use $_GET[] for get requests and use $_POST[] for post requests. Also, there is one more problem involving the $paycheck variable. You set the x variable to be the value of paycheck, then you do some calculations, but only show the original paycheck variable. Did you mean to set paycheck to the value of the calculations and then show that variable?
Also, I believe the problem was that instead of $links , %links as used, but your solution provides a safety if there are no rows returned (Which is a good thing to have!).
The proper use of this would be declaring the $link variable as an array before you start looping through the array results of the query. Your code should look like this:
$result = mysql_query($sql); //Make sure to use a semicolon at the end of a line, not a colon!
$link = array(); //Here is where you declare the array
while ($row = mysql_fetch_array($res))
{
//$row[tableLegends]; <---Not sure what this is for, and since it doesn't appear to have a function I commented it out
$link[] = $row; //You were right by approaching this with $link[] and not $link, if you used $link then link would only contain the value of row and won't be an array of the rows
}
Also looking at your foreach code, what is %links . Don't you mean $links :
Ok this makes much more sense! I'll take a look and see if I can come up with something. One question: When you click submit, do you want PHP to show the background and text combined together into an image, or replicate the table on the forum page?
Koyel555, I think you might need to give your controls some more thought. You say there are several divs and several factors (font, color, size) but you only mention one control - a select menu.
It depends exactly what you want but you may need one menu per div, and possibly one menu per factor per div.
Another way of diing it would be to have a single set of controls (one per factor) and a way to signify an "active" div on which they operate - one div at a time.
It is certainly hard to see how a single select menu would offer adequate control.
Airshow
Ahh....yes! I clumsily forgot to place .style before the .color or any other property. :D Wouldn't be the first time... ;)
It would help if you could use some clear and exact English so that we can all understand your question. First off, why are you placing DIV's inside of a table? TR's belong in a table and TD inside that. Why not use a bunch of DIV's? How is the background being changed? Couldn't you just create in invisible text field containing the address of the background image since it is dynamically created?
Ok this makes much more sense now!! :D Your problem is this: Everything is correct with the checkboxes. Your checkbox method does work, but your PHP script is not receiving these values correctly. I have fixed the problem area so that it should now work!
if($_POST['delete']=="Delete") {
$checkbox = $_POST['checkbox'];
if($checkbox) {
foreach($checkbox as $box) {
$result = mysql_query("DELETE FROM $tbl_name WHERE id='".mysql_real_escape_string($box)."'") or die("Query Error!");
}
}
if($result){
//Here a code to reload change.php
}
}
Also as a note, the changing file doesn't require admin authentication. If anyone knew the location of this change file they could access it and delete rows in tables without loging into the admin panel you have built. I would suggest checking the admin username and password before deleting.
The header() function sends a raw HTTP header to a client.
It is important to notice that header() must be called before any actual output is sent (In PHP 4 and later, you can use output buffering to solve this problem):
<html>
<?php
// This results in an error.
// The output above is before the header() call
header('Location: http://www.example.com/');
?>
I apoligize for suggesting this, but I noticed that you use 'echo "<meta http-equiv=\"refresh\" content=\"0;URL=change.php\">";' and 'echo "<meta http-equiv=\"refresh\" content=\"0;URL=admin.php\">";'. There is a better method of redirecting users it also has less over head I think on both sides of processing the web page/there is less typing involved which makes development easier. I thought I could pass this on for future reference for you.
As example:
header("Location: admin.php");
By the way have a great day. :)
Matthewl beings up a good point. The only problem is that once you have placed content on a page (He has already echoed a table), neither the meta tag (which belongs in the head) nor the header() function (which only works before there is any text outputted) will work. This also made me realize that not only is this script incorrect in PHP, but also is not a valid HTML document (There is no HTML, HEAD, or BODY tags!).
First off, Your Welcome! :D About that line: I didn't realize that you used that result variable so here is the corrected version with explanation.
$result = mysql_query("DELETE FROM $tbl_name WHERE id='".mysql_real_escape_string($del_id)."'") or die("Query Error!");
First off, never echo the actual MySQL Error to a page visible to anyone. This could cause the integrity of your database to be sacrificed (I always test a script and if it doesn't work, run the problem queries through the PphMyAdmin Sql section). Next, the mysql_real_escape_string() is so that malicious users can't perform a MySQL Injection attack on your script (Looking at it now it seems pretty vulnerable). And the reason for the quotes is basic PHP. When you put a function result in a string you have to put a quote that started the string (In you case the ", double quote, from "DELETE) to tell PHP execute this and then a period, the function, and another period and quote. In your case you also need to include the MySQL ' and ' (single) quotes so that MySQL determines that $del_id . Beyond that I just realized that your implementation with the checkboxes it incorrect. $_POST[] variables are not arrays, they are strings (or integers, etc.). So in this case you will need to change the code that creates the checkboxes. I also determined that their are problems way beyond the scope of this post. For one, you add a meta tag to a page that has content. (In PHP you …
$checkbox = $_POST[checkbox]; //Problem #1
if($_POST['delete']){ //Problem #2
echo "test"; //Nice Debugging But This Should Fix It So We'll Remove It
for($i=0;$i<$count;$i++){
$del_id = $checkbox[$i]; //Problem #3
$sql = "DELETE FROM $tbl_name WHERE id='$del_id'"; //And to save a line We'll condense this
$result = mysql_query($sql);
Problem #1 This line should be:
$checkbox = $_POST['checkbox[]'];
This is because the name of the checkbox is checkbox[] and $_POST['checkbox[]'] holds the value.
Problem #2 This line should be:
if($_POST['delete']=="Delete"){
This is because the submit button is named delete and when clicks sends the post variable $_POST['delete'] with the value "Delete".
Problem #3 You don't need this line. As I mentioned in Problem #1, the variable $checkbox will contain the id that you seek in this line.
Housekeeping I always hate when people make a variable to pass to a mysql function. Why not just put it in the function like this:
$result = mysql_query("DELETE FROM {$tbl_name} WHERE id='".$del_id."'");
You should also know that your script is vulnerable to a MySQL Injection. So I would protect against that (See below I did it for you).
The End Result So fixing all those lines would make the problem area look like this:
if($_POST['delete']) {
$checkbox = $_POST[checkbox];
for($i=0;$i<$count;$i++){
mysql_query("DELETE FROM $tbl_name WHERE id='".mysql_real_escape_string($del_id)."'");
Try that out and see if it works. You also have a few other little things that are making your code longer. If you absolutely want to, post …
The problem is with the descriptions. It's just that IE will stop working on errors while most other browsers will try to continue. Specifically, the problem with your code is with line 248. Specifically the line that contains:
description = this.getDescription();
I would delete that line if the image descriptions are not important (which they probably are). I'll investigate more and see what the specific problem is with this line. IE claims it as: "Object doesn't support this property or method." Did you write this code yourself or do you know if there is any documentation that we can examine? It would help to have some kind of explanation for the code.
Well at first glance it looks like your JS file contains some JQuery, so I would start by making sure you have the JQuery library included in your page. It would also help if you could give us the address of the gallery page as it might aid in finding the problem.
You're welcome. You know you can also create your own contact form (In a language such as PHP or by using a free service) so that no one would know your email, but that is totally up to you. I ask though if you have no further questions to mark this thread as solved.
Ok I didn't realize that...whoops! Anyway I believe bgeisel meant for you to get the author name first and replace the ... with the_author(). So something like this:
Correct, unless these bots can register on your sites and access the page with your email. You can always protect your email simply by making it an image and putting that image on your page instead of using a plugin like this.
Here is your code for the specific user variable. You must understand that for each user you need to create a new folder with an index.php containing the same code each time.
That is true sDJh, but wouldn't this mean that spank would have to create a folder and index.php for every user the registers? Still though, I like your solution (My other plan was using almost exactly your idea except the URL would be /user.php/username but that wasn't exactly what was required).
Make sure that you have no html or other text written with echo before that or any html or text before the <?php tag. This could be something as simple as a space:
<?php //Notice the space here
header("Location: thiswillnotwork.html");
?>
And can easily be fixed by removing the extra space or characters:
<?php //No spaces so the header call should work along with the use of session vars
header("Location: thiswillwork.html");
?>
I also find it to be a good practice to begin the session right after the <?php tag if I am using it in that script.
It looks like you need htaccess code. This will make a rule on your webserver to call a server file (I'll call it user.php) with details from your url.
RewriteEngine On
RewriteRule ^/user/([A-Za-z0-9]+)/?$ user.php?username=$1
With this code in user.php username would be a GET variable you could use. Also, I would suggest making the URL's /user/username/ instead because lets say a user called themself index.php. Then this code would redirect that to their page. There is a way around this but I would require me to know every page on your site. It's just easier to use the /user/username structure.
And that's exactly what I'm saying. It could be done with JavaScript but it won't be compatible everywhere and there's no guarantee it will work. It would be something like this...
<html>
<head><titleDynamic iFrame Height</title>
<script type='text/javascript'>
var divName = 'testDiv'; //ID of the div you want to adjust height for
function dynamicIframe(name)
{
if(name=='') name = divName;
document.getElementById('iFrame').style.height = document.getElementById('iFrame').document.getElementById(name).style.height;
}
</script>
</head>
<body>
<iframe src="thispage.html" height="100" width="100%" id="iFrame" onload="javascript:dynamicIframe();"></iframe>
</body>
</html>
Simply replace divName with the name of the tag you would like to adjust height of iFrame for.
Probably not, there is no simple way (or any way at all for that matter) to determine content height unless it is a style on the page (Ex. something like maybe:
.myDivThatIWantTheHeightFrom {
height: 50px;
}
I'm not sure though even in that case if you could determine height.
It probably is a problem with a file being used for the installation while you are trying to run the installer. Make sure to only install one program at one time and close all other applications when installing. You may also want to run the installer in Administrator mode (if you have these privileges, if not ask your system administrator). This can be done by right clicking on the installer program and clicking Run As Administrator... and then clicking allow when prompted by the UAC.
Sure here's a suggestion. Find a host with MySQL (or ask your host to install it for you) and create a database for your site. Then create a table and call it years. In this table create three columns: a name column, a year column, and a code column (all of them can be varchar). The name column should contain the names, the year column the year (first, second, third, etc.), and the code column the abbreviation (sci, etc). Then use this code:
<div class='year'>
<a href='#' onfocus="showIt('firstyear');" onclick="showIt('firstyear');" onblur="hideIt('firstyear');" taborder=1>First Year</a> <!-- onfocus onclick allows for keyboard tab as well as mouseclick-->
<div id='firstyear' style='visibility:hidden'> <!-- initially invisible -->
<?php
$con = mysql_connect("HOST", "USERNAME", "PASSWORD") or die("Cannot connect to MySQL");
mysql_select_db("DB", $con);
$result = mysql_query("SELECT * FROM years WHERE year='first'");
while($row=mysql_fetch_array($result)
echo "<input type='checkbox' value='".$row['code']."'>".$row['name']."<br>\n";
}
?>
</div></div>
<div class='year'>
<a href='#' onfocus="showIt('secondyear');" onclick="showIt('secondyear');" onblur="hideIt('secondyear');" taborder=2>second Year</a>
<div id='secondyear' style='visibility:hidden'>
<?php
$result = mysql_query("SELECT * FROM years WHERE year='second'");
while($row=mysql_fetch_array($result)
echo "<input type='checkbox' value='".$row['code']."'>".$row['name']."<br>\n";
}
?>
</div></div>
Well, I would start out by making a rectangle that is the same color as the background (Important: Make sure the shape's color has a transparency of 1%) to cover the whole stage with the rectangle tool. Next, convert the new rectangle to a shape (Make it a button). Then add AFLAX to the button that sends a message to a server page on a click. Note: If that seems too complicated you may also try another method I though of (I'm not sure if it works!!). Add the parameter border=0 to the Flash Element and then surround the whole object in <a href='javascript:void(0);' onclick='sendMessage()'> ... </a> . Then simply, using regular AJAX, write a javascript function that calls the server page and call it 'sendMessage()'. Good Luck!!
Try AFLAX. It is an AJAX for Flash. You can use some ActionScript to call an AFLAX command to send a message to a Server Page (I prefer PHP, but you can use anything that you are comfortable with) which processes the request and updates the record in a file or MySQL table. If you have any more questions please post them here or PM me! Good Luck! :)
