Ingolme

Why not use the external URL all the time? It should work even if you're inside the network. If you can do that, it's much better than any other solution. Other solutions are complicated and less reliable.

This probably has something to do with the Spry menu. Maybe you're using an old version of dreamweaver. You should look into CSS menus because currently your menu could be working exactly the same as it is now except without any Javascript needed. One problem with your page is that you're using an XHTML Doctype but your page is written in HTML (and has a few HTML errors as well). http://validator.w3.org/check?uri=http%3A%2F%2Fwww.todd-holland.com%2F&charset=(detect+automatically)&doctype=Inline&group=0

XPath can't do that because "Harry" is not part of the document structure. If you want it to be accessible to XPath you would have to put "Harry" and "Potter" in two separate nodes. <title> <word>Harry</word> <word>Potter</word></title>

The link you're providing doesn't work. And it's actually impossible to tell using Javascript which programs are installed on a user's computer.

The language used to program a game isn't relevant to how well the game is made. Once a game is compiled it's pretty much impossible to tell what language it was programmed in. What's important in game development is knowing software engineering. The whole process behind making software which involves a lot more than just programming. During the development process a company decides which tools are the best to build the game in based on many factors. If you want a cheap way to develop across multiple platforms, Java is the best language. C++ is an efficient language but it's basic and not specialized for making games which means you need to do more work than if you used a game-oriented language. Game-oriented languages trade cusomizability for faster development times. All the languages have their purposes and depending on your project's budget and time constraints you may choose one over another.

It's right there in the W3Schools tutorial: http://w3schools.com/php/php_file_upload.asp

Check to see that $rows['decision'] has the value you expected it to have.

You have to decide the length of the array before hand and then create arrays as elements of the parent one. var a = [];a[0] = [];a[1] = [];a[2] = [];a[3] = [];a[4] = [];

This is from the MDN reference: I don't know if all browsers behave this way, you can test to make sure.

What kind of help are you looking for? This is a very large task. Proper design is important before you start even writing code. Decide in theory what every part of the program does, then design the database tables it needs, then decide the structure of the code.

Google provides translation for websites. I recommend hiring a human translator for your website if you're managing a professional website. Being bilingual myself I can atest to having seen some pretty stupid translations by automatic translators.

That looks like a hack, it's incorrect CSS. I wouldn't recommend using it. It's meant to take advantage of mistakes in older browsers' CSS parsers.

Most likely the problem occured at {$courseid}. Check that $courseid has a valid value. I assume that in your original code query actually is $query or it would have given a syntax error rather than a MySQL error.

You should show your code. I can't know why it's not working if you don't show what you did.

A CSS pixel refers to a screen pixel. There's no difference.

The way you had it should work, but I'd initialize the object first before trying to set properties. var x = 'key';var data = {}; // Initialize the object first.data[x] = 'value';console.log(data);

Theoretically, DTDs tell mark-up readers what is the proper structure for an XML or SGML document. In practise, the browser just chooses a rendering engine based on whether the DTD is a current standard or not. You should use one of the standard DTDs, there are eight of them which you can find here: http://w3schools.com/tags/tag_doctype.asp If you don't use one of those DTDs, the browser will render your page using an old rendering engine which will cause it to look very different in different browsers.

Just setting the image as the background of the <body> and setting background-position to top center should work. Copyright is as strict in web design as anywhere else, it's a law that applies to everybody regardless of what field their publications are in. However, until the company actually files a copyright claim against you, you aren't caught and won't have any trouble. Large companies don't usually go around the internet taking down websites because it's counterproductive.

Is the scene called "Map" or is it something else, like "map" or "MAP"?

They're all pointing to the same array. What you need to do is make a copy of that array before shuffling it.

Object doesn't have any properties or methods. It has prototype and constructor, but so do all the other objects.

An object, in Javascript, is a data type. It's "name", like other variable names, isn't accessible from its value. Since everything is a property of the window object, what you can do is this: var x = {a: 1};getName(x);function getName(obj) { for(i in window) { if(window[i] == obj) { alert("The object's name is " + i); } }} But really, I see absolutely no need for this in Javascript. At most you would need it for debug purposes.

The question is why you want to display something like that. There's no practical purpose. Objects aren't the only ones. Every single value has that problem. Take this, for example: var a = "String";var b = 5;var c = false;fnshow(a);fnshow(;fnshow(c);function fnshow(data) { var out=document.getElementById('out'); var o = '?'; out.innerHTML += typeof data +' '+ o + ' = '+ data;}

The object could be anonymous. You don't need the name, just a reference to the object itself. Take this, for example: for (i in {a: 1, b: 2 } ) { document.write(i);} What's the "name" of that object?

You actually do have a syntax error here which is visible using a syntax highlighter. function test(value){value = document.getElementById("colorOption").selectedIndex;alert("document.getElementById("colorOption")[x].value);} The other problem is that you're not using the options property of the <select> element. function test() { var element = document.getElementById("colorOption"); var index = element.selectedIndex; var option = element.options[index]; alert(option.value);}