Ingolme
-
Posts
14,897 -
Joined
-
Last visited
-
Days Won
176
Content Type
Profiles
Forums
Events
Everything posted by Ingolme
-
You just need a set of <div> elements one after another, each with height:100% Be sure that the <body> and <html> element heights are also set to 100%.
-
It doesn't exist. Check that the path is correct and be sure that uppercase and lowercase letters are taken into account.
-
Gradients are usually done with 1px thin images that are repeated vertically or horizontally as CSS.
-
Applying the background-size property on the html or body element will work. Background-size was introduced in CSS3 and won't work in older browsers.
-
array_unique() is working just fine. Just type this so that you can see that the $product array only has unique items: print_r($product);
-
Java is not better or worse than PHP.Your problem here is that you don't seem to have an idea of what data is going where in your system.
-
Which tags aren't working for you?
-
What you need is something like this: var arrPos = [[37.772323, -122.214897], [21.291982, -157.821856], [-18.142599, 178.431], etc. ]; var flightPlanCoordinates = [];var temp;for(var i = 0; i < arrPos.length; i++) { flightPathCoordinates.push( new google.maps.LatLng(arrPos[i][0], arrPos[i][1]);} The push() method adds an element to the end of an array.
-
Your document structure is really unorganized. You should fix the validation errors before continuing.
-
Not all the tags work yet, though a lot of them do.
-
Are you thinking about linux servers? I'm looking up "crontab" and there's hardly a mention of apache.
-
Is there a particular reason you need to know? There might be other ways to solve your problem.
-
Or var cellCount = document.getElementsByTagName("table")[0].getElementsByTagName("tr")[0].getElementsByTagName("td").length;
-
You've added a [2] to the person. Most likely if you add [0] to people you won't need the for-each loop.
-
There is no way to retrieve login credentials. If it were so easy any website could easily be hacked.
-
That's the URL, but what shows up in the window?
-
Oh, well. I mentioned how to access your table using getElementsByTagName in one of my earlier posts.document.getElementsByTagName("table") returns a list of every single table element in your document. The first one being 0, the second 1 and so on. Figure out which index your table has and then access it from the list of table nodes that are returned. var tables = document.getElementsByTagName("table");var myTable = tables[0]; // 0 is the first table of the list
-
The question is why you think you need getElementsByTagName(). The getElementsByTagName() method is meant to return a list of nodes and is not the best option to get a single node.
-
No, I cannot help any further if the parser is reading a different document than the one you're giving it.
-
We do have a thread concerning your issue: http://w3schools.inv...showtopic=44106 Another problem seems to be the misuse of the $count variable here: $count = "select count(*)-1 from core_store_group";$i = '';$sql = '';for($i=0;$i<$count;$i++){
-
You still have a <body> tag before the <?xml ?> declaration.
-
You have an unneeded square bracket. at the end of that line.And before the error shows up, you need another ]> after that <!ELEMENT> entity to close the <!DOCTYPE> that's at the beginning of the document.
-
If it says Not Found it means Not Found. Make sure the files are actually in the right directory and that you have written the file names correctly. File names are case sensitive. I notice some of the requests are to .jpg files but the actual files are .JPG
-
The <?xml ?> declaration has to be the very first thing in the document. Where did that <BODY> tag come from?
-
Here's the first thing I spotted. <?php while($data=mysql_fetch_array($query)){<option value="<?php echo $data['id'] ?>"><?php echo $data['name'] ?></option><?php } ?> You should read the error messages. They're very descriptive and they tell you exactly where the error is occurring so that you can find it easily.
