Reputation Activity
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Ryan_Belfast reacted to FoxIII in A FooterJust for info, you have the stylesheet being loaded twice in the header. Also, I don't think you need to have two footers. You seem to have one inside of another. Just stick with one.
#footer { width: 100%; height: 50px; margin: 0px; padding: 0px;
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Ryan_Belfast reacted to Faevilangel in A FooterThe reason for your borders are because you haven't reduced the added browser padding and margins. Add the following to your body
margin:0; padding:0;
Your footer code can be streamlined and fixed by changing to:
#footer { width:100%; height:50px; clear:both; background: url("./bg.jpg") repeat-x scroll 0 0 transparent; }
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Ryan_Belfast reacted to StuartPB in A FooterI've just been looking at the HTML, and you have a load of paragraph tags that are unclosed as well as the footer problem. To resolve the footer problem, you need to add the following to your body tag:
margin:0;
padding:0;
Cheers,
Stu
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Ryan_Belfast reacted to Lev in Alternative to onblur ?http://www.brightcherry.co.uk/scribbles/2008/12/04/html-form-effect-remove-default-value-on-focus/ -
Ryan_Belfast reacted to pat24 in Cant click on form elementsyou will have to use position:relative and z-index:1on your #form
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Ryan_Belfast reacted to Jay Gilford in Frustrating Error$query = "SELECT `MemberName`, `Age` FROM `Member` WHERE `BandName` = '".$value."'";
Try that
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Ryan_Belfast reacted to webdeveloper93 in Frustrating ErrorUse the below code
mysqli_query($conn,$query) or die(mysqli_error($conn));
Should fix the prob please give me a +1 if this helps
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Ryan_Belfast reacted to nathan@attaindesign.co.uk in Retrieval QueryOnce you have connected to the database preform a simple SELECT as shown below:
<? $sql = "SELECT BandName FROM Member"; $result = mysql_query($sql); //Loop through the results and create the select ?><select name="band_name"><? while($data = mysql_fetch_assoc($result)){ ?><option value="<?=$data['BandName'];?>"><?=$data['BandName'];?></option><? } ?></select><? ?>
Then on the page that will recieve the post
if(isset($_POST['band_name'])){ //This should be validated and checked to prevent SQL injections $sql = "SELECT BandName FROM Member WHERE BandName = '".$_POST['band_name']."'"; $result = mysql_query($sql); while($data = mysql_fetch_assoc($result)){ echo $data['MemberName']; } }
This was very quickly wrote and very basic, but it should help you on your way.
Hope your studying goes well
Nathan
Attain Design
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Ryan_Belfast reacted to bocaj in Images wont displayIt can't find the images.
Right click where it says monogram > Open image in new tab
It won't be there. Double check the file names, capitals and extensions. You've put jpeg, are they jpg?