Search Results

Search found 4 results on 1 pages for 'mxmastamills'.

Page 1/1 | 1 

• Macro To Choose "Don't Save" on pop-up window (Mac OS X)

  - by MxmastaMills

  I was wondering if there is a keyboard shortcut to choose "Don't Save" when an alert comes up. I'm talking specifically about Photoshop but the same action happens in many applications. It's the standard pop up that shows when you are canceling out of a window and haven't saved the document or project. To be clear it says "Don't Save", "Cancel" and "Save". Save is generally the default but I was just curious if there's a way to choose "Don't Save" without clicking. I know this may be a pretty simple question but I was curious about it and it's surprisingly hard to Google... Thanks!

  Read the article

• Cannot Resize DIV to anything smaller than maximum width

  - by MxmastaMills

  Quick CSS Question. I cannot seem to figure out a very simple CSS/HTML problem. Here's a JSFiddle. http://jsfiddle.net/BmM6t/ Basically, these two divs are not resizing to anything less than 100% width: <div id="main_pic" class="main_div"> THIS IS WHERE THE MAIN PIC WILL GO. </div> <div id="signin" class="main_div"> SIGN IN TO THE SITE! <form> <label for="name">EMAIL/USERNAME:</label> <input type="text" name="name" id="name"> <label for="email">PASSWORD:</label> <input type="text" name="email" id="email"> <br /> <input type="submit" value="Let's Play!"> </form> </div> You'll see that the two divs with class main_div actually take up the entire width of the screen for some reason (you can see that if you inspect the element) even though their widths are clearly written. When you try to change the width of one of the main_div's it changes the width but if you inspect the element, it just changes the width of the object and adds padding around it to make sure it still takes up the entire width of the screen. I'm sure it's some little error I've made but I can't find it. Thanks for the help!

  Read the article

• Syntax Problems of if Statement (php)

  - by MxmastaMills

  I need a little help with an if statement in php. I'm trying to set a variable called offset according to a page that I am loading in WordPress. Here's the variable: $offset = ($paged * 6); What it does is it loads the first page, which is: http://example.com/blog and $offset is thus set to 0 because $paged is referring to the appending number on the URL. The second page, for example is: http://example.com/blog/2/ which makes $offset set to 12. The problem is, I need the second page to define $offset as 6, the third page to define $offset as 12, etc. I tried using: $offset = ($paged * 6 - 6) which works except on the first page. On the first page it defines $offset as -6. SO, I wanted to create an if statement that says if $paged is equal to 0 then $offset is equal to 0, else $offset is equal to ($paged * 6 - 6). I struggle with syntax, even though I understand what needs to be done here. Any help would be greatly appreciated. Thanks!

  Read the article

• Cannot Display Data from MySQL table

  - by MxmastaMills

  I've got a pretty standard call to a MySQL database and for some reason I can't get the code to work. Here's what I have: $mysqli = mysqli_connect("localhost","username","password"); if (!$mysqli) { die('Could not connect: ' . mysqli_error($mysqli)); } session_start(); $sql = "SELECT * FROM jobs ORDER BY id DESC"; $result = $mysqli->query($sql); $num_rows = mysqli_num_rows($result); Now, first, I know that it is connecting properly because I'm not getting the die method plus I added an else conditional in there previously and it checked out. Then the page displays but I get the errors: Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result, boolean given in blablabla/index.php on line 11 Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given in blablabla/index.php on line 12 I've double-checked my database and there is a table called jobs with a row of "id" (it's the primary row). The thing that confuses me is this is code that I literally copied and pasted from another site I built and for some reason the code doesn't work on this one (I obviously copy and pasted it and then just changed the table name and rows accordingly). I saw the error and tried: $num_rows = $mysqli_result->num_rows; $row_array = $mysqli_result->fetch_array; and that fixed the errors but resulted in no data being passed (because obviously $mysqli_result has no value). I don't know why the error is calling for that (is it a difference in version of MySQL or PHP from the other site)? Can someone help me track down the problem? Thanks so much. Sorry if it's something super simple that I'm overlooking, I've been at it for a while.

  Read the article

1