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DarkKnight2011

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About DarkKnight2011

  • Birthday 07/02/1983

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    Male
  • Location
    UK
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    29

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  1. hi, To get an additional window to appear that will show content from another URL then yes, you could use something like window.open(url, "Popup Window", "width=400,height=300,resizable=yes"); Although, As gristoi stated, this is a very old (and also annoying) way of showing information even if it's not intended for advertisements, the code provided by gristoi can be used to popup a form or some other small information on the current page the user is viewing, Which is much preferrred. And yeah, Inline javascript.... eeeeeewww
  2. Did this work ok for you? could you please mark as answered if so, Thanks
  3. try one of these or maybe best to buy a book, but be prepared to put alot of time into it, You can't master programming languages easily www.php.net http://www.codecademy.com/tracks/php The topic of your post concerns me though, it wont be fast lol
  4. Try this... <script> $(function() { $( "#delDate" ).datepicker({ dateFormat: 'd MM yy', changeMonth: true, changeYear: true, minDate: "+1D", constrainInput: true }); var currentdate = new Date(); if (currentdate.getHours() >= 17) { currentdate.setDate(currentdate.getDate() + 2); $( "#delDate" ).datepicker( "option", "minDate", currentdate); } }); </script>
  5. I would suggest that you use an associative array, ie. $arr = array( 'value1' => 1, 'value2' => 2 ); echo $arr['value1']; // outputs 1 or given your latest comment... $arr = array( 'tableName' => 'Table1', 'tableData => array( 'value1' => 1 ) );
  6. Yeah I agree, This does sound like a fairly big project, if your new i would personally suggest that you start off with something smaller first, There are alot of considerations regarding which technologies to use for a project, and varies for each one. Personally my choice for that application would probably be Backend - PHP (Symfony2/Zend Framework 2) FrontEnd - ExtJs (www.sencha.com) or maybe something like BackboneJS with Twitter Bootstrap But to use those things would take a lot of learning, Maybe try a smaller project and see how you progress, You could of course hire an experienced developer to build this on your behalf but it would be expensive
  7. Like Jessica has mentioned, when a user is registered or changes their password, store the date this occured in the user table, Then in your script you can check to see if that was last done more than 90 days ago, if it was more then show a change password screen otherwise success
  8. your welcome mate, could you mark the question as answered, Good luck with the project
  9. sorry I'm really missing something here I think, maybe this helps... $color = ''; switch($severity) { case 'Extreme': $color = 'red'; break; case 'Severe': $color = 'amber'; break; default: $color = 'white'; } echo "<span style='color: $color;'>$severity</span>"; Does that help?
  10. that is the var_dump($sql) line we added earlier, just remove it, was to try and find out what is happening with the query
  11. sorry what do you mean? the query looks ok? I have never used on duplicate key to be honest so not sure about that
  12. SELECT requests.*, users.*, areas.* FROM requests LEFT JOIN users ON requests.requesterID = users.userid LEFT JOIN areas ON requests.areaID = areas.areaID try that, you had brackets in there which weren't needed
  13. in this case your using a variable that hasn't been defined before, so either your calculate function isn't working or you need to initialise it first, ie. add $calories = 0; near the top of your code before you use it in a calculation
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