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Faracus

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About Faracus

  • Birthday 05/18/1989

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    HTML, CSS, PHP

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    http://shackguys.siteburg.com
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    A small world in my head.

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  1. Faracus

    Is this possible?

    I'm looking to start up a website where a user can build their own computer, and then check out the stats, the issue is that some parts can be used in multiple slots (e.g cases can support multiple motherboard types) so I'm wondering can I do something like Comer Separated Values in the database, then have PHP check each value individually to check if that is in there. For example this Case can support the following motherboards ATX Micro ATX Baby AT E-ATX So I would like it to only show up when the user has selected one of those types of motherboards.
  2. thanks for that, it was a semi-colon from the previous line. But now I'm getting an error saying that the 64 character database table is not long enough for the final password, how long is it going to be?
  3. it was set to E_ALL, but display errors was set to off, now after turning them on I get a Parse error: syntax error, unexpected T_VARIABLE pointing to the link of my cyrpt, Ihave fixed the . before the salt but it's still giving me this error $encpassword = crypt($password, '$6$rounds=5000$'.$salt);
  4. which is why I updated the code to $encpassword = hash("sha512",$password.$salt); I'll try the crypt function. *EDIT* After switching my code to $encpassword = crypt ($password, '$6$rounds=5000$'$salt); I am still getting the error 500 with still nothing showing in my error logs.
  5. It looks like I have to upgrade my PHP, thanks for the point in the right direction. I thought it was a mod in apache that I needed, not PHP itself. After looking I do have SHA512, but I am still getting the 500 error, and after looking in my error logs there is nothing there about it. I've updated my hash code to:$encpassword = hash("sha512",$password.$salt);
  6. When I try and hash passwords on my site using SHA512 it goes straight in to a 500 error, but as soon as I take the hash out, then it loads properly. $encpassword = hash(sha512($password.$salt)); that is the line of code that is causing the problem. I'm running the web server off my computer for development, I have Apache 2.2.22 with PHP 5.3.14. I've tryed a google search, but SHA512 is a popular checksum for downloads so I get a lot of hits with that.
  7. Faracus

    Broken Style Sheet

    I'm sorry but just saying, well it's not the right way, isn't really telling me how to fix it. I am thankful for the help here, don't get me wrong.I have tryed to use other positioning elements, and I found that this one carry's over better through multiple browsers, and is more accurate. But with both absolute, and relative positioning the table still ignores the CSS, and I don't want to have to write in a new block of CSS just to position a table, in case I want to continue the text below it.
  8. Faracus

    Broken Style Sheet

    From my screen, and everyone else I get to look at the page, the only issue is the table, would a relative positioning be the solution to this then? EDIT: Even when switching to relative, the table breaks CSS and starts all over again.
  9. Faracus

    Broken Style Sheet

    If it works for me (most of the time), then whats the problem?? I don't see how any of these answers help me with my problem. I'm not trying to be rude here, but I only asked for help with this problem, not to be criticized on my work.
  10. Faracus

    Broken Style Sheet

    i have no problems with overlapping, and I use absolute, as I find it the easiest, and also most convenient to use.
  11. Faracus

    Broken Style Sheet

    Hi there, I have a external style sheet, and every now and then I run in to a problem where a html tag breaks the css formatting and the broken part, just goes to the top left of the page. The ones that are the most common are <form> and <table>, the table one is what I am working on right now. Is there any way I can get it to not break the CSS and continue to display as normal? CSS: p.main{position:absolute;left:5%;top:10%;} HTML/PHP echo "<p class='main'>Weclome to Precinct 1's Information Page. This precinct is made up of 6 districts as shown on the map.<br /><br />The following people are currently online in this district.<br /><table border='1'><tr><th>Officers Currently Online</th></tr><tr><td>Members would show up here</td></tr></table></p>"; result: http://raccooncitypd.clanteam.com/RPD-Computer/precinctinformation.php?precinct=1
  12. Thanks for all the help.
  13. if I put in "You have chosen " . $query1 . ", and updated the total to " . $number1 . ".<br /> it prints out the right values. and if I print out sql3 I get UPDATE lottomax SET 5=1
  14. sadly, I'm still getting the error "Error: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '5=1' at line 1"
  15. I'm trying to make this code work $sql3= "UPDATE lottomax SET " . $query1 . "='" . $number1 . "'";if (!mysql_query($sql3,$con)){die('Error: ' . mysql_error());} and I keep getting this error:Error: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '5='1'' at line 1
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