Craig Hopson
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Posts posted by Craig Hopson
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hi guys i cant see where im going wrong there must be an easier way to get rounded corners in IE any way...trying this https://github.com/malsup/cornerwith this code..
<script type="text/javascript" src="js/jquery.corner.js"></script><script type="text/javascript">$('#header').corner();</script>
does this look correct?
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hi guys well the question is in the topic i found this http://strd6.com/2011/09/html5-javascript-pasting-image-data-in-chrome/ and this http://www.pasteshack.net/ proving it can be done!! The reason i want this is because i want users of my website to be able to upload photos from there mobiles and file cant be uploaded via http on mobile device (i read somewhere)if there is a better way or and example code for doing this then please help Thanks.
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hi guys how do i supply website information?(see pic) i am guessing its done with meta tags hence the topic title but i have googled but with no luck
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Encrypting
in PHP
thanks
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Encrypting
in PHP
hi i have this function to encrypt data in cookies is this the correct way?
<?phpdefine('SALT', 'hyDap4XQ6TDu');function encrypt($text){ return trim(base64_encode(mcrypt_encrypt(MCRYPT_RIJNDAEL_256, SALT, $text, MCRYPT_MODE_ECB, mcrypt_create_iv(mcrypt_get_iv_size(MCRYPT_RIJNDAEL_256, MCRYPT_MODE_ECB), MCRYPT_RAND))));}function decrypt($text){ return trim(mcrypt_decrypt(MCRYPT_RIJNDAEL_256, SALT, base64_decode($text), MCRYPT_MODE_ECB, mcrypt_create_iv(mcrypt_get_iv_size(MCRYPT_RIJNDAEL_256, MCRYPT_MODE_ECB), MCRYPT_RAND)));}?>
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hi i'm stuck again i want a simple image slider but the problem is this
var finalslide=''leftrightslide[0]='<img src="dynamicbook1.gif" border=1></a>'leftrightslide[1]='<img src="dynamicbook2.gif" border=1></a>'leftrightslide[2]='<img src="dynamicbook3.gif" border=1></a>'leftrightslide[3]='<img src="dynamicbook4.gif" border=1></a>'leftrightslide[4]='<img src="dynamicbook5.gif" border=1></a>'
as you can see i need to tell it what images to display problem is my images come from this
$result = mysql_query("SELECT * FROM files ORDER BY RAND() LIMIT 20");while($row = mysql_fetch_array($result)) {$img = 'uploads/'.$row['album'].'/'.$row['filename'];echo '<span style="padding:10px;"><img src="'.$img.'" width="50px" height="50px" style="border:1px solid black;"/></span>';}
so how can i integrate them?
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start session
in PHP
i have many pages of scripts across many files all starting
<?session_start();include("config.php");
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start session
in PHP
hi I use sessions now and then i get this error
once i get this error the only way to get rid is to delete a cookie called PHPSESSID with value of = 53ff81c3601f65a11a9026a079284d89. what is this error why do i get (not often)?Warning: session_start() [function.session-start]: open(/tmp/sess_53ff81c3601f65a11a9026a079284d89, O_RDWR) failed: Permission denied (13) in /home/chsitesc/public_html/test/index.php on line 2Warning: session_start() [function.session-start]: Cannot send session cache limiter - headers already sent (output started at /home/chsitesc/public_html/test/index.php:2) in /home/chsitesc/public_html/test/index.php on line 2Warning: Cannot modify header information - headers already sent by (output started at /home/chsitesc/public_html/test/index.php:2) in /home/chsitesc/public_html/test/config.php on line 14 -
ok new problem
$('#div1, #div2, #div3, #div4, #div5, #div6, #div7').hide('slow');
any think i put in hide like 'slow', 'fast', or 'fade', {}, 1000dont work any idea?
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SOLVED like this
<script type="text/javascript"> $('#test').addClass('js');$(function() { var timer = setInterval( showDiv, 1000); var counter = 0; function showDiv() { if (counter ==0) { counter++; return; } $('#div1, #div2, #div3, #div4, #div5, #div6, #div7') .stop() .filter( function() { return this.id.match('div' + counter); }) .show('fast'); counter == 10? counter = 0 : counter++; if(counter == 10){ $('#div1, #div2, #div3, #div4, #div5, #div6, #div7').hide('slow'); counter = 0; } }});</script>
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hi guys i'm using this
<script type="text/javascript"> $('#test').addClass('js');$(function() { var timer = setInterval( showDiv, 2000); var counter = 0; function showDiv() { if (counter ==0) { counter++; return; } $('#div1, #div2, #div3, #div4, #div5, #div6, #div7') .stop() .filter( function() { return this.id.match('div' + counter); }) .show('fast'); counter == 7? counter = 0 : counter++; }});</script>
to show #div1, #div2 ect One after another Question:- How do i make it hide them all 20 seconds after the last one shows and restart? Thanks guys
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My page is a simple gallery users upload pictures mainly from mobile devices, if user takes photo and phone is sidewise picture is sidewise in gallery so need a option to rotate the image also saving image so next time they look at the gallery it's correct
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Thanks for the quick reply how would i do it with AJAX?
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hi guys i have a few problems when i use this function to rotate image
function rotate($img,$rot){global $id; //get the detail of the image $imageinfo=getimagesize($img); switch($imageinfo['mime']) { //create the image according to the content type case "image/jpg": case "image/jpeg": case "image/pjpeg": //for IE $src_img=imagecreatefromjpeg("$img"); break; case "image/gif": $src_img = imagecreatefromgif("$img"); break; case "image/png": case "image/x-png": //for IE $src_img = imagecreatefrompng("$img"); break; } //rotate the image according to the spcified degree $src_img = imagerotate($src_img, $rot, 0); //output the image to a file imagejpeg ($src_img,$img);}rotate($_GET['img'],$_GET['rot'])
PROBLEMS1.......When using browsers cache images so browser needs refreshing after use(I have used httaccess to stop cache'ing)2.......Calling the function. At the moment the function is in a file and i use the GET method from a href (<a href="rotate.php?img='.$Image.'&rot=90">Rotate</a> which flashes a white blank page before coming back to the gallary3.......Could i use jQuery to rotate (It has to save the image after rotation)4.......IS THERE A BETTER WAY TO DO THIS?????? Thanks guys
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So simple THANKS!!
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hi guys i cant get this working am i missing something simple?
<script src="js/jquery-1.7.2.min.js"></script><script type="text/javascript">$('.show-element', '.Item').hover(function() { var container = $(this).attr("id").replace('show_', 'info_'); $('#' + container).show();}, function() { var container = $(this).attr("id").replace('show_', 'info_'); $('#' + container).fadeOut(200);});</script><style>.hidden{ display: none;}</style><div class='Item'> <div class='show-element' id='show_1'>Show</div> <div id="info_1" class="hidden">some stuff</div> <div class='show-element' id='show_2'>Show</div> <div id="info_2" class="hidden">some stuff</div></div>
(does not show on hover)
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PROBLEM SOLVED THANKS BIRBAL YOU REALLY ARE THE MANyou should have to use imagecreatetruecolor() instead of imagecreate(). -
function converttojpg($originalimg){$filetype = @end(explode(".", $originalimg));if (strtolower($filetype) == 'jpg' or strtolower($filetype) == 'jpeg') { $srcImg = imagecreatefromjpeg($originalimg); imagejpeg($srcImg,$originalimg,75); imagedestroy($srcImg); $thumb_img = imagecreatefromjpeg($originalimg); $origw=imagesx($thumb_img);$origh=imagesy($thumb_img);$new_w = 150;$diff=$origw/$new_w;$new_h=$new_w;$dst_img = imagecreate($new_w,$new_h);imagecopyresampled($dst_img,$thumb_img,0,0,0,0,$new_w,$new_h,imagesx($thumb_img),imagesy($thumb_img));$ONLYname = str_replace('uploads/'.$_COOKIE['id'].'/',"",$originalimg);imagejpeg($dst_img, 'uploads/'.$_COOKIE['id'].'/thumbnail/'.$ONLYname,100); } if (strtolower($filetype) == 'png') { $srcImg = imagecreatefrompng($originalimg); imagejpeg($srcImg,$originalimg,100); imagedestroy($srcImg); $NewName = str_replace(".png",".jpg",$originalimg); rename($originalimg,$NewName); $thumb_img = imagecreatefromjpeg($NewName); $origw=imagesx($thumb_img);$origh=imagesy($thumb_img);$new_w = 150;$diff=$origw/$new_w;$new_h=$new_w;$dst_img = imagecreate($new_w,$new_h);imagecopyresampled($dst_img,$thumb_img,0,0,0,0,$new_w,$new_h,imagesx($thumb_img),imagesy($thumb_img));$ONLYname = str_replace('uploads/'.$_COOKIE['id'].'/',"",$NewName);imagejpeg($dst_img, 'uploads/'.$_COOKIE['id'].'/thumbnail/'.$ONLYname,100); } if (strtolower($filetype) == 'gif') { $srcImg = imagecreatefromgif($originalimg); imagejpeg($srcImg,$originalimg,100); imagedestroy($srcImg); $NewName = str_replace(".gif",".jpg",$originalimg); rename($originalimg,$NewName);$thumb_img = imagecreatefromjpeg($NewName); $origw=imagesx($thumb_img);$origh=imagesy($thumb_img);$new_w = 150;$diff=$origw/$new_w;$new_h=$new_w;$dst_img = imagecreate($new_w,$new_h);imagecopyresampled($dst_img,$thumb_img,0,0,0,0,$new_w,$new_h,imagesx($thumb_img),imagesy($thumb_img));$ONLYname = str_replace('uploads/'.$_COOKIE['id'].'/',"",$NewName);imagejpeg($dst_img, 'uploads/'.$_COOKIE['id'].'/thumbnail/'.$ONLYname,100); } }
Thanks guys for taking the time to look into this for me
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Yes I have sorted that out the only problem is the thumbnail quality
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ok this is original imageAnd this is after now this is not a usable thumbnail what am i doing wrong?
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i have done this but still the same any other ideas??posting the original image with that one would be more helpful. though you should use imagecropresampled() rather than imagecopyresized(). imagecopyresized() deform when it crop which could make image defective. -
Like this
imagejpeg($dst_img, 'uploads/'.$_COOKIE['id'].'/thumbnail/'.$ONLYname,100);
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like a cron job?
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This is the quality i'm getting
jquery corners
in JavaScript
Posted
strange it just started working dont know why....I dislike I.E. Thanks for the reply tho