floridaflatlander
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Posts posted by floridaflatlander
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Have you put any other scripts on this server and had them to work?
Can you build a bare bones script to test your server?
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If your doing this on your local computer you need to have an email service set up.
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Is this happening on your local computer or online?
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The activation code is being logged in the database but still won't let me activate the account.
But your not getting your emails right?
Also test with different email address.
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Is this
id() . "$y=$a";
a typo above in your activation email?
My problem is it isn't sending the code to the email specified
Try an echo somewhere on your register page to see if your assigning the email address to the $email variable like you think you are.
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... someone had a link on here that went to a site explaining it's use as well as encrypting it with md5 instead of using an ip. Hopefully who ever provided the link will chime in it was a good article.
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If someone uses AOL, which I only know a few people that do, can't their IP change during use?
Some people use HTTP_USER_AGENT, someone had a link on here that went to a site explaining it's use as well as encrypting it with md5 instead of using an ip. Hopefully who ever provided the link will chime in it was a good article.
My set up is basically
if ((!isset($_SESSION['mem_id'])) OR (!isset($_SESSION['user_agent']) OR ($_SESSION['user_agent'] != md5($_SERVER['HTTP_USER_AGENT'])))){ Not a logged in member, redirect } else {do whatever} -
You'r deleting it from phpmyadmin? I think this will do it if you put it in your sql of phpmyadmin fot that table
DELETE FROM table_name WHERE id != 16430; if 16430 is the record you want to keep
OR
DELETE FROM table_name WHERE id != 16430 And cupID = 45; for a little insurance and if you want to get rid of only the 45's
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thanks
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Using strip_tags with the second parameter opens up XSS attacks. ...
I can see how strip_tags($input, <a>); could be used for a XSS attack but how could strip_tags($input, <i>); be used for a XSS attack?
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http://php.net/manual/en/function.strip-tags.php
$input = strip_tags($input);
or echo strip_tags($input);
or to allow a tag
or in your case $input = strip_tags($input, <br/ >);
$input = strip_tags($input, <h1>); ?? I use <i> & <b> but I don't know how safe this is.
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Is this tutorial about this certain question or similar? if so I will have a look
Similar
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They use to have a good tutorial here at freaks, the link is somewhere on the front page or there about
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The answer to your questions is yes, depending on the category(url aka get statement) I call different css scripts.
Are you sure you have your file paths right?
Are you sure your style actually effects the web page?
If they're correct the code Muddy_Funster gave you will work. (I like single quotes)
echo '<head>
<title>You should always have a title in the header</title>
<link rel="stylesheet" type="text/css" href="rcm/stylesheet.css">
</head>';
The code Muddy_Funster had had "rcf/stylesheet.css", you didn't check that? Again is the path correct, are there any typos here? Are you sure the css effects the page?
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The stuff I googled said it could and pbs is right but some people break this rule if the images are small, grouped with other images and they are links to larger images or more info.
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And why would someone want the redundancy of the statement the way I had originally copied it?
Are you talking about the "if (isset($_POST['action']) && $_POST['action'] == 'submitform')". ? I don't know, I do this with my sessions but not my submit POST
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if (isset($_POST['action']) && $_POST['action'] == 'submitform')",
isset checks to see if $POST['action'] has a value
so you're basiclly saying, if (action has a value) & (that value equals submitform) run the following script
heres how I do my submits
f (isset($_POST['submit'])){ $roastturkey = $_POST['roastturkey']; $broccoli = $_POST['broccoli']; $brisket = $_POST['brisket']; $carrots = $_POST['carrots']; $sql = "INSERT INTO pass (roastturkey,broccoli,brisket,carrots) VALUES ('$roastturkey','$broccoli','$brisket','$carrots')"; $result=mysql_query($sql); if($result){ echo "Successful"; }else { echo "ERROR"; } } // end of itthis is my button
<input type="submit" name="submit" value="Submit" />
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Where was your update script at in relation to the above?
And it may be just me but I'd put all that code in your "if (isset($_POST['action']) && $_POST['action'] == 'submitform')", that way it only runs if the if is true.
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I would think something changed, can you post your sql code? What about the input to the variables for the sql, have you changed that?
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Phpfreaks; however, YOU, as one of the noobies, are responsible for bringing the brownies every other Friday.
That's true because I had to bring them every other Friday until I had 100 post.
As an added note I think phpfreaks is great for php and sql help.
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Thanks kickstart & mikosiko.
This forum is the best.
And this explains why I was getting 90 plus results from 15 records with some of the code I played with last night
This should still work, but what it will do is join every row from those 2 tables (a CROSS JOIN), irrespective of the id_mem fields. This is one way to get a massive number of records returned (if each table had 100 records, you would get 10000 rows brought back).Thanks again
S
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Thanks, that works and looks better than this,
$q = "SELECT photos.thumb, product.id_prod, product.title, photos.thumb_width, photos.thumb_height FROM members, product LEFT JOIN photos ON product.id_prod = photos.id_prod AND photos.main_photo = '1' WHERE product.publish = '1' AND product.id_mem = members.id_mem AND members.mem_group >=1 AND members.mem_group <100 ORDER BY product.id_prod DESC";
which I got to work last night, the odd thing is when I take the where clause out it doesn't work. I'll look at the links you gave me and keep playing with these and see if I can understand it better.
Thanks again
S
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mikosiko, I added the LEFT JOIN and took out the OR
$q = "SELECT photos.thumb, product.id_prod, product.title, photos.thumb_width, photos.thumb_height FROM members, product LEFT JOIN photos WHERE product.publish = '1' AND product.id_mem = members.id_mem AND mem_group >=1 AND mem_group <100 AND product.id_prod = photos.id_prod AND photos.main_photo = '1' ORDER BY product.id_prod DESC";
I kept 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 'WHERE sn_product.publish = '1' AND sn_product.id_mem = sn_members.id_mem A' at line 3
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When I run it I get one picture for every item, whether the item has a main picture assigned to it or not.
So if I have 16 tems and 4 have a main picture I get 16 items all with the same picture.
As a note an item can have no pictures, pictures but no main photo or pictures with a main photograph in the photos table.
Account Activation not working
in PHP Coding Help
Posted
I guess I should have said email scripts.