AyKay47

MHMM

as i stated previously.. you need to make two spearate ajax requests here..one for each url

1.you can either output it via php using either echo or print.. 2. the method that I use is by switching from PHP to html like so.. <?php if($_GET['via']=="security_question"){?> <div>html test</div> <?php } ?> <?php if($_GET['via']=="previous_password"){?> <div>same thing here</div> <?php } ?>

1. your jquery library should be referenced in the head of the page.. 2. what errors are you receiving..?

1. what errors? 2. I would declare this function in the head section of the page, as this is where it should go.. 3. also since you are using the name attribute here and not the id attribute.. it should be written like this.. function submitForm() { document.myform.submit(); }

worthless post by WebStyles. Coolness!

well hopefully my post helped you as well..

1. no errors triggered by the mysql_query..? 2. the parenthesis around the WHERE clause statement are not needed.. 3. try outputting $qry_result4 to make sure your query is what you expect.. 4. you can also use mysql_num_rows() to see how many rows your query is grabbing.. 5. place quotes around your indices.. good practice.. $row['film'] not $row[film]

it explains this example on the website.. which part don't you understand? the query selects the fields lastname and firstname from the table persons.. and orderno from the table orders..now you can't reference two tables without using a join..that is what they are for.. now the condition on which theses two tables are joined is the field p_id from the table persons must be equal to the p_id field from the persons table..the resulting fields are to be ordered by the lastname field from the persons table..

one table, will save you time and memory in the long run.. I would always try to have as few tables as possible in your database..

I recommend using jquery's AJAX API for this.. you should be using the onblur event here.. or in jquery.. .blur()

well your user is wrong, because it doesn't have access..

the user for localhost is "root".. can we see where you changed the settings..

1. yes 2. no

what is the issue here..? Why not write it how you have it written?

i was referring to using something like radio buttons to allow the user to choose which images they want..

I like to get into the habit of setting variables with indices into alternative variables to avoid syntax errors, and for the overall appeal of the code.. something like $valid_user = $_SESSION['valid_user'];//proceed using $valid_user

im thinking something along the lines of this.. foreach ($array as $tag) { if(end($array) == $tag){ echo $tag; }else{ echo "$tag,"; } } Edit I like the implode method here

to my knowledge, you cannot have two open() methods in one ajax request, could be mistaken.. I believe that you would have to create another AJAX call in the same function to get your desired functionality.. also, I recommend using jquery's AJAX API instead of standard js

looks like it is establishing an asynchronous connection to the server using AJAX method.. sends data to the given url, and places the response into the element with id = ajaxDivVideo.. also applies a custom function..

i believe that earlier versions of IE have trouble with the <dl> tag, what version of IE are you using..? and what exactly is "messing up" in your code?

1. you will want to add the name attribute to the submit button which will allow you to check for it being set. 2. can you post the "additional script" that you are referring to

1. the first example you have should work.. you are missing the $ call in your second bit of code.. 2. why would you need to code this inline, I personally would rather use a function..

why not use drop boxes or something along those lines to allow user to select what pictures they want

the parent function needs to return false, not your ajax child function