fobos

first off, you cannot put session_start() at the top of a login page, because there is no session yet. I would just take out $_SERVER["REQUEST_METHOD"] == "POST") unless you know that the variables are being posted. Try this because i have a feeling you got this from another website.. Also, …

Hello all, i am stuck on this.. i want to be able to play a video from a mysql database. Currently i can upload the video to mysql (name, type, size, content) and download the file. i have a media player that shows a video, but i have to use …

Google is a really useful tool. http://www.w3schools.com/js/js_form_validation.asp

Try this. In this case, you already have the errors being posted in the div, so this will scroll to the div with the errors. <?php if(isset($_POST["myform"]) && !empty($errors)){?> $('html, body').animate({scrollTop: $("#error").offset().top}, 2000); <?php } ?>

Once you hash a password, it cannot be undon. Therefore, you cannot cross check a text password with a hashed one. If this was possible, no password woule be safe on the web.

Lets see

@SimonIoa, the onfocus event will remove the "Search". With that being said, SimonIoa still has a point. You should have that in your input if you are asking to verify if its blank or not. Instead of calling a $_REQUEST, use $_POST instead. $_REQUEST is the default if you didnt …

This post is dead, dont bring it back. Besides, dont hijack peoples post.

Hello all, im having some problem, i have a function call dbconn which i run after my session. i can successfully connect to the database and run the mysql statement outside of the function with no problems. However, when i put it in the function and return it, nothing happens. …

Try close your first echo with a quote and semi colon

To display your form in an iframe, you must first put the form onto a page and save it.. let call it form.html. Then you must link that page to the iframe. <iframe id="something" border="0" width="500px" height="500px" src="form.html"></iframe> Hope this helps

Have you tried calling the function with onblur then get the value to pass? <input type="text" name="txtSearchPolicyNumber" id="txtSearchPolicyNumber" onblur="showResults();" /> When you perform the "onBlur", it will call the function, get the value and check to see if its blank. function showResults() { var policynumber = document.getElementById("txtSearchPolicyNumber").value; if ( policynumber …

LastMitch, this thread is old. Please dont raise the dead

- ok, extract xampp (the .zip version) to a drive or directory (D: or C:) and it will create a folder. - Go into the folder and click on control(beta).exe and after that pops up, install the services for http and mysql and start the services. Installing the services will …

Here you go [w3schools.com](http://www.w3schools.com/php/func_mail_mail.asp)

When i run this statement, it works fine for me. What does the error say? $sw11 = "hi"; $symp39=$sw11.";"; echo $symp39;

Have you tried putting session_start() at the top of the page?

The only thing i can think of is, 1) your hardrive is spinning up, and 2) some programs have to get loaded, so its slowing down the time to get to the desktop.

try this website, it will help you to use mime content type. Hope that helps

Whats the code for the _partialview page? That will be better to look at than a 2 line code that no one can help you with.

Well is the person running this page on his own server? Also, did he change "table1" to what ever the table name is, in the SQL statement?

Try just comparing the two variables of the dates. var strtdate = document.getElementById("x_START_DATE").value; var enddate = document.getElementById("x_END_DATE").value; if(strtdate >= enddate) { window.scrollTo(0,0); alert("End date should not be less than start date."); return false; } else { alert("Good"); }

in your select input, have an onchange function that will call a function that will reload the page (or direct to another) with the value in the url. <select name="cars" onchange="myfunction()"> then call for the redirect function. function myfunction(){ var sel = document.getElementById("cars").value; window.location = "yourpage.php?id="+sel; } Then use the …

Do you have 4 ram slots (Channel A and B)? I think you have 4 slots. You have can up to 4GB total, but will only be able to utilize the 3GB rule. You must have either 2 or 4 ram sticks installed, but not have 2 in channel A …

That because something is using port 80. in the control panel, there is a button called port check. It will tell you what program is using that port.

This is a really good website that i learned from. [W3Schools](http://www.w3schools.com/ajax/default.asp)

you could try this to detect the browser. if( /Android|webOS|iPhone|iPad|iPod|BlackBerry/i.test(navigator.userAgent) ) { // some code.. }

Ibrahim Hossain, dont hijack this post. If you have a problem, start your own thread, otherwise delete your comment.

Can you please post your php code, because this is a problem with your mysql statement. Thanks

You can try: if(!mysql_num_rows($result) || mysql_num_rows($result) == 0) { echo "<b>Not Found</b>"; } or (mysql_num_rows($result) < 1) If that still doesnt work, then the problem is before that, with your select statement.

Or just pull the data using ajax

I believe because you didnt have script.onload=scriptLoaded; on your first post. you could also include the script <script type="text/javascript" src="https://ajax.googleapis.com/ajax/libs/jquery/1.8.0/jquery.min.js"></script> <script type="text/javascript"> $(document).ready(function () { $('#main').click(function(){ alert('hi'); }); }) </script>

Here is a website that tells you alot of useful things [Click Here](http://www.htaccess-guide.com/). But when in doubt, go to the apache website [Click Here](http://httpd.apache.org/docs/2.4/howto/htaccess.html).

it looks to me that $checked = a variable with an extra ' in it. It says it in the message (private='0'') with the extra aposrophe. Maybe thats why?

Other thoughts: - re-install W7 and perform updates (without installing drivers). W7 usually provides drivers in update. - Do you have the right RAM installed? if you didnt change then no worries. - Did you have any problems with W7U on your old hard drive?

You might have to re-install your OS. What were you "working on" when this happened? Heres a link if you get your keyboard working. [Click Here](http://support.microsoft.com/kb/927392)

yes, since i know php any mysql, i would use that. just display that data on one page have have them echo a link that uses a keyword or id to specifically display the info on that person. Ex Members.php <?php //database connection // select statement for database //from w3schools.com …

First off did you set the content as a blob in mysql. Second, go into your http conf file and just like what veedeoo said, change the values to something like that. Third, no need to have a hidden field for the upload size, just have the file upload and …

When using the insert statement, there is no need to include the "ID" field. Since your database has auto encriment, it will add it for you automatically. So take out the "ID" in the INSERT statement and '' in the VALUES.

So what is your php problem?

you can try using the LIKE operator in your mysql statement. SELECT fields FROM database WHERE field LIKE 'variable' ex SELECT lastName, firstName FROM employees WHERE firstName LIKE 'a%'

AleMonteiro, this post is dead, please dont bring it back from the grave.

you cannot use $result twice. try making one $results and the other $result

ill see if i can help.

Hello, i redid your coding and took out the $nt in the action and added a name tag to ur option. Also, just noting, you hae to have a ? mark when passing variables through a URL. <?php $sql = "SELECT DISTINCT dept FROM sw_lic WHERE dept != ''"; // …

Well you are echoing something that either should be wrapped in quotes or need to make it a variable. You have <td><input type='text' name='<?php echo mark.$y; ?>' id='textfield' /></td> Possibilities <td><input type='text' name='<?php echo "mark". $y; ?>' id='textfield' /></td> <td><input type='text' name='<?php echo "mark $y"; ?>' id='textfield' /></td>

Well on the form part on the first set of code, i changed the php code so i echos all variable into a select element, rather than what you had. Your method make it echo all variables into 1 select element. <form> <select name="state" onChange="display(this.value)"> <option value="">-- Select state --</option> …

So whats your problem?

are you sure you wrote the php statement? if so, then you would know about html. put the input tag into the form