tcollins412

hello, i am making a website and i need some help. I am writing a table, here is the code:[CODE]<table style='float: left;'> <tr><td style='width: 200px; height: 500px; background-color: red; valign: top; align: top;'><a class='mainlink' href='about.html'>About Me (under construction)</a><br /> </td></tr> </table>[/CODE] now, i want the link to align to the …

i am trying to submit a form in jquery without refreshing it. for my jquery code i have this: [CODE] $(function() { $(".submitbuttonsend").click(function() { var dataString = 'reqtoid=<?php echo $uid; ?>&reqfromid=<?php echo $id; ?>'; $.ajax({ type: "POST", url: "proccess.php", data: dataString, success: function() { $('#freekk').html("<div id='message' style='color: #4f4f4f; font-size: 12px; …

i am trying to put all of my cookies into an array. heres my code [ICODE] <? setcookie('failn','fail'); setcookie('goodn','good'); setcookie('han','ha'); $allcookies=explode(' ',$_COOKIE); print_r($allcookies); ?> [/ICODE] but the page only returns Array ( [0] => Array ) please help

i am writing a login script in a perl and i set a cookie. here it is: [CODE] if ($V::testsubmit eq 'yes') { ($userid)=$dbh->selectrow_array("SELECT id from logins WHERE username='$V::username' and password='$V::password'"); } if ($userid > 0){ @alphanumeric = ('a'..'z', 'A'..'Z', 0..9); $sessionid = join '', map $alphanumeric[rand @alphanumeric], 0..50; print<<"EOF"; …

lets say i have a get variable:[CODE]$goto=$_GET['goto'];[/CODE] and i have a array [CODE]$sections= array('fail' => fail());[/CODE] how would i make it so when the $goto=fail, it does the function fail from the array?

i have a script that sends an email to cellphones: [CODE=html] <form name='celltest' method=post> <input type=hidden name=cellsubmit value=yes> <tr><td rowspan=1>Mobile Phone: <input type='text' class='efield' id='phone' name='phonetest' style='width: 200px;'></td><td style='padding-bottom: 15px; padding-left: 3px;'>&nbsp;<a href="?ref=ea&dotest=1" class='eap' style='width: 50px;'>Test</a></td></tr></table></td></tr><tr><td style='font-size: 12px; text-align: left; padding-top: 2px; padding-left: 35px;'>Provider: <select name=provider class=efield style='width:258px;' id='provider'> <option …

i have a pop up div that i want to close onmouseout. but it keeps closing when it rolls over a child element. Can anyone help me? thanks in advanced

i have two sub-domains pointing at the same file.the sub-domain's are test.test.com and fail.test.com and the file is index.php. Now i want to be able to work with both of those domains within the same file so i used this code: [CODE] if($_SERVER['HTTP_HOST']=='test'){ echo"test"; }else if($_SERVER['HTTP_HOST']=='fail'){ echo"fail"; } [/CODE] but …

i need somebody's help with the style of navigation bar i should use. if anybody can help please tell me.

this will not work on change. also the innerhtml wont display. please help. here is the js: [CODE]function checkemail(){ if (document.suform.email.value && document.suform.emailcheck.value) { if (document.suform.email.value == document.suform.emailcheck.value) { document.suform.emailcheck.style.backgroundColor='#FFFFFF'; document.getElementById('wm').innerHTML=='Emails Match!'; }else{ document.suform.emailcheck.style.backgroundColor='#F75B53'; document.getElementById('wm').innerHTML=='Emails Dont Match!'; } } }[/CODE] and here is the html: [ICODE] Re-type Email: <input type='text' …

i am making a register page on my website. i have a password and a re-enter password text fields. how would i validate those two by making sure they are the same in ajax? and how would i make that happen on the re-enters onblur or onchange? thank you

i am submitting a form in javascript and i want it to submit on key press submit, but it wont work. here is my code: [CODE] <script type='text/javascript'> function login(evt){ if (evt){var keyCode = evt.which ? evt.which : evt.keyCode;}else{keyCode='';} if ((evt == '')||(keyCode == 13)){ if (document.forms["login"].email.value=='') { alert('You must …

i am using the following code to submit a form: [CODE] <script> function marktaskcomplete(){ document.forms["markcomplete"].submit(); } </script> <? if ($_POST['markcompletevalue']=='yes'){ include "dbconnect.php"; $taskid=$_POST['taskid']; $query="UPDATE tasks SET status='0' WHERE uid='$id' AND id='$taskid'"; $result=mysql_query($query); if ($result){ echo "<script>location.href='?goto=taskmanager'</script>"; }else{ echo"fail"; } mysql_close; } ?> [/CODE] and when i try to submit more …

Hi there, I have added a validation script to this page [url]http://antobbo.webspace.virginmedia.com/webediting/documents.htm[/url] but it is not working the way I want it to, I am sure I made some silly mistake. Basically I want to make sure that people actually input something in the form below before submitting it. If …

i am writing javascript for something and it wont work on google chrome. it will work on everything else but chrome. can you please help? here is the code: [CODE] var title='CrazieToday - Best Event Organizer Ever'; var url='http://crazietoday.com'; function AddToFavorites(){ if (window.sidebar) { window.sidebar.addPanel(title, url,""); } else if( window.external …

i am wring a website and i need help with some javascript. first ill give you the code [CODE] <!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd"> <html> <head> <script type='text/javascript'> function showdiv(name){ var divname=name; show(); } function show(){ document.getElementById(divname).style.display='block'; document.getElementById(divname).style.overflow='visible'; var lastdiv=divname } function closediv{ if (lastdiv) { document.getElementById(lastdiv).style.display='none'; …

i am trying to submit a form in javascript, and then process it in php. and it will not work. can you please help? this is the code: [CODE] <?php if ($_POST['markcomplete']=='yes'){ $taskid=$_POST['taskid']; $query="UPDATE tasks SET status='0' WHERE uid='$id' AND id='$taskid'"; $result=mysql_query($query); if ($result){ echo "<script>location.href='?goto=taskmanager'</script>"; }else{ echo"fail"; } } …

i need help with a project im working on. im writing a task manager for school, and lets say i have a section for the due-date. How would i write a calendar that pops up like in Google tasks or phpmyadmin when you go to insert a date. thank you.

i need help with something lets say i had a url: [url]http://domain.com[/url] and it was supposed to go to the page [url]http://domain.com/loggedin.php?goto=hi[/url] how would i make it so in the browsers url bar, it only says [url]http://domain.com/?goto=hi[/url]

i am trying to download an application i got in an email from my friend, but i keep getting: "Windows found this file to be potentially harmful. To help protect your computer, windows has blocked access to this file." i tried turning off my firewall but that didnt seem to …

i am writing code and i keep getting the error: mysql_fetch_array: supplied arguement is not a valid mysql resource. here is the code: [CODE] <?php include"dbconnect.php"; $item=$_POST['item']; $query3="SELECT * FROM selling WHERE name LIKE %item% AND state='$state' LIMIT 0,30"; $result3 = mysql_query($query3); while($item = mysql_fetch_array($result3)){ echo "<img src='$item[pic]' alt='pic'><a style='color: …

i am writing a website that loges somebody in by cookies. i want to make it so if the person goes to the main page, (index) but did not log out from their last session it redirects them to the members page. how would i do that? also, how would …

i am writing something in php and i keep getting the error: mysql_fetch_array not a valid sql argument. here is the code: [CODE=php] <?php if ($_GET['goto']=='requests'){ ?> <table> <?php $query = "SELECT * FROM friends WHERE toid='$id'"; $result = mysql_query($query); while($row = mysql_fetch_array($result)){ $query = "SELECT * FROM members WHERE …

how would i do something like facebooks friends in php?

lets say im making a dating site. and i want people to be able to insert their own pictures. how would i do that? i have a mysql database. and i want this to happen in php. and then how would i make it display that persons image when someone …

i have this code: [CODE]<?php session_start(); $id=$_SESSION['uid']; include "dbconnect.php"; $query = "SELECT FROM members WHERE id='$id'"; $result = mysql_query($query); $row = mysql_fetch_array($result); ?>[/CODE] and it says mysql_fetch_array is not a valid argument. help please

i am making a website in php all on one page. here is my script [CODE]<a href='?goto=register'>register</a> <?php if ($_GET['goto']=='register'){ ?> <form name='register' method='POST' action='?goto=reg'> <table style='margin-left: auto; margin-right: auto; color: white; font-weight: bold;'> <tr><td style='text-align:center;'>First Name: <input type='text' name='first'></td></tr> <tr><td style='text-align:center;'>Last Name: <input type='text' name='last'></td></tr> <tr><td style='text-align:center;'>Email: <input type='text' …

how do you change the timeout of a session? can you do something in the php.ini?

hi I'm having trouble with php and mysql I'm getting this error [CODE] [Sun Jan 30 17:43:40 2011] [error] [client 127.0.0.1] PHP Warning: mysql_query(): Access denied for user 'root'@'localhost' (using password: NO) in /var/www/t/view_tg.php on line 29, referer: http://localhost/t/ [Sun Jan 30 17:43:40 2011] [error] [client 127.0.0.1] PHP Warning: mysql_query(): …

i am writing a php code that displays text from a database. how would i make the text auto break `<br />` after a certain amount of chars? please help asap. thanks