adam84

You want to select a certain option in a drop down box via javascript?

Maybe sleep/wait function....?

<input onClick="javascript:window.open('<%=shareURL%>','','resizable=1,toolbar=no,width=550,height=250');" type="submit" name="submit" value="terminate it" class="fancy"> I think window.open takes in 1 or 3 agurements. 1 -> url 3 -> url, window name, window attributes

<frame src="title.html" noresize scrolling="no"> http://www.tizag.com/htmlT/frames.php

Awesome, that is the second greatest news I have heard all day!

I was just wondering if this would work. I have my php files, where I have all my html code and stuff. At the top of the page I have about five or six different include files. What I was wondering is if I make a master file, that I put all my includes in and on my html page, I just included the master file (instead of all of the other includes), would that work correctly?

awesome thanks for you help

I have a number of pages that a user (not logged in), can visit. *each page have a form that allows a user to login What I want to do is, say a user is on a page and they decide to log in. I want to reload that same page that user was on prior to them being logged in, but now with them being logged in. * I hope that make sense, it is too early to think. Sorry Would I store the current page url in a session variable, valid the username and password and I would do something like: header("Location: " . $_SESSION['currentPage']); Am I on the right track?

Ok, I am trying get out of the habit of tables and using div tags. But I haven't had a whole lot of luck. The page I am working on is a register page. I have my main div tag called 'mainContent' where all my site information gets loaded into. Inside that div, I have another div tag called 'subContainer', which I refer it to be like a <tr> Inside that div, I have two more div tags, 'fieldTitle', which formats the field title ('user name', 'password', 'email', etc) The next tag is 'fieldElement', which just holds the textfield, select, etc. What I want to do is have the fieldTitle div span 50% of the row( 'subContainer' ), then have the 'fieldElement' span the remaining 50%. But for some reason, it is not doing that. HTML <div class="mainContent"> <div class="subContainer"> <div class="fieldTitle"> User Name: </div> <div class="fieldElement"> <input class="textField_Effects" type="textfield" name="userName" id="userName" maxlength="25" size="25"> </div> </div> </div> CSS .mainContent{ border: 1px solid #000000; margin-left: 15%; width: 100%; height: 100%; } .fieldTitle{ width: 50%; border: 1px solid #000000; font-weight: bold; font-size: 12px; text-align: right; display: inline; } .subContainer{ width: 100%; margin: 0; border: 1px solid #000000; } .fieldElement{ border: 1px solid #000000; width: 50%; text-align: left; }

yea, thats good. Thanks

I have a div inside a div. What I want to do is have the loginMenu div at the bottom of the leftMenu div. But I do not know how to do that. Thanks <div class="leftMenu"> Left Menu <div class="loginMenu"> Login Menu </div> </div> .leftMenu{ border: 1px solid #333233; width: 15%; height: 65%; margin:0; padding:0; float:left; } .loginMenu{ border: 1px solid #333233; width: 100%; height: 25%; margin:0; padding:0; float:left; }

Maybe try something like this..... if($sort = "") { $query = "SELECT * FROM dw_customers ORDER BY customer_bride ASC" ; } else { $query = "SELECT * FROM dw_customers WHERE SUBSTRING(customer_bride,1) = '$sort' ORDER BY customer_bride ASC" ; }

Awesome, thanks for the help!

On my site, I have an area where users (logged in) can comment on stuff (pictures, etc). I was just wondering is it considered to be good practice to allow users that have not created an account to post comments? Cause I was thinking of allowing non-registered user's to post comments in certain area's, but since I cannot track them like a registered user. I was thinking of validating their comments, so their comment's won't get out of hand(langauge wise). What do you think? This is just a thought as of now, I have not implemented anything yet. Thanks!

I want to add a feature to my site, but I have no idea what it is called. So I cannot even google it. What I want is when a user clicks on a certain textfield, I want to populate a menu under the textfield with different things. If the user clicks on one of the options, I want the value of that option to be inserted into the textfield, or if the user do not want any of the options, they can enter type in what they want. Any Ideas?

if you have all your images stored in an array, why dont you use the next and previous buttons to change the .src of the <IMG> tag. Create a variable that stores the current picture index and either increase it or decrease the index depending on what button was clicked by the user. This way the page would need to be refreshed everytime a button was clicked.

hmmm will if($_POST){ } ever evaluate to true?

Can you show me you html source code, I would be able to get a better idea of what the problem is by looking at the.

I dont know if you are on the right track or what. But this method should work $sql= "SELECT * FROM DISPLAY_VALUES WHERE PID = '".$p_num."'"; $results = mysql_query( $sql ); echo "<div class='redbg'> <b><big>TESTING:</big></b></div>"; echo "<br>"; if( mysql_num_rows( $results ) < 1) echo "No values exist for this parcel!"; else{ while( $row = mysql_fetch_array( $results ) ){ //do whatever you are trying to achieve } echo "</table>"; }

Couldn't you get put all the information you want to access in the 'radio count:' span tag and just used the .innerHTML property to get the text found in the span tag?

Instead of $index = $row[0]; Just leave it at 0 and increase the value $index = 0; while{ .. $index++; }

in the JS function remove the line gTotal += parseFloat(document.getElementById('total'+i).innerHTML); and try alert( document.getElementById('total'+i).innerHTML ); See you if are getting a value?

SELECT * From db_table ORDER BY month desc

What the error means is you are giving six column names 'article_id, article_name, entry, poster, date, time', but you are only supplying four values '$name , $posted, today, 4:00pm' You need to give six values $sql = "INSERT INTO `articles` (article_id, article_name, entry, poster, date, time) VALUES(null, '$name' , $entry, '$posted', 'today', '4:00pm')";

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