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Hi all, I think this would be done using javascript, Im not really sure. Im trying to work out a way to split a value between 3 select boxes. For example if the max value was 9 I only want a total value of 9 to be selectable between the 3 select boxes e.g box 1 could have a value of 2, box 2 could have a value of 5 and box 3 could have a value of 2. Hope that makes sense, Does anyone know how I could do this please ? Many thanks, Scott

Problem solved. I used this code to check the fields before they were submitted <script language="javascript" type="text/javascript"> function checker() { var myForm = this.document.myForm; if(myForm.from.value == '') { alert("Please Select the pick up area"); myForm.from.focus(); return false; } if(myForm.to.value == '') { alert("Please Select the droping area"); myForm.to.focus(); return false; } if(myForm.from.value == myForm.to.value) { alert("Pick up area and droping area should not be same"); myForm.to.focus(); return false; } } </script> Thanks alot for your help, its most appreciated

nope at the very top

sorry i meant to put if(isset($_POST['submit'])){ if (($_POST['from'] == "") || ($_POST['to'] == "")){ ?> <script type="text/javascript"> window.alert("You message goes here!") </script> <?php }else{ header( 'Location:quote_part1.php?from='.$_POST['from'].'&to='.$_POST['to'] ) ; } }

ok hes just told me that a normal alert box is fine but when i do this it messes up the styling on the page if(isset($_POST['submit'])){ if (($_POST['from'] == "") || ($_POST['to'] == "")){ }else{ header( 'Location:quote_part1.php?from='.$_POST['from'].'&to='.$_POST['to'] ) ; } } any ideas why ? thanks.

Ok so i put something like this ? if (isset($_POST['submitted'])) { $errors = array(); if (($_POST['from'] == "") || ($_POST['to'] == "")){ } The data will be sent to another page as example.php?from=example&to=example

Thanks il give it a try. Sorry for being a newbie but what would I put in the : // do all your error checking All it needs to check is 2 fields Thanks again.

Thanks, Unfortunatly the person im doing the site for wants a facebook style popup for some strange reason

Hi, I'm trying to work out how I can trigger a javascript popup type box in an if statement, to trigger it normally from a button it would be : <input type="button" value="Display" onclick="javascript: formFunction();" class="submit" /> But I want to use it when people submit a form without filling in some of the fields so something like this : if (($_POST['first_name'] == '')) { run javascript popup } Any ideas ? Many thanks in advance. Scott

Hey, thanks very much for the reply. Just pasted your code and for some reason it still skips the image when using tab. Would it be easy to convert it into a proper button ? Sorry Im a bit of a newbie if you hadnt already guessed lol

Hi, Im having troubles setting up the tab order for a small form. When I click in the first input box and press tab it goes to the next input box but when I press tab again it skips the submit button. Any ideas ? Heres the code : <form action="quote_part1.php" method="post"> <div align="center"> <p align="center" class="style1"> <br /> <span class="style3">Pick-up Location : <span class="style5">e.g Gatwick Airport </span></span></p> <p> <select name="from" id="from" class="editable-select" tabindex="1" > <option></option> <option>Gatwick Airport Terminal 1</option> <option>Gatwick Airport Terminal 2</option> <option>Gatwick Airport Terminal 3</option> <option>Heathrow Airport Terminal 1</option> <option>Luton Airport</option> </select> <br /> </p> <p align="center" class="style4">Drop-off Location : <span class="style5">e.g Bournemouth </span> </p> <p> <select name="to" tabindex="2" id="to" class="editable-select" onKeyPress="return submitenter(this,event)" > <option></option> <option>Gatwick Airport Terminal 1</option> <option>Gatwick Airport Terminal 2</option> <option>Gatwick Airport Terminal 3</option> <option>Heathrow Airport Terminal 1</option> <option>Luton Airport</option> </select> </p> <p> </p> <p> <img src="images/quote_button.png" tabindex="3" OnMouseOver="this.style.cursor='pointer';" onclick="document.forms[0].submit()" /> </p> </div> </form> Many thanks, Scott

Hi, I have been using the following function on my server and it works fine but when I try and use it on a different server it wont seem to work. Any ideas on why it wont work on the new server and any ideas on how I can do some error checking to figure out why ? Thanks in advance, Scott <?php function get_driving_information($start, $finish, $raw = false) { # Convert any lon / lat coordingates if(preg_match('@-?[0-9]+\.?[0-9]*\s*,\s*-?[0-9]+\.?[0-9]@', $start)) { $url = 'http://maps.google.co.uk/m/local?q='.urlencode($start); if($data = file_get_contents($url)) { if(preg_match('@<div class="cpfxql"><a href="[^"]*defaultloc=(.*?)&ll=@smi', $data, $found)) { $start = trim(urldecode($found[1])); } else { throw new Exception('Start lon / lat coord conversion failed'); } } else { throw new Exception('Start lon / lat coord conversion failed'); } } if(preg_match('@-?[0-9]+\.?[0-9]*\s*,\s*-?[0-9]+\.?[0-9]@', $finish)) { $url = 'http://maps.google.co.uk/m/local?q='.urlencode($finish); if($data = file_get_contents($url)) { if(preg_match('@<div class="cpfxql"><a href="[^"]*defaultloc=(.*?)&ll=@smi', $data, $found)) { $finish = trim(urldecode($found[1])); } else { throw new Exception('Finish lon / lat coord conversion failed'); } } else { throw new Exception('Finish lon / lat coord conversion failed'); } } if(strcmp($start, $finish) == 0) { $time = 0; if($raw) { $time .= ' seconds'; } return array('distance' => 0, 'time' => $time); } $start = urlencode($start); $finish = urlencode($finish); $distance = 'unknown'; $time = 'unknown'; $url = 'http://maps.google.co.uk/m/directions?saddr='.$start.'&daddr='.$finish.'&hl=en&oi=nojs&dirflg=d'; if($data = file_get_contents($url)) { if(preg_match('@<span[^>]+>([^<]+) (mi|km)</span>@smi', $data, $found)) { $distanceNum = trim($found[1]); $distanceUnit = trim($found[2]); if($raw) { $distance = $distanceNum.' '.$distanceUnit; } else { $distance = number_format($distanceNum, 2); if(strcmp($distanceUnit, 'km') == 0) { $distance = $distanceNum / 1.609344; } } } else { throw new Exception('Could not find that route'); } if(preg_match('@<b>([^<]*)</b>@smi', $data, $found)) { $timeRaw = trim($found[1]); if($raw) { $time = $timeRaw; } else { $time = 0; $parts = preg_split('@days?@i', $timeRaw); if(count($parts) > 1) { $time += (86400 * $parts[0]); $timeRaw = $parts[1]; } $parts = preg_split('@hours?@i', $timeRaw); if(count($parts) > 1) { $time += (3600 * $parts[0]); $timeRaw = $parts[1]; } $time += (60 * (int)$timeRaw); } } return array('distance' => $distance, 'time' => $time); } else { throw new Exception('Could not resolve URL'); } }

Hi, Im having some troubles with a date picker on a form. It enters the date into the date field like this : 21 Sep 2011 I want to be able to insert it into a date field in the mysql database but it just enters it as 0000-00-00 Any ideas ? Many thanks, Scott.

works perfect. Thanks again.

Thanks alot webstyles, il give it a try

Hi, I have a database table which is full of books e.g book_title, book_author, book_date etc. Im trying to work out a way to list all the authors without duplicating them, if that makes sense. If I do a simple query and then display the book_author in a loop it will show multiple entries for each author. I just want it to display each author once. Hope that makes sense lol sorry Im a newbie. Anyone know how I can do it ? Thanks in advance, Scott.

Ive just realised i've made a stupid mistake. It was the rest of my code that was messing me up and Aykay's code did work. Thanks very much for your help

Anyone know how I can fix this ?, its driving me crazy

maybe change to this : <?php echo 'Title: <a href="Job-Details.php?id='.$row['id'].'">'.$row['JobTitle'].'</a><br />';

Thanks, unfortunatly that gives the same results

it seems like its just the end part that isnt working. The AND part doesnt seem to affect the results

Thanks but getting this error with that code : Parse error: syntax error, unexpected T_STRING

Hi all, Anyone know why this query is giving me the wrong results ? $books2 = mysql_query("SELECT * FROM books WHERE (extra1 LIKE '%$cat%') OR (extra2 LIKE '%$cat%') OR (extra3 LIKE '%$cat%') AND type = 'public'"); Its giving me results that dont have the 'type' 'public' Any ideas ? Many thanks in advance. Scott

oh right thats odd. il double check lol. thanks for your help by the way its appreciated.

you might have to scroll right