Search Results

Search found 10 results on 1 pages for 'bsandrabr'.

Page 1/1 | 1 

  • pagination with css

    - by bsandrabr
    Hi I've tried every combination I can think of but I can't get this to work. Can you help? I'm trying to put some css onto my pagination and have read all the examples but they all contain so many backslashes and concatenation that I just dont know how to apply it Here is my pagination code (which works fine) along with my feeble attempt at styling it if ($st > 0) { $st3=$st; print "< Previous Page "; } $f=$st+3; for($i2=$st+1;$i2<=$f;$i2++) { $i3=$i2-3; if ($i3 0) { print "$i3 "; if($i2 % 3 == 0) { print ""; } } } $g=$st+3; for($i=$st+1;$i<=$g;$i++) { print "$i "; if($i % 3 == 0) { print ""; } } $st2=$st+2; print " Next Page "; Here is the css that I took from the website /* CSS Document */ body { background: #2D2D2D; font-family:Verdana, fantasy; font-size:13px; color: white; scrollbar-base-color: black; scrollbar-arrow-color: red; scrollbar-DarkShadow-Color: black; } a:visited,a:active,a:link { color: white;text-decoration: none; } a:hover { color: red;text-decoration: overline underline;background: none; } table,tr,td { font-family:Palatino Linotype; color: #FFFFFF;font-size: 12px; } .button { font-family:Verdana, fantasy; font-size:13; color:#FFFFFF; background-color: red; } input,textarea,dropdown{ font-family:Verdana, fantasy; font-size:13; color: #FFFFFF; background-color: #000000; border: 1px solid; } textarea,.submit input{ font-family:Verdana, fantasy; font-size:13; color:#ffffff; background-color: black; } .table { background-color:#000000; } .table3 { background-color:#000000; } .table td { color: #000000; background-color:#DEDEDE; height:22px; } .table3 td { background-color:#CCCCCC; } td .alt { background-color:#EEEEEE; height:22px; } td .h { background-image:url(tablehgrad.png); background-repeat:repeat-x; font-weight: bold; background-color: #D6D6D6; } .table th { background-image:url(tablehgrad.png); background-repeat:repeat-x; color: #000000; font-weight: bold; background-color: #D6D6D6; } .menu th { font-font-size: 12px; color: silver; background-image:url(th.png); background-repeat:repeat-x; font-weight: bold; background-color: #4B4B4B; } .stats td { font-font-size: 12px; color: white; font-weight: bold; } .menu td { font-size: 12px; text-align: center; color: white; background-image:url(tdover.png); background-repeat:repeat-x; font-weight: bold; background-color: #4B4B4B; } .menu td:hover{ color: white; background-image:url(td.png); div.pagination { padding: 3px; margin: 3px; } div.pagination a { padding: 2px 5px 2px 5px; margin: 2px; border: 1px solid #AAAADD; text-decoration: none; /* no underline */ color: #000099; } div.pagination a:hover, div.pagination a:active { border: 1px solid #000099; color: #000; } div.pagination span.current { padding: 2px 5px 2px 5px; margin: 2px; border: 1px solid #000099; font-weight: bold; background-color: #000099; color: #FFF; } div.pagination span.disabled { padding: 2px 5px 2px 5px; margin: 2px; border: 1px solid #EEE; color: #DDD; } thanks

    Read the article

  • javascript select box hanging on second select in ie7

    - by bsandrabr
    I have a drop down select box inside a div. When the user clicks on change, a dropdown box appears next to the change/submit button and the user makes a selection which then updates the db and the selection appears instead of the dropdown. All works fine in IE8 and firefox but in IE7 it allows one selection (there are several identical dropdowns) but the second time a selection is made it hangs on please wait. This is the relevant code <td width=200> <input type="button" onclick="startChanging(this)" value="Change" /></td> <script type="text/javascript"> var selectBox, isEditing = false; var recordvalue; if( window.XMLHttpRequest ) { recordvalue = new XMLHttpRequest(); } else if( window.ActiveXObject ) { try { recordvalue = new ActiveXObject('Microsoft.XMLHTTP'); } catch(e) {} } window.onload = function () { selectBox = document.getElementById('changer'); selectBox.id = ''; selectBox.parentNode.removeChild(selectBox); }; function startChanging(whatButton) { if( isEditing && isEditing != whatButton ) { return; } //no editing of other entries if( isEditing == whatButton ) { changeSelect(whatButton); return; } //this time, act as "submit" isEditing = whatButton; whatButton.value = 'Submit'; var theRow = whatButton.parentNode.parentNode; var stateCell = theRow.cells[3]; //the cell that says "present" stateCell.className = 'editing'; //so you can use CSS to remove the background colour stateCell.replaceChild(selectBox,stateCell.firstChild); //PRESENT is replaced with the select input selectBox.selectedIndex = 0; } function changeSelect(whatButton) { isEditing = true; //don't allow it to be clicked until submission is complete whatButton.value = 'Change'; var stateCell = selectBox.parentNode; var theRow = stateCell.parentNode; var editid = theRow.cells[0].firstChild.firstChild.nodeValue; //text inside the first cell var value = selectBox.firstChild.options[selectBox.firstChild.selectedIndex].value; //the option they chose selectBox.parentNode.replaceChild(document.createTextNode('Please wait...'),selectBox); if( !recordvalue ) { //allow fallback to basic HTTP location.href = 'getupdate.php?id='+editid+'&newvalue='+value; } else { recordvalue.onreadystatechange = function () { if( recordvalue.readyState != 4 ) { return; } if( recordvalue.status >= 300 ) { alert('An error occurred when trying to update'); } isEditing = false; newState = recordvalue.responseText.split("|"); stateCell.className = newState[0]; stateCell.firstChild.nodeValue = newState[1] || 'Server response was not correct'; }; recordvalue.open('GET', "getupdate.php?id="+editid+"&newvalue="+value, true); recordvalue.send(null); } } </script> If anyone has any idea why this is happening I'd be very grateful

    Read the article

  • ignoring saturday and sunday

    - by bsandrabr
    I am pulling in all the records from my customer database(mysql) for the last ten days $offset1 =strtotime("-10 day"); $date3=date("Y-m-d",$offset1); SELECT * FROM customers WHERE date between '$date3' and '$date' AND customer.custid = '$custid' ORDER by date DESC I would like to leave out the dates falling on a saturday or sunday and would like to put this in my query rather than the php If you can help thanks

    Read the article

  • How to organise a php based website

    - by bsandrabr
    I am putting my php /mysql website up and this is my scenario The users are grouped into sites each site with their own unique database. There will be about 40 users per site. the two options I'm trying to decide between are have a central website running the php and directing the users off to their own database using sub domains for each user each with their own php in htdocs I dont even know if 2 is possible/stupid but if it was, would it make any difference to performance as they're all being run by the same server. Any other ideas/ advice much appreciated as I want to organise it the best way from the start

    Read the article

  • Can I do this in one Mysql query?

    - by bsandrabr
    Hi I have a table with two columns: column A column B 1 2 1 2 2 1 I want to return total of ones = 3 total of twos = 3 The best I can come up with is two queries like so: SELECT sum( CASE WHEN columnA =1 THEN 1 ELSE 0 END ) + sum(CASE WHEN columnB =1 THEN 1 ELSE 0 END ) SELECT sum( CASE WHEN columnA =2 THEN 1 ELSE 0 END ) + sum(CASE WHEN columnB =2 THEN 1 ELSE 0 END ) Can this be done in one query? Thanks

    Read the article

  • making a combined sum of two columns

    - by bsandrabr
    I have a table (apples) containing: cid date_am date_pm ---------------------- 1 1 1 2 2 1 3 1 3 1 1 2 I asked a question earlier (badly) about how I would rank the customers in order of the number of ones(1) they had. The solution was (based on one column): SELECT cid, sum( date_pm ) AS No_of_ones FROM apples WHERE date_am =1 GROUP BY cid ORDER BY no_of_ones DESC This works great for one column but how would I do the same for the sum of the two columns. ie. SELECT cid, sum( date_pm ) AS No_of_ones FROM apples WHERE date_am =1 add to SELECT cid, sum( date_am ) AS No_of_ones FROM apples WHERE date_pm =1 GROUP by cid ORDER by no_of_ones(added) hope I've managed to make that clear enough for you to help -thanks

    Read the article

  • formatting mysql data for ouptut into a table

    - by bsandrabr
    Following on from a question earlier today this answer was given to read the data into an array and separate it to print vehicle type and then some data for each vehicle. <?php $sql = "SELECT * FROM apparatus ORDER BY vehicleType"; $getSQL = mysql_query($sql); // transform the result set: $data = array(); while ($row = mysql_fetch_assoc($getSQL)) { $data[$row['vehicleType']][] = $row; } ?> <?php foreach ($data as $type => $rows): ?> <h2><?php echo $type?></h2> <ul> <?php foreach ($rows as $vehicleData):?> <li><?php echo $vehicleData['name'];?></li> <?php endforeach ?> </ul> <?php endforeach ?> This is almost perfect for what I want to do but I need to print out two columns from the database ie ford and mondeo before going into the second foreach loop. I've tried print $rows['model'] and all the other combinations I can think of but that doesn't work. Any help much appreciated

    Read the article

  • getting rid of repeated customer id's in mysql query

    - by bsandrabr
    I originally started by selecting customers from a group of customers and then for each customer querying the records for the past few days and presenting them in a table row. All working fine but I think I might have got too ambitious as I tried to pull in all the records at once having heard that mutiple queries are a big no no. here is the mysqlquery i came up with to pull in all the records at once SELECT morning, afternoon, date, date2, fname, lname, customers.customerid FROM customers LEFT OUTER JOIN attend ON ( customers.customerid = attend.customerid ) RIGHT OUTER JOIN noattend ON ( noattend.date2 = attend.date ) WHERE noattend.date2 BETWEEN '$date2' AND '$date3' AND DayOfWeek( date2 ) %7 >1 AND group ={$_GET['group']} ORDER BY lname ASC , fname ASC , date2 DESC tables are customer-customerid,fname,lname attend-customerid,morning,afternoon,date noattend-date2 (a table of all the days to fill in the blanks) Now the problem I have is how to start a new row in the table when the customer id changes My query above pulls in customer 1 morning 2 customer 1 morning 1 customer 2 morning 2 customer 2 morning 1 whereas I'm trying to get customer1 morning2 morning1 customer2 morning2 morning1 I dont know whether this is possible in the sql or more likely in the php

    Read the article

  • Mysql filling in missing dates

    - by bsandrabr
    I have the following query SELECT * FROM attend RIGHT OUTER JOIN noattend ON attend.date = noattend.date2 WHERE attend.date BETWEEN '2010-02-01' AND '2010-04-01' AND attend.customerid =1 ORDER BY date DESC LIMIT 0 , 30 Attend is the table with customerid noattend is the table with a row for each date (date2) I followed the advice in other questions to right outer join it to create values where there is no record in attend but it still isn't filling in the empties any help much appreciated

    Read the article

1