Search Results

Search found 20755 results on 831 pages for 'click counting'.

Page 20/831 | < Previous Page | 16 17 18 19 20 21 22 23 24 25 26 27  | Next Page >

  • Counting the most tagged tag with MySQL

    - by Jack W-H
    Hi folks My problem is that I'm trying to count which tag has been used most in a table of user-submitted code. But the problem is with the database structure. The current query I'm using is this: SELECT tag1, COUNT(tag1) AS counttag FROM code GROUP BY tag1 ORDER BY counttag DESC LIMIT 1 This is fine, except, it only counts the most often occurence of tag1 - and my database has 5 tags per post - so there's columns tag1, tag2, tag3, tag4, tag5. How do I get the highest occurring tag value from all 5 columns in one query? Jack

    Read the article

  • problem in counting children category

    - by moustafa
    I have this table: fourn_category (id , sub) I am using this code to count: function CountSub($id){ $root = array($id); $query = mysql_query("SELECT id FROM fourn_category WHERE sub = '$id'"); while( $row = mysql_fetch_array( $query, MYSQL_ASSOC ) ){ array_push($root,$row['id']); CountSub($row['id']); } return implode(",",$root); } It returns the category id as 1,2,3,4,5 to using it to count the sub by IN() But the problem is that it counts this: category 1 category 2 category 3 category 4 category 5 Category 1 has 1 child not 4. Why? How can I get all children's trees?

    Read the article

  • Drupal - Counting data in nodes, creating custom statistics

    - by hfidgen
    Hiya, I'm building some custom content types to capture customer data on a website. Admins will enter the data, users will be able to view it, but I also need to be able to bolt on some statistics and infographics to the data. The problem I have is that I can't see any simple way of doing this within Drupal. Are there modules which can produce simple stats on selected node types or will I have to write a complete custom module using the data abstraction layer? Thanks for any insights!

    Read the article

  • counting twice in a query, once using restrictions

    - by Andrew Heath
    Given the following tables: Table1 [class] [child] math boy1 math boy2 math boy3 art boy1 Table2 [child] [glasses] boy1 yes boy2 yes boy3 no If I want to query for number of children per class, I'd do this: SELECT class, COUNT(child) FROM Table1 GROUP BY class and if I wanted to query for number of children per class wearing glasses, I'd do this: SELECT Table1.class, COUNT(table1.child) FROM Table1 LEFT JOIN Table2 ON Table1.child=Table2.child WHERE Table2.glasses='yes' GROUP BY Table1.class but what I really want to do is: SELECT class, COUNT(child), COUNT(child wearing glasses) and frankly I have no idea how to do that in only one query. help?

    Read the article

  • Counting words in a collection using LINQ

    - by icemanind
    Guys, I have a StringCollection object with 5 words in them. 3 of them are duplicate words. I am trying to create a LINQ query that will count how many unique words are in the collection and output them to to the console. So, for example, if my StringCollection has 'House', 'Car,'House','Dog', 'Cat', then it should output like this: House -- 2 Car -- 1 Dog -- 1 Cat -- 1 any ideas on how to create a LINQ query to do this?

    Read the article

  • Counting HTML images with Python

    - by user2537246
    I need some feedback on how to count HTML images with Python 3.01 after extracting them, maybe my regular expression are used properly. Here is my code: import re, os import urllib.request def get_image(url): url = 'http://www.google.com' total = 0 try: f = urllib.request.urlopen(url) for line in f.readline(): line = re.compile('<img.*?src="(.*?)">') if total > 0: x = line.count(total) total += x print('Images total:', total) except: pass

    Read the article

  • Counting divs for pagination in Jquery

    - by Craig Ward
    I want to create a nice pagination in Jquery for a number of divs I have. EG: <div class="container"> <div id="one">content</div> <div id="two">content</div> <div id="three">content</div> <div id="four">content</div> </div> The number will not always be the same so I need to count the divs, and display a pagination like the one below. 1|2|3|4 Clicking on the page number would display the relevant div. I know how to show and hide elements using Jquery and css and have figured out I can count the divs using var numPages = $('.container').size(); but I can't work out how I can display the pagination. Any pointers?

    Read the article

  • Counting amount of items in Pythons 'for'

    - by Markum
    Kind of hard to explain, but when I run something like this: fruits = ['apple', 'orange', 'banana', 'strawberry', 'kiwi'] for fruit in fruits: print fruit.capitalize() It gives me this, as expected: Apple Orange Banana Strawberry Kiwi How would I edit that code so that it would "count" the amount of times it's performing the for, and print this? 1 Apple 2 Orange 3 Banana 4 Strawberry 5 Kiwi

    Read the article

  • Counting characters in an Access database column using SQL

    - by jzr
    I have the following table col1 col2 col3 col4 ==== ==== ==== ===== 1233 4566 ABCD CDEF 1233 4566 ACD1 CDEF 1233 4566 D1AF CDEF I need to count the characters in col3, so from the data in the previous table it would be: char count ==== ===== A 3 B 1 C 2 D 3 F 1 1 2 Is this possible to achieve by using SQL only? At the moment I am thinking of passing a parameter in to SQL query and count the characters one by one and then sum, however I did not start the VBA part yet, and frankly wouldn't want to do that. This is my query at the moment: PARAMETERS X Long; SELECT First(Mid(TABLE.col3,X,1)) AS [col3 Field], Count(Mid(TABLE.col3,X,1)) AS Dcount FROM TEST GROUP BY Mid(TABLE.col3,X,1) HAVING (((Count(Mid([TABLE].[col3],[X],1)))>=1)); Ideas and help are much appreciated, as I don't usually work with Access and SQL.

    Read the article

  • Jquery cant get facebox working inside ajax call

    - by John
    From my main page I call an ajax file via jquery, in that ajax file is some additional jquery code. Original link looks like this: <a href="/page1.php" class="guest-action notify-function"><img src="/icon1.png"></a> Then the code: $(document).ready(function(){ $('a[rel*=facebox]').facebox(); $('.guest-action').click( function() { $.get( $(this).attr('href'), function(responseText) { $.jGrowl(responseText); }); return false; }); $('.notify-function').click( function() { $(this).find('img').attr('src','/icon2.png'); $(this).attr('href','/page2.php'); $(this).removeClass('guest-action').removeClass('notify-function').attr('rel','facebox'); }); }); So basically after notify-function is clicked I am changing the icon and the url of the link, I then am removing the classes so that the click wont be ran again and add rel="facebox" to the link so that the facebox window will pop up if they try to click the new icon2.png that shows up. The problem is after I click the initial icon everything works just fine except when I try to click the new icon2.png it still executes the jgrowl code from the guest-action. But when I view the source it shows this: <a href="/page2.php" rel="facebox" class=""><img src="/icon2.png"></a> So it seemed that should work right? What am I doing wrong? I tried adding the facebox code to the main page that is calling the ajax file as well and still same issue.

    Read the article

  • Date check for counting 30 days

    - by sijith
    Hi, I want to count the days passed with respect to a given date. I have a predefined date with me and i want to check the days passed, once the day pass 30 days with respect to the given time i want to get a message. example given date is25/03/2010 and when my system date reaches 25/04/2010 i have to get a message. How can i implement it. please give some help

    Read the article

  • Programming Contest Question: Counting Polyominos

    - by Martijn Courteaux
    Hi, An example question for a programming contest was to write a program that finds out how much polyominos are possible with a given number of stones. So for two stones (n = 2) there is only one polyominos: XX You might think this is a second solution: X X But it isn't. The polyominos are not unique if you can rotate them. So, for 4 stones (n = 4), there are 7 solutions: X X XX X X X X X X XX X XX XX XX X X X XX X X XX The application has to be able to find the solution for 1 <= n <=10 PS: Using the list of polyominos on Wikipedia isn't allowed ;) EDIT: Of course the question is: How to do this in Java, C/C++, C#

    Read the article

  • counting number of new entries in guest book

    - by Lucka
    I want to display the number of new(unseen) guest entries in the guest book of a user. I was thinking to count it so that, total number of entries in guestbook minus entries in guestbook at time of last login. However, I think that is not a good approach, because if the user logs in but does not go to his guestbook, in that case, the entries should be still "new", also if some new entry is posted in the user while he is online, it does not work in that case too. Any suggested please?

    Read the article

  • Counting combinations in c or in python

    - by Dennis
    Hello I looked a bit on this topic here but I found nothing that could help me. I need a program in Python or in C that will give me all possible combinations of a and b that will meet the requirement n=2*a+b, for n from 0 to 10. a, b and n are integers. For example if n=0 both a and b must be 0. For n=1 a must be zero and b must be 1, for n=2 a can be 1 and b=0, or a=0 and b=2, etc. I'm not that good with programming. I made this: #include <stdio.h> int main(void){ int a,b,n; for(n = 0; n <= 10; n++){ for(a = 0; a <= 10; a++){ for(b = 0; b <= 10; b++) if(n == 2*a + b) printf("(%d, %d), ", (a,b)); } printf("\n"); } } But it keeps getting strange results like this: (0, -1079628000), (1, -1079628000), (2, -1079628000), (0, -1079628000), (3, -1079628000), (1, -1079628000), (4, -1079628000), (2, -1079628000), (0, -1079628000), (5, -1079628000), (3, -1079628000), (1, -1079628000), (6, -1079628000), (4, -1079628000), (2, -1079628000), (0, -1079628000), (7, -1079628000), (5, -1079628000), (3, -1079628000), (1, -1079628000), (8, -1079628000), (6, -1079628000), (4, -1079628000), (2, -1079628000), (0, -1079628000), (9, -1079628000), (7, -1079628000), (5, -1079628000), (3, -1079628000), (1, -1079628000), (10, -1079628000), (8, -1079628000), (6, -1079628000), (4, -1079628000), (2, -1079628000), (0, -1079628000), ideone Any idea what is wrong? Also if I could do this for Python it would be even cooler. :D

    Read the article

  • Counting distinct and duplicate attribute values in an array

    - by keruilin
    I have an array of users that's sorted in descending order based on total_points. I need to find the rank of each user in that array. The issue is that more than one user can have the same total points and, thus, the same rank. For example, three users could be in 3rd place with 200 Points. Here's my current code: class Leader < ActiveRecord::Base def self.points_leaders all_leaders = all_points_leaders # returns array of users sorted by total_points in desc order all_leaders_with_rank = [] all_leaders.each do |user| rank = all_leaders.index(user)+1 all_leaders_with_rank << Ldr.new(rank, user) # Ldr is a Struct end return all_leaders_with_rank end end How must I modify the code so that the correct rank is returned, and not just the value of the index position?

    Read the article

  • problem in counting two fields in one query

    - by Mac Taylor
    hey guys i need to count new private messages and old one from a table so first thing come to mind is using mysql_num_rows and easy thing to do // check new pms $user_id = $userinfo['user_id']; $sql = "SELECT author_id FROM bb3privmsgs_to WHERE user_id='$user_id' AND (pm_new='1' OR pm_unread='1')"; $result = $db->sql_query($sql) ; $new_pms = $db->sql_numrows($result); $db->sql_freeresult($result); // check old pms $sql = "SELECT author_id FROM bb3privmsgs_to WHERE user_id='$user_id' AND (pm_new='0' OR pm_unread='0')"; $result = $db->sql_query($sql) ; $old_pms = $db->sql_numrows($result); $db->sql_freeresult($result); but how can i count these two fields just in one statement and shorter lines ?~

    Read the article

  • Counting in R data.table

    - by Simon Z.
    I have the following data.table set.seed(1) DT <- data.table(VAL = sample(c(1, 2, 3), 10, replace = TRUE)) VAL 1: 1 2: 2 3: 2 4: 3 5: 1 6: 3 7: 3 8: 2 9: 2 10: 1 Now I want to to perform two tasks: Count the occurrences of numbers in VAL. Count within all rows with the same value VAL (first, second, third occurrence) At the end I want the result VAL COUNT IDX 1: 1 3 1 2: 2 4 1 3: 2 4 2 4: 3 3 1 5: 1 3 2 6: 3 3 2 7: 3 3 3 8: 2 4 3 9: 2 4 4 10: 1 3 3 where COUNT defines task 1. and IDX task 2. I tried to work with which and length using .I: dt[, list(COUNT = length(VAL == VAL[.I]), IDX = which(which(VAL == VAL[.I]) == .I))] but this does not work as .I refers to a vector with the index, so I guess one must use .I[]. Though inside .I[] I again face the problem, that I do not have the row index and I do know (from reading data.table FAQ and following the posts here) that looping through rows should be avoided if possible. So, what's the data.table way?

    Read the article

  • Counting Sublist Elements in Prolog

    - by idea_
    How can I count nested list elements in prolog? I have the following predicates defined, which will count a nested list as one element: length([ ], 0). length([H|T],N) :- length(T,M), N is M+1. Usage: ?- length([a,b,c],Out). Out = 3 This works, but I would like to count nested elements as well i.e. length([a,b,[c,d,e],f],Output). ?- length([a,b,[c,d,e],f],Output). Output = 6

    Read the article

  • Counting vowels

    - by user74283
    Can anyone please tell me what is wrong with this script. I am a python newb but i cant seem to figure out what might be causing it not to function. def find_vowels(sentence): """ >>> find_vowels(test) e """ count = 0 vowels = "aeiuoAEIOU" for letter in sentence: if letter in vowels: count += 1 print count if __name__ == '__main__': import doctest doctest.testmod()

    Read the article

  • question about counting sort

    - by davit-datuashvili
    hi i have write following code which prints elements in sorted order only one big problem is that it use two additional array here is my code public class occurance{ public static final int n=5; public static void main(String[]args){ // n is maximum possible value what it should be in array suppose n=5 then array may be int a[]=new int[]{3,4,4,2,1,3,5};// as u see all elements are less or equal to n //create array a.length*n int b[]=new int[a.length*n]; int c[]=new int[b.length]; for (int i=0;i<b.length;i++){ b[i]=0; c[i]=0; } for (int i=0;i<a.length;i++){ if (b[a[i]]==1){ c[a[i]]=1; } else{ b[a[i]]=1; } } for (int i=0;i<b.length;i++){ if (b[i]==1) { System.out.println(i); } if (c[i]==1){ System.out.println(i); } } } } // 1 2 3 3 4 4 5 1.i have two question what is complexity of this algorithm?i mean running time 2. how put this elements into other array with sorted order? thanks

    Read the article

  • Excel - Counting unique values that meet multiple criteria

    - by wotaskd
    I'm trying to use a function to count the number of unique cells in a spreadsheet that, at the same time, meet multiple criteria. Given the following example: A B C QUANT STORE# PRODUCT 1 75012 banana 5 orange 6 56089 orange 3 89247 orange 7 45321 orange 2 apple 4 45321 apple In the example above, I need to know how many unique stores with a valid STORE# have received oranges OR apples. In the case above, the result should be 3 (stores 56089, 89247 and 45321). This is how I started to try solving the problem: =SUM(IF(FREQUENCY(B2:B9,B2:B9)>0,1)) The above formula will yield the number of unique stores with a valid store#, but not just the ones that have received oranges or bananas. How can I add that extra criteria?

    Read the article

  • Java Counting # of occurrences of a word in a string

    - by Doug
    I have a large text file I am reading from and I need to find out how many times some words come up. For example, the word "the". I'm doing this line by line each line is a string. I need to make sure that I only count legit "the"'s the the in other would not count. This means I know I need to use regular expressions in some way. What I was trying so far is this: numSpace += line.split("[^a-z]the[^a-z]").length; I realize the regular expression may not be correct at the moment but I tried without that and just tried to find occurrences of the word the and I get wrong numbers to. I was under the impression this would split the string up into an array and how many times that array was split up was how many times the word is in the string. Any ideas I would be grateful.

    Read the article

  • Code Golf: Counting XML elements in file on Android

    - by CSharperWithJava
    Take a simple XML file formatted like this: <Lists> <List> <Note/> ... <Note/> </List> <List> <Note/> ... <Note/> </List> </Lists> Each node has some attributes that actually hold the data of the file. I need a very quick way to count the number of each type of element, (List and Note). Lists is simply the root and doesn't matter. I can do this with a simple string search or something similar, but I need to make this as fast as possible. Design Parameters: Must be in java (Android application). Must AVOID allocating memory as much as possible. Must return the total number of Note elements and the number of List elements in the file, regardless of location in file. Number of Lists will typically be small (1-4), and number of notes can potentially be very large (upwards of 1000, typically 100) per file. I look forward to your suggestions.

    Read the article

  • MySQL help, counting information on last records

    - by ee12csvt
    I need some advice I have two tables, one holds unique serial numbers of items (items) and the other holds status changes and other information for these items (details) The Tables are set up as follows Item itemID itemName itemDate details detID itemID modlvl status detDate All items have at least one record in the details table, but over time the status has changed or the modification level has changed (Both of these are identified by numbers which are held in other appropriate tables) and a new record is created each time the status/modlvl changes I want to display a table on my webpage using php that identifies the different mod levels of the items and shows a count of each of the current status of the items EDIT Hi Ronnis, This is an example of the data in the tables and what I want to achieve The current Mod Levels range from 1 to 3 Status representations are 1 In Use 2 In Store 3 Being repaired 4 In Transit 5 For Disposal 6 Disposed 7 Lost Item itemID OrigMod created 1000 1 2009-10-01 22:12:12 1001 1 2009-10-01 22:12:12 1002 1 2009-10-01 22:12:12 1003 1 2009-10-01 22:12:12 1004 1 2009-10-01 22:12:12 1005 1 2009-10-01 22:12:12 1006 1 2009-10-01 22:12:12 1007 1 2009-10-01 22:12:12 1008 1 2009-10-01 22:12:12 1009 1 2009-10-01 22:12:12 1010 1 2009-10-01 22:12:12 Details detID itemID modlvl detDate status 1 1000 1 2009-10-01 1 2 1001 1 2009-10-01 1 3 1002 1 2009-10-01 1 4 1003 1 2009-10-01 1 5 1004 1 2009-10-01 1 6 1005 1 2009-10-01 1 7 1006 1 2009-10-01 1 8 1007 1 2009-10-01 1 9 1008 1 2009-10-01 1 10 1009 1 2009-10-01 1 11 1010 1 2009-10-01 1 12 1001 1 2010-02-01 2 13 1001 1 2010-02-03 4 14 1001 1 2010-03-01 3 15 1000 1 2010-03-14 2 16 1001 2 2010-04-01 4 17 1006 1 2010-04-01 2 18 1001 2 2010-04-03 2 19 1006 1 2010-04-14 4 20 1006 1 2010-05-01 5 21 1002 1 2010-05-02 2 22 1003 1 2010-05-10 2 23 1010 1 2010-06-01 2 24 1006 1 2010-06-18 6 25 1010 1 2010-07-01 7 26 1007 1 2010-07-02 2 27 1007 1 2010-07-04 4 28 1003 1 2010-07-10 2 29 1007 1 2010-07-11 3 30 1007 2 2010-07-12 4 31 1007 2 2010-07-15 2 32 1001 2 2010-08-31 1 33 1001 2 2010-09-10 2 34 1001 2 2010-10-01 4 35 1008 1 2010-10-01 2 36 1001 2 2010-10-05 3 37 1008 1 2010-10-05 4 38 1008 1 2010-10-10 3 39 1001 3 2010-10-20 4 40 1001 3 2010-10-25 2 Using the tables above I want to get this result MoLvl Use Store Repd Transit Displ Dispd Lost Total 1 3 3 1 0 0 1 1 9 2 0 1 0 0 0 0 0 1 3 0 1 0 0 0 0 0 1 Total 3 5 1 0 0 1 1 11

    Read the article

< Previous Page | 16 17 18 19 20 21 22 23 24 25 26 27  | Next Page >