I am building a training platform for work. I have created the requirements for a user to be trained based on a role given to them. If that role is aligned to a document it will sit against the user. I have managed to get most of the way but am struglling on the best way to finish the where statement within mysqli.
tbldocfiles is a list of my files. I am looking at docid (could be multiple files associated to the document)
tbltrainingaccess sets the roles (driver, warehouseman, customer services) and shows which role (by id) is associated to the document in docfiles.
tblusertraining is the list of users and what role they have associated to them. (driver, warehouseman, customer services).
I am listing the documents associated to the user so have thought the following is the best way:
Look at the user and how many roles he/she is allocated
Look at the roles returned in point 1 (where function)
Identify and match the documents that have the same roles as the user (Join function)
create the list, then look at the unique values for docid. (distinct value)
Example User Bri has the driver and warehouseman role.
There are 5 documents in the db, 3 of them are associated to the driver role (docid 1,2,3) and 2 of them are associated to the warehouseman role (docid 2,4) the 5th document is associayted to customerservice.
My query should do this:
List all documents associated to the roles, that are associated to the user Bri
1
2
3
2
4
Now select unique values (using docid) from the above list:
1,2,3,4.
So my answer will be a used as a count function at the end using mysql_fetch_rows
SELECT DISTINCT tbldocfiles.docid FROM tbldocfiles LEFT JOIN tbltrainingaccess ON (tbldocfiles.docid = tbltrainingaccess.docid) where groupid='1' or groupid='9'
The above code works. but i've got myself confused.
The where statement needs to be the result of a query similar to :
select * from tblusertrainingrole where userid='1' (1 will be a variable based on page selection)
the result in this would be 1, 9 which are the groupid results.
Basically any help would be appreciated! I am sure it will be simple but have burnt myself out on this for a while and most answers in here helped with joining but not the where statement (that I could find)
Thank you in advance everyone!
You can do a select statement in the where. Since it is an or statement you can use in for the results. Please replace * with the column name for the value you need. Should look like
where groupid in (select * from tblusertrainingrole where userid = '1')
CPC is a kind of measurement, we ran different CPCs in every day and get second data. So I have a long list of CPC files and one location file. In here, I only use CPC1 in 4/27/2017 as an example (the following first jpg). Location file has the longitude and latitude of CPCs everyday of the entire experiment (the second example table in the following jpg).
Because the location only has date, ID, and long/lat, I want to link ID (CPC name) column in location file to file name (also CPC name) of every CPC files.
I am not sure how it can be done, but I guess it should be done in SQL or VBA. I hope there is a simpler way. Although I only know VBA in Excel, but if it has to be done in SQL or VBA, please get me some ideas. Thank you so much.
Example of CPC1
Example of location file
Perhaps this query will accomplish:
SELECT location.*, CPC1.* FROM CPC1 INNER JOIN CPC1.Date ON location.date WHERE location.ID = "CPC1";
Assumes same dates are in both tables, hence the INNER JOIN.
I am not that experience in sql so please forgive if its not a good question to ask,but i researched around almost for 3-4 days but no able to solve.
My problem is i have a table which have multiple image names in it,so what i have to do is whoever is the follower of a particular user i have to get the imaged from this table,so one user there can be multiple followers,so i have to fetch the images posted by all the followers.
Here is the subquery code snippet i am using.
SELECT id,
outfit_image,
img_title,
description
FROM outfitpic_list r2
WHERE Email=ANY(SELECT being_followed
FROM follower_table
WHERE follower='test#gmail.com')
So the inner query here returns multiple values,for each value(being_followed) i have to fetch all the images and display it,but with this query each time i get only one image.I tried IN also but didnot work out.
Table structure:-
Outfitpic_list table
id|outfit_image|datetime|Email|image_title|description
Follower_table
bring_followed|follower
Please help,I am stuck..!!
Thank you..!!
I think your problem may be the = sign between "E-mail" and "Any". Try this statement:
SELECT
id,
outfit_image,
img_title,
description
FROM outfitpic_list r2
WHERE Email IN
(
SELECT being_followed
FROM follower_table
WHERE follower='test#gmail.com'
)
It's the same statement, without the = sign, and the ANY keyword replaced with IN. (I cleaned it up a little to make it more readable)
Hey guys,
This is a follow-up to a question that I asked earlier. It is my first time using a relational database and I need help with a quick search string to bring up desired results.
Background information: I'm making a database for my photo portfolio and want to be able to retrieve image links/data via their categories. Each image can be listed in multiple categories.
My database is set-up as follows :
TABLE tbl_images (image_id, image_title, image_location, image_descrip,image_url)
TABLE tbl_categories (category_id,category_name,category_descrip)
TABLE tbl_image_categories (image_id,category_id)
Where one of my images (image_id=1) has two categories (Desert [category_id=1] and Winter [category_id=2]). Which I defined in tbl_image_categories as 1,1 and 1,2.
I also have a few other images that I defined as Desert images [category_id=1].
How would I go about getting which images should be loaded based on the Desert Category?
I tried:
SELECT tbl_images.image_url
FROM tbl_images,
tbl_image_categories,
tbl_categories
WHERE tbl_categories.category_id = 1
Try this:
SELECT DISTINCT tbl_images.image_url
FROM tbl_images,
tbl_image_categories,
tbl_categories
WHERE chad_categories.category_id = 1 //category_id=1 for Desert
AND chad_images.image_id = chad_image_categories.image_id
AND chad_image_categories.category_id = chad_categories.category_id
Wow, makes your head spin!
I am about to start a project, and although my mySql is OK, I can't get my head around what required for this:
I have a table of web addresses.
id,url
1,http://www.url1.com
2,http://www.url2.com
3,http://www.url3.com
4,http://www.url4.com
I have a table of users.
id,name
1,fred bloggs
2,john bloggs
3,amy bloggs
I have a table of categories.
id,name
1,science
2,tech
3,adult
4,stackoverflow
I have a table of categories the user likes as numerical ref relating to the category unique ref. For example:
user,category
1,4
1,6
1,7
1,10
2,3
2,4
3,5
.
.
.
I have a table of scores relating to each website address. When a user visits one of these sites and says they like it, it's stored like so:
url_ref,category
4,2
4,3
4,6
4,2
4,3
5,2
5,3
.
.
.
So based on the above data, URL 4 would score (in it's own right) as follows: 2=2 3=2 6=1
What I was hoping to do was pick out a random URL from over 2,000,000 records based on the current users interests.
So if the logged in user likes categories 1,2,3 then I would like to ORDER BY a score generated based on their interest.
If the logged in user likes categories 2 3 and 6 then the total score would be 5. However, if the current logged in user only like categories 2 and 6, the URL score would be 3. So the order by would be in context of the logged in users interests.
Think of stumbleupon.
I was thinking of using a set of VIEWS to help with sub queries.
I'm guessing that all 2,000,000 records will need to be looked at and based on the id of the url it will look to see what scores it has based on each selected category of the current user.
So we need to know the user ID and this gets passed into the query as a constant from the start.
Ain't got a clue!
Chris Denman
What I was hoping to do was pick out a random URL from over 2,000,000 records based on the current users interests.
This screams for predictive modeling, something you probably wouldn't be able to pull off in the database. Basically, you'd want to precalculate your score for a given interest (or more likely set of interests) / URL combination, and then query based on the precalculated values. You'd most likely be best off doing this in application code somewhere.
Since you're trying to guess whether a user will like or dislike a link based on what you know about them, Bayes seems like a good starting point (sorry for the wikipedia link, but without knowing your programming language this is probably the best place to start): Naive Bayes Classifier
edit
The basic idea here is that you continually run your precalculation process, and once you have enough data you can try to distill it to a simple formula that you can use in your query. As you collect more data, you continue to run the precalculation process and use the expanded results to refine your formula. This gets really interesting if you have the means to suggest a link, then find out whether the user liked it or not, as you can use this feedback loop really improve the prediction algorithm (have a read on machine learning, particularly genetic algorithms, for more on this)
I did this in the end:
$dbh = new NewSys::mySqlAccess("xxxxxxxxxx","xxxxxxxxxx","xxxxxxxxx","localhost");
$icat{1}='animals pets';
$icat{2}='gadget addict';
$icat{3}='games online play';
$icat{4}='painting art';
$icat{5}='graphic designer design';
$icat{6}='philosophy';
$icat{7}='strange unusual bizarre';
$icat{8}='health fitness';
$icat{9}='photography photographer';
$icat{10}='reading books';
$icat{11}='humour humor comedy comedian funny';
$icat{12}='psychology psychologist';
$icat{13}='cartoons cartoonist';
$icat{14}='internet technology';
$icat{15}='science scientist';
$icat{16}='clothing fashion';
$icat{17}='movies movie latest';
$icat{18}="\"self improvement\"";
$icat{19}='drawing art';
$icat{20}='latest band member';
$icat{21}='shop prices';
$icat{22}='recipe recipes food';
$icat{23}='mythology';
$icat{24}='holiday resorts destinations';
$icat{25}="(rude words)";
$icat{26}="www website";
$dbh->Sql("DELETE FROM precalc WHERE member = '$fdat{cred_id}'");
$dbh->Sql("SELECT * FROM prefs WHERE member = '$fdat{cred_id}'");
#chos=();
while($dbh->FetchRow()){
$cat=$dbh->Data('category');
$cats{$cat}='#';
}
foreach $cat (keys %cats){
push #chos,"\'$cat\'";
push #strings,$icat{$cat};
}
$sqll=join("\,",#chos);
$words=join(" ",#strings);
$dbh->Sql("select users.id,users.url,IFNULL((select sum(scoretot.scr) from scoretot where scoretot.id = users.id and scoretot.category IN \($sqll\)),0) as score from users WHERE MATCH (description,lasttweet) AGAINST ('$words' IN BOOLEAN MODE) AND IFNULL((SELECT ref FROM visited WHERE member = '$fdat{cred_id}' AND user = users.id LIMIT 1),0) = 0 ORDER BY score DESC limit 30");
$cnt=0;
while($dbh->FetchRow()){
$id=$dbh->Data('id');
$url=$dbh->Data('url');
$score=$dbh->Data('score');
$dbh2->Sql("INSERT INTO precalc (member,user,url,score) VALUES ('$fdat{cred_id}','$id','$url','$score')");
$cnt++;
}
I came up with this answer about three months ago, and just cannot read it. So sorry, I can't explain how it finally worked, but it managed to query 2 million websites and choose one based on the history of a users past votes on other sites.
Once I got it working, I moved on to another problem!
http://www.staggerupon.com is where it all happens!
Chris