$queryread = "SELECT Properties.ID, Properties.Name,
Properties.Description, County.County
FROM Properties
INNER JOIN County ON Properties.County_ID = County.ID
AND Properties.Ban = 0";
This is the query to display properties on the public website. When an owner is signed in how would i display only the properties that the owner owns.
Owner table is called "RegisteredOwners" and id is "id", there is also a "Owner_ID" in the Properties table.
Thanks
You will simply need to add a WHERE clause to your query. To keep the answer simple and contained, I'll just use a subquery for the where clause. The query below is identical to yours, with an added line for the where clause.
SELECT Properties.ID, Properties.Name, Properties.Description, County.County
FROM Properties INNER JOIN County ON (Properties.County_ID = County.ID AND Properties.Ban = 0)
WHERE Properties.Owner_ID = (SELECT id FROM RegisteredOwners WHERE Name=?)
You will need to replace the Name=? portion with the appropriate method of identifying the user's RegisteredOwners row.
Related
I want to apply conditions with my query like i want to interchange one column's value to another column. Here is my query and now i am not getting how to apply
mysqli_query($connect, "
select
user_pro.User_No,
user_pro.Fist_Name,
user_pro.Last_Name,
user_pro.Designation,
user_pro.Profile,
user_pro.ProfileDP,
user_fndrst.Frnd_SNo,
user_fndrst.Rqst_Sender,
user_fndrst.Rqst_Receiver,
user_fndrst.Rqst_Status
from
user_profile INNER JOIN user_fndrst ON user_pro.User_No = user_fndrst.Rqst_Sender
WHERE
user_fndrst.Rqst_Status='1' AND
user_fndrst.Rqst_Receiver='$user_id'
ORDER BY Frnd_SNo DESC
");
Here "user_pro" table contains user's details, "user_fndrst" table contains user's friend request status and $user_id is user's logined id. Here I want that IF user_fndrst.Rqst_Receiver='$user_id' THEN user_fndrst.Rqst_Receiver value change to user_fndrst.Rqst_Sender.
For this I have stuck with user friend request status fetching from table "user_fndrst".
It seems your query has some problems. Use this:
SELECT
user_pro.User_No, user_pro.Fist_Name, user_pro.Last_Name, user_pro.Designation, user_pro.Profile, user_pro.ProfileDP, user_fndrst.Frnd_SNo, user_fndrst.Rqst_Sender, user_fndrst.Rqst_Receiver, user_fndrst.Rqst_Status
FROM
user_profile user_pro
INNER JOIN user_fndrst ON user_pro.User_No = user_fndrst.Rqst_Sender
WHERE
user_fndrst.Rqst_Status='1' AND user_fndrst.Rqst_Receiver='$user_id'
ORDER BY
user_fndrst.Frnd_SNo DESC
I am trying to echo both the employee name and manager name
SQL QUERY:
SELECT *
FROM `form`
INNER JOIN `emp` AS employee
ON `form`.emp_ID = employee.emp_ID
INNER JOIN `emp` AS manager
ON `form`.manager_ID = manager.emp_ID
ECHO:
while($row = $result->fetch_assoc()){
echo $row['emp_name'];
}
Always outputs the managers name.
have tried the following:
$row['employee.emp_name']
$row['employee']['emp_name']
which all don't work.
any help is appreciated
When the columns have the same name, their values will overlap when retrieving by fetch_assoc(). You can either use fetch_array() and reference the columns by numeric index (not recommended with SELECT *, since you can't easily guarantee the order of columns), or you will have to list the column names explicitly and alias them. For example:
SELECT emp.emp_name emp_emp_name, manager.emp_name manager_emp_name, ...
This would give you separate distinctly named fields in the result, that you could then access from what fetch_assoc() returns.
I've been using Grocery Crud to develop a simple local application that allows users to register themselves and like bands and rate them and select people they know that are also registered in the application.
Entities:
Person(person_id,person_URL, fullname, hometown)
Band(band_id,band_URL,band_name,country,genre)
Relationships:
Likes(person_id,band_id,rate)
Knows(person_id,known_person_id)
My questions are:
1) I want to return a table of person and known person like below:
KNOWS
person_id | fullname | known_person_id | known_fullname
but I can't use *set_relation_n_n* function 'cause the relationship is (Person -> Likes -> Person), so it's giving me error. The other solution I came up with is making a custom table making a query to return the values I want and show it in the table (code below). The custom table returned is correct but when I render it to my Grocery Crud table, I need to specify $crud->columns('person_id', 'fullname', 'known_person_id', 'fullname'), and it cannot differentiate the fullname of the person and the fullname of the known person. How would I make it in order to be able to show the table that way?
2) I have the same issue in another table but could manage that using the function *set_relation_n_n* 'cause it's a relationship (Person -> Likes -> Band), so since it's 2 different entities it didn't return me a error. The problem is that the query (code below) returns me the whole table and I want only 25 records per page. When I try to use "LIMIT 25" in the query, it returns me ONLY 25 records and the "next page" button doesn't work. Any solutions?
Below, all the information:
CODE for question 1:
function knows_management()
{
$crud = new grocery_CRUD();
$crud->set_model('model_socialnetwork');
$crud->set_table('knows');
$crud->set_subject('Known');
$crud->basic_model->set_query_str('SELECT tb1.person_id, tb1.fullname, tb1.known_person_id, person.fullname FROM (SELECT person.person_id, person.fullname, knows.known_person_id FROM person INNER JOIN knows ON person.person_id = knows.person_id) tb1 INNER JOIN person ON tb1.known_person_id = person.person_id');<br>
$crud->columns('person_id','fullname','known_person_id','fullname');
$output = $crud->render();
$this->_socialnetwork_output($output);
}
CODE for question 2:
function likes_management()
{
$crud = new grocery_CRUD();
$crud->set_model('model_socialnetwork');
$crud->set_table('likes');
$crud->set_subject('Like');
$crud->columns('person_id','fullname','band_id','band_name', 'rate');
$crud->basic_model->set_query_str('SELECT tb2.person_id, tb2.fullname, tb2.band_id, band.band_name, tb2.rate FROM(SELECT tb1.person_id, person.fullname, tb1.band_id, tb1.rate FROM(SELECT person.person_id, likes.band_id, likes.rate FROM person INNER JOIN likes ON person.person_id = likes.person_id) tb1 INNER JOIN person ON tb1.person_id = person.person_id) tb2 INNER JOIN band ON tb2.band_id = band.band_id');
$output = $crud->render();
$this->_socialnetwork_output($output);
}
Question 1) What if you use an alias name in your query, for example
SELECT tb1.person_id, tb1.fullname as Tb1fullName, tb1.known_person_id, person.fullname as PersonFullName
Question 2) I would not recommend you to add LIMIT directly / manually in your query. In the file application/config/grocery_crud.php, you have two options directly related to pagination
You should use and configure them properly
// The default per page when a user firstly see a list page
$config['grocery_crud_default_per_page'] = 25;
....
//Having some options at the list paging. This is the default one that all the websites are using.
//Make sure that the number of grocery_crud_default_per_page variable is included to this array.
$config['grocery_crud_paging_options'] = array('10','25','50','100');
I have two tables 'accounts_transactions' and 'accounts_bills_transactions'.
I have to left join these two using active record of codeigniter.But the names of key columns used to join are different.So I am not getting the key column from the left table in the output .What query should I write to get the key column from the left table included in the result.
My code is
$this->db->select('*');
$this->db->from('accounts_transactions');
$this->db->join('accounts_bills_transactions', 'accounts_transactions.id = accounts_bills_transactions.transaction_id','left');
$query = $this->db->get();
So, as you see the key columns used to join here are , id from left table and transaction_id from second table.The problem is that I am not getting the id from left table in the result.But I am getting all other columns.I assume the problem is because of difference in column names used to join.ie both the column names are not named 'id' .So how can I get the id from left table included in the result.
You could alias them:
$this->db->select('accounts_transatctions.*, account_transactions.id AS a_id,
accounts_bills_transactions.*,
account_bills_transactions.id AS ab_id');
$this->db->from('accounts_transactions');
$this->db->join('accounts_bills_transactions', 'accounts_transactions.id = accounts_transactions.transaction_id','left');
$query = $this->db->get();
The two IDs will now be available as a_id and ab_id (or whatever alias you choose)
Note: I'm not sure if you can alias in AR without avoiding escaping (haven't been using CI for a while). Should you get any error for that reason, just pass false as second parameter of $this->db->select():
$this->db->select('...', false);
you can try this if you confuse of using $this->where or $this->join
$query = $this->db->query("select ......");
return $query;
You problem is so simple. You can use this query
$query = $this->db
->select('at.*')
->select('abt.id as abt_id');
->from('accounts_transactions at');
->join('accounts_bills_transactions abt', 'at.id = abt.transaction_id','left');
->get()
->result();
When same column are used in join it selects only one. You need to give alise to the other column in second table. The best practice is to use a structure like this
accounts_transatctions
--------------------------
accounts_transatctions_id
other_columns
accounts_bills_transactions
---------------------------
accounts_bills_transactions_id
accounts_transatctions_id
other_columns
I have two primary MySQL tables (profiles and contacts) with many supplementary tables (prefixed by prm_). They are accessed and manipulated via PHP.
In this instance I am querying the profiles table where I will retrieve an Owner ID and a Breeder ID. This will then be referenced against the contacts table where the information on the Owners and Breeders is kept.
I received great help here on another question regarding joins and aliases, where I was also furnished with the following query. Unfortunately, I am having huge difficulty in actually echoing out the results. Every single site that deals with Self Joins and Aliases provide lovely examples of the queries - but then skip to "and this Outputs etc etc etc". How does it output????
SELECT *
FROM (
SELECT *
FROM profiles
INNER JOIN prm_breedgender
ON profiles.ProfileGenderID = prm_breedgender.BreedGenderID
LEFT JOIN contacts ownerContact
ON profiles.ProfileOwnerID = ownerContact.ContactID
LEFT JOIN prm_breedcolour
ON profiles.ProfileAdultColourID = prm_breedcolour.BreedColourID
LEFT JOIN prm_breedcolourmodifier
ON profiles.ProfileColourModifierID = prm_breedcolourmodifier.BreedColourModifierID
) ilv LEFT JOIN contacts breederContact
ON ilv.ProfileBreederID = breederContact.ContactID
WHERE ProfileName != 'Unknown'
ORDER BY ilv.ProfileGenderID, ilv.ProfileName ASC $limit
Coupled with this is the following PHP:
$owner = ($row['ownerContact.ContactFirstName'] . ' ' . $row['ownerContact.ContactLastName']);
$breeder = ($row['breederContact.ContactFirstName'] . ' ' . $row['breederContact.ContactLastName']);
All details EXCEPT the contacts (gender, colour, etc.) return fine. The $owner and $breeder variables are empty.
Any help in settling this for me would be massively appreciated.
EDIT: My final WORKING query:
SELECT ProfileOwnerID, ProfileBreederID,
ProfileGenderID, ProfileAdultColourID, ProfileColourModifierID, ProfileYearOfBirth,
ProfileYearOfDeath, ProfileLocalRegNumber, ProfileName,
owner.ContactFirstName AS owner_fname, owner.ContactLastName AS owner_lname,
breeder.ContactFirstName AS breeder_fname, breeder.ContactLastName AS breeder_lname,
BreedGender, BreedColour, BreedColourModifier
FROM profiles
LEFT JOIN contacts AS owner
ON ProfileOwnerID = owner.ContactID
LEFT JOIN contacts AS breeder
ON ProfileBreederID = breeder.ContactID
LEFT JOIN prm_breedgender
ON ProfileGenderID = prm_breedgender.BreedGenderID
LEFT JOIN prm_breedcolour
ON ProfileAdultColourID = prm_breedcolour.BreedColourID
LEFT JOIN prm_breedcolourmodifier
ON ProfileColourModifierID = prm_breedcolourmodifier.BreedColourModifierID
WHERE ProfileName != 'Unknown'
ORDER BY ProfileGenderID, ProfileName ASC $limit
Which I could then output by:
$owner = ($row['owner_lname'] . ' - ' . $row['owner_fname']);
Many Thanks to All!
I guess you're using the mysql_fetch_array or the mysql_fetch_assoc-functions to get the array from the result-set?
In this case, you can't use
$row['ownerContact.ContactFirstName']
as the PHP-Docs read:
If two or more columns of the result have the same field names, the
last column will take precedence. To access the other column(s) of the
same name, you must use the numeric index of the column or make an
alias for the column. For aliased columns, you cannot access the
contents with the original column name.
So, you can either use an AS in your SQL-query to set other names for the doubled rows or use the numbered indexes to access them.
This could then look like this:
Using AS in your Query
In your standard SQL-query, the columns in the result-set are named like the columns which their values come from. Sometimes, this can be a problem due to a naming-conflict. Using the AS-command in your query, you can rename a column in the result-set:
SELECT something AS "something_else"
FROM your_table
This will rename the something-column to something_else (you can leave the ""-quotes out, but I think it makes it more readable).
Using the column-indexes for the array
The other way to go is using the column-index instead of their names. Look at this query:
SELECT first_name, last_name
FROM some_table
The result-set will contain two columns, 0 ==> first_name and 1 ==> last_name. You can use this numbers to access the column in your result-set:
$row[0] // would be the "first_name"-column
$row[1] // would be the "last_name"-column
To be able to use the column-index, you'll need to use mysql_fetch_row or the mysql_fetch_assoc-function, which offers an associative array, a numeric array, or both ("both" is standard).
you need to replace the * with the data you need , and the similar ones you have to make aliases too :
ownerContact.ContactFirstName as owner_ContactFirstName
and
breederContact.ContactFirstName as breeder_ContactFirstName .
like this :
select ownerContact.ContactFirstName as owner_ContactFirstName , breederContact.ContactFirstName as breeder_ContactFirstName from profiles join ownerContact ... etc
in this way you will write :
$owner = ($row['owner_ContactFirstName'] . ' ' . $row['owner_ContactLastName']);
$breeder = ($row['breeder_ContactFirstName'] . ' ' . $row['breeder_ContactLastName']);
You cannot specify table alias when you access row using php. Accessing it by $row['ContactFirstName'] would work if you didn't have 2 fields with the same name. In this case whatever ContactFirstName appears second overwrites the first.
Change your query to use fields aliases, so you can do $owner = $row['Owner_ContactFirstName'].
Another option I'm not 100% sure is to access field by index, not by name(e.g. $owner=$row[11]). Even if it works I don't recommend to do so, you will have a lot of troubles if change your query a bit.
On outer select You have only two tables:
(inner select) as ilv
contacts as breederContact
there is no ownerContact at all