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.
Related
So I'm trying to implement a JOIN in my PHP but I don't know how to include the WHERE clause with JOIN in my query.
I'm trying to do:
SELECT SpecialFacts.ConservationStatus, SpecialFacts.Reproduction, SpecialFacts.Length, Habitat.Type, Habitat.located
FROM SpecialFacts
INNER JOIN Habitat ON SpecialFacts.SpeciesID=Habitat.SpeciesID
WHERE SpeciesID="G. cuvier"
Basically I'm trying to make sure the join happens on the tables where the SpeciesID is G. cuvier but everything I have tried so far doesn't work and the error it is giving me now is "Column SpeciesID in where clause is ambiguous".
Here is my relevant PHP code:
<?php
include 'connect.php';
$result = mysqli_query($connect,"SELECT SpecialFacts.ConservationStatus, SpecialFacts.Reproduction, SpecialFacts.Length, Habitat.Type, Habitat.located FROM SpecialFacts INNER JOIN Habitat ON SpecialFacts.SpeciesID=Habitat.SpeciesID WHERE SpeciesID='G. cuvier'") or die("fail" . mysqli_error($connect));
$i = 0;
while($row = mysqli_fetch_array($result))
{
echo $row[$i];
$i = $i + 1;
}
Column SpeciesID exists in both tables, so it doesn't know which need to be compared to value given as G. cuvier. As you are writing every column name as table name with dot(.), the same should be in where condition column SpecialFacts.SpeciesID = "G. cuvier".
So query should be like:
SELECT SpecialFacts.ConservationStatus, SpecialFacts.Reproduction, SpecialFacts.Length, Habitat.Type, Habitat.located
FROM SpecialFacts
INNER JOIN Habitat ON SpecialFacts.SpeciesID=Habitat.SpeciesID
WHERE SpecialFacts.SpeciesID="G. cuvier"
It's not giving you that error because you're using WHERE incorrectly. It's giving you that error because there are multiple tables in your query that have a column with that name, hence the "ambiguous". You just need to disambiguate it by adding the table name to the identifier.
WHERE SpecialFacts.SpeciesID="G. cuvier"
or
WHERE Habitat.SpeciesID="G. cuvier"
Since you're inner joining the tables on that column, either table should work for the WHERE clause. I would suggest using the smaller table for performance reasons, but honestly I'm not 100% certain if it will matter or not. You can do EXPLAIN on each one to see how they compare.
Use below query instead
SELECT SpecialFacts.ConservationStatus, SpecialFacts.Reproduction, SpecialFacts.Length, Habitat.Type, Habitat.located
FROM SpecialFacts INNER JOIN Habitat
ON SpecialFacts.SpeciesID=Habitat.SpeciesID
WHERE SpecialFacts.SpeciesID='G. cuvier'
I have used single quote and used column name with table name.
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
Basically I have 2 SQL queries from 2 different databases and I am trying to compare where they are equal and then join together the other information for that value. My first query contains an id and a product name, my second query contains a product name and components. So I'm trying to join them on the product name and then show the other two bits of information with them. The db I selected is being used in the second query. Any idea what I should do?
So far I have this, which only seems to show one result:
$catid = mysql_query("Select a.entry_id, b.cat_name from blog.exp_category_posts a inner join blog.exp_categories b on a.cat_id=b.cat_id where b.Group_ID = 3");
$results = mysql_query("Select a.name, c.product from wr_scientific a inner join wr_scientific_products b on a.id=b.scienc_id Join xcart_products c on b.prod_id=c.productid LIMIT 1000");
while($row1 = mysql_fetch_array($catid)){
$row2= explode("™", $row1['cat_name']);
$row3= explode("®", $row2[0]);
while($row = mysql_fetch_array($results)){
$rowpro = explode("™", $row['product']);
$rowprod = explode("®", $rowpro[0]);
if($rowprod[0] == $row3[0]){
echo $rowprod[0].$row['name'].$row1['entry_id'];
}
}
}
Thanks for any help
If the two databases are located on the same instance of MySQL (~ same machine), then you can refer to a table in say db2 from (say) db1 by prefixing the table name with the database name.
E.g:
USE db1 ;
SELECT
db1.table_in_db1.id, -- you can specify the database name here, but
table_in_db2.id, -- there is no ambiguity, the database name is optional
field_in_table_in_db2 -- the same way the table name is optionnal when there is no ambiguity
FROM
db1.table_in_db1 -- database name is optionnal here, current database is assumed
JOIN
db2.table_in_db2
ON ...
I have made an SQL query from two tables. Everything works good but problem is that this two tables have the same field names and after I do not know how to display them correct, how to tell that $data['aaa'] is from table 1 and the same $data['aaa'] from table 2
here is my SQL query :
$query_str = "SELECT
cm.id,
cm.global_category_id,
cm.num,
cm.menu_lv,
cm.menu_ru,
cm.menu_en,
u.id,
u.menu_lv,
u.menu_ru,
u.menu_en
FROM products_category cm, products_global_category u
WHERE cm.global_category_id = u.id
";
and display data
<? foreach ($sub_category_list as $line) : ?>
<tr>
<td><?=$line['menu_lv']?></td> <---- here I want to display data from products_global_category u
<td><?=$line['sub_menu_lv']?></td> <---- products_category cm
<td><?=$line['sub_menu_ru']?></td> <---- products_category cm
<td><?=$line['sub_menu_en']?></td> <---- products_category cm
</tr>
<? endforeach; ?>
As a solution you can change your SQL query to give an alias to the fields that have the same name.
For example:
SELECT somefield AS othername FROM table.
In this case, the somefield field will be available through the alias othername.
In your case:
$query_str = "SELECT
cm.id AS cm_id,
cm.global_category_id,
cm.num,
cm.menu_lv AS cm_menulv,
cm.menu_ru AS cm_menuru,
cm.menu_en AS cm.menuen,
u.id as u_id,
u.menu_lv AS u_menulv,
u.menu_ru AS u_menuru,
u.menu_en AS u_menuen
FROM products_category cm, products_global_category u
WHERE cm.global_category_id = u.id
";
And then in your PHP:
$line['u_menulv'] //Access field menu_lv from products_global_category table
$line['cm_menulv'] //Access field menu_lv from products_category table
EDIT: In mysql_fetch_array documentation page:
If two or more columns of the result have the same field names, the
last column will take precedence. To access the other column(s) of the
same name, you must use the numeric index of the column or make an
alias for the column. For aliased columns, you cannot access the
contents with the original column name.
In other words, either create an alias like shown above or access the fields by the numeric index of the array.
Simply alias the records you wish to print out which conflict with one another, for example:
SELECT
table1.pid AS page_id,
table2.pid AS product_id
FROM
table1
LEFT JOIN
table2
ON
table1.id = table2.id
And then within your PHP you can echo them out as follows:
while ($row = mysql_fetch_assoc($res)) {
echo $row['page_id'];
echo $row['product_id'];
}
Hope this helps!
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