I have multiple tables and I want to access the fields in a specific table. Here are the table schemas:
Each tbl_acct has one tbl_unit. A tbl_unit has many tbl_groups and a tbl_groupcontact consist of the contacts and the related group. Every time I log in, I set a session where $_SESSION['unitid'] = $row['unitid'];.
What I wanted to do is to access the fields in the tbl_contacts. For now my code is here:
$data = $conn->prepare("SELECT * FROM tbl_groups LEFT JOIN tbl_groupcontact ON tbl_groups.id=tbl_groupcontact.group_id WHERE tbl_groups.unitid = ?");
$data->execute(array($_SESSION['unitid']));
foreach ($data as $row) {
echo $row['fname'] . " " . $row['lname']. "<br />";
}
As you can see, I can related the tbl_groups and tbl_groupcontact in my code however, I can't get the fields in the tbl_contacts. Am I missing something here? Any help would be much appreciated.
You need to join another table.
SELECT tbl_contacts.*
FROM tbl_groups
INNER JOIN tbl_groupcontact ON tbl_groupcontact.group_id = tbl_groups.id
INNER JOIN tbl_contacts ON tbl_contacts.id = tbl_groupcontact.contact_id
WHERE tbl_groups.unitid = ?
No need for (slower) LEFT JOIN btw - use it when you want to retrieve records from (left-side) table even when there's no match found in joined (right-side) table (it's columns will be filled with nulls in this case).
Related
I have 2 tables. One called local_requestors and one called local_aid. They both have different sets of data, but they both share a column, the column in local_requestors called suser_scheme is the same as the data in the local_aid table column called scheme_id.
I want to query my tables to get all data from the local_requestors table, where the ID matches that of the row in the local_aid table.
I've used the following...
$stmt = $conn->prepare('SELECT *
FROM local_requestors
INNER JOIN local_aid
WHERE local_requestors.user_scheme=local_aid.scheme_id
AND local_requestors.id = :aid_id');
$stmt->execute(['aid_id' => $user_id]);
while($row = $stmt->fetch()) {
echo '<p>' . $row['user_name'] . '</p>';
}
This is all on a local page that I can only access, but to query the table I post some $_GET[] variables...
$user_id = $_GET['id'];
$scheme_id = $_GET['scheme_id'];
No matter what I try, I keep getting the same row from the local_requestors table, repeated around 10 times.
My goal is to get print the name of the user in the record from local_aid and a number of related records from local_requstors.
echo 'Hello ' . $aid_name . ', there is ' . $rowCount . ' people that need help in your area.;
I think you may have ment
SELECT *
FROM local_requestors
INNER JOIN local_aid ON local_requestors.user_scheme=local_aid.scheme_id
WHERE local_requestors.id = :aid_id
And from your description of what you want you may want a LEFT JOIN instead of a INNER JOIN
I have a query that grabs a bunch of diffrent tables using LEFT JOIN, and I was wondering if I could incorporate another table, but only count the values in it.
The problem
When I try to use COUNT(row_id) in the query with everything else, it only returns the count and nothing anything else.
$query = "
SELECT COUNT(subscriptions.sub_id) AS total_subscriptions,
postings.posting_id,
postings.posting_headline,
postings.posting_body,
postings.timestamp,
users.user_name
FROM postings
LEFT JOIN users ON postings.user_id = users.user_id
LEFT JOIN subscriptions ON postings.group_id = subscriptions.group_id
WHERE postings.group_id='" . $group['group_id'] . "'
ORDER BY postings.posting_id DESC
";
How can I fix this?
Add an explicit GROUP BY clause to the query.
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
Ok, here's a database.
http://i.stack.imgur.com/j05AB.png
Say I've inserted values into the database for each of these tables, although the IDs would be auto incrementing. There are many BVALUES from each AVALUE, thus the AB table. I have all the AVALUEs from TABLE A in a drop-down list. A user selects an AVALUE which I put into a variable using
$AVALUE = $_POST['AVALUE']
Then I do an sql statement to get the AVALUEs from TABLE A that equal $AVALUE.
$sql = "SELECT AVALUE FROM TABLEA WHERE" . $AVALUE . " = AVALUE";
How do I then get the NAMEID from TABLEA for that AVALUE, then reference to AB where TABLEANAMEID = NAMEID from TABLEA? Then I want to get all the BVALUES by getting all the TABLEBNAMEIDs that correspond to the TABLEANAMEIDs.
I then want those BVALUES in a drop-down list on a seperate HTML page. After a bit of Googling the solution I think would be to do some sort of a loop putting the BVALUES into a variable as all the NAMEIDs from TABLE B increment where the variable would be in an $BVALUE loop and the list values would show with all the BVALUES.
I hope I explained that right. I think I know what I'm trying to do but I have no idea how to actually implement it. Please help guys.
You need to join those table together. What you are describing is an m:n relation. In this case you have do use 2 joins like this:
SELECT * FROM TableA AS a WHERE a.avalue = $AVALUE
JOIN TableAB AS a2b ON a.namevalue = a2b.id_a
JOIN TableB AS b ON a2b.id_b = b.id
Maybe u means, u want to get all BVALUE which has relation with selected AVALUE in table AB
$sql = "SELECT B.NAMEID as id, BVALUE as value FROM TABLEA A
LEFT JOIN AB ON TABLEANAMEID=A.NAMEID
LEFT JOIN TABLEB B ON TABLEBNAMEID=B.NAMEID
WHERE AVALUE = $AVALUE";`
get mysql result
$result = mysql_query($sql);
iterate
echo "<select>";
foreach ($r = mysql_fetch_object($result)) {
echo '<option value="'.$r->id.'">'.$r->value.'</option>';
}
echo "</select>";
I have 3 tables I need to join. The contracts table is the main table, the 'jobs' and 'companies' table are extra info that can be associated to the contracts table.
so, since I want all entries from my 'contracts' table, and the 'jobs' and 'companies' data only if it exists, I wrote the query like this....
$sql = "SELECT * FROM contracts
LEFT JOIN jobs ON contracts.job_id = jobs.id
LEFT JOIN companies ON contracts.company_id = companies.id
ORDER BY contracts.end_date";
Now how would I output this in PHP? I tried this but kept getting an undefined error "Notice: Undefined index: contracts.id"...
$sql_result = mysql_query($sql,$connection) or die ("Fail.");
if(mysql_num_rows($sql_result) > 0){
while($row = mysql_fetch_array($sql_result))
{
$contract_id = stripslashes($row['contracts.id']);
$job_number = stripslashes($row['jobs.job_number']);
$company_name = stripslashes($row['companies.name']);
?>
<tr id="<?=$contract_id?>">
<td><?=$job_number?></td>
<td><?=$company_name?></td>
</tr>
<?
}
}else{
echo "No records found";
}
Any help is appreciated.
The column names will not be prefixed like this - and with each table having a column called "id" you could be in trouble. You should explicitly identify the columns you want returned rather than using "select *", and you then just retrieve the column by name un prefixed (e.g. $row['job_number']).
The below would solve you problem.
$sql = "SELECT contracts.id AS contract_id, jobs.job_number, companies.name FROM contracts
LEFT JOIN jobs ON contracts.job_id = jobs.id
LEFT JOIN companies ON contracts.company_id = companies.id
ORDER BY contracts.end_date";
Your problem is likely to be realted to the fact you are using two tables with the field id this is why you should select them as an alias using the mysql as clause.
You may also want to look into using a naming convention for your fields and sticking with it. For example, check out the theory of Hungarian Notation, this would stop issues like this from arrissing.