hope my title is clear. I am trying to retrieve results from two tables.
So I want everything (hence *) from the table called 'albums'.
and I want only all matching (with album_id) results from table 'contributors'.
$result = mysql_query("SELECT * FROM albums LEFT JOIN contributors ON albums.album_id =
contributors.album_id ORDER BY albums.datum DESC; ") or die(mysql_error());
$aantal_rijen = mysql_num_rows($result);
if ($aantal_rijen > 0) {
for ($i = 0; $i < $aantal_rijen; $i++){
$contributors[] = mysql_result($result, $i, 'contributors');}
but i get a list of similar errors:
contributor not found in MySQL result index ...
Joining of two tables is totally new to my, and maybe it's not the way to go, but maybe it's just a plain simple error in this code, anyway, I'm stuck here,
all help is welcome
thx S
http://php.net/manual/en/function.mysql-result.php
FIELD
The name or offset of the field being retrieved.
It can be the field's offset, the field's name, or the field's table dot field
name (tablename.fieldname). If the column name has
been aliased ('select foo as bar from...'), use the alias instead of
the column name. If undefined, the first field is retrieved.
You're just using the table name contributors, so either you need to specify a single field or limit your select to only the fields you want
$contributors[] = mysql_result($result, $i, 'contributors.name');
OR
$result = mysql_query("
SELECT contributors.*
FROM albums
LEFT JOIN contributors ON albums.album_id = contributors.album_id
ORDER BY albums.datum DESC;
") or die(mysql_error());
Related
I'm trying to display a HTML table with information from 3 different tables in my mysql DB. However I am unsure on how to display the information from the third table.
Currently what I am using is:
$SQL = "SELECT members.*, exp.*, lvl.*
FROM members
INNER JOIN exp ON members.id = exp.member_id
INNER JOIN lvl ON members.id = lvl.member_id
ORDER BY lvl.level DESC,
lvl.total DESC, xp.total DESC";
$result = mysql_query($SQL) or die(mysql_error());
$count = 1;
while ($row = mysql_fetch_assoc($result)) {
$level = $row['level'];
$exp = $row['exp.overall'];
}
the $level is from the second table which grabs correctly, and the $exp is what I want to grab from the third table which is "exp" but it doesn't return anything
How can I change this because at the moment it just seems to be focusing on the data from the "lvl" table when using $row[]
Edit: Both the lvl and exp tables have a row in called 'overall' which is why using $row['overall'] doesn't return what I want as it returns the data from lvl table rather than exp.
First off I believe you have a typo in your last order column: should be exp.total DESC.
Secondly, unless you specify the columns to be named with dot notation explicitly they will retain their column names so try changing the last line to:
$exp = $row['overall'];.
Also consider using mysqli or PDO.
I have updated my original post based on what I learned from your comments below. It is a much simpler process than I originally thought.
require '../database.php';
$pdo = Database::connect();
$pdo->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
$sql = "SELECT * FROM Orders WHERE id = 430";
$q = $pdo->prepare($sql);
$q->execute(array($id));
$data = $q->fetch(PDO::FETCH_ASSOC);
echo 'Order Num: ' . $data['id'] . '<br>';
$sql = "SELECT * FROM Order_items
JOIN Parts ON Parts.id = Order_Items.part_id
WHERE Order_Items.orders_id = 430";
$q = $pdo->prepare($sql);
$q->execute(array($line_item_id));
$data = $q->fetch(PDO::FETCH_ASSOC);
while ($data = $q->fetch(PDO::FETCH_ASSOC))
{
echo '- ' . $data['part_num'] . $data['qty'] . "<br>";
}
Database::disconnect();
Unfortunately, only my first query is producing results. The second query is producing the following ERROR LOG: "Base table or view not found: 1146 Table 'Order_items' doesn't exist" but I am expecting the following results.
Expected Results from Query 1:
Order Num: 430
Expected Results from Query 2:
- Screws 400
- Plates 35
- Clips 37
- Poles 7
- Zip ties 45
Now that I understand where you are coming from, let's explain a couple of things.
1.PDO and mysqli are two ways of accessing the database; they essentially do the same things, but the notation is different.
2.Arrays are variables with multiple "compartments". Most typical array has the compartments identified by a numerical index, like:
$array[0] = 'OR12345'; //order number
$array[1] = '2017-03-15'; //order date
$array[2] = 23; //id of a person/customer placing the order
etc. But this would require us to remember which number index means what. So in PHP there are associative arrays, which allow using text strings as indexes, and are used for fetching SQL query results.
3.The statement
$data = $q->fetch(PDO::FETCH_ASSOC)
or
$row = $result->fetch_assoc()
do exactly the same thing: put a record (row) from a query into an array, using field names as indexes. This way it's easy to use the data, because you can use field names (with a little bit around them) for displaying or manipulating the field values.
4.The
while ($row = $result->fetch_assoc())
does two things. It checks if there is a row still to fetch from the query results. and while there is one - it puts it into the array $row for you to use, and repeats (all the stuff between { and }).
So you fetch the row, display the results in whatever form you want, and then loop to fetch another row. If there are no more rows to fetch - the loop ends.
5.You should avoid using commas in the FROM clause in a query. This notation can be used only if the fields joining the tables are obvious (named the same), but it is bad practice anyway. The joins between tables should be specified explicitly. In the first query you want the header only, and there is no additional table needed in your example, so you should have just
SELECT *
FROM Orders
WHERE Orders.Order_ID = 12345
whereas in the second query I understand you have a table Parts, which contains descriptions of various parts that can be ordered? If so, then the second query should have:
SELECT *
FROM Order_items
JOIN Parts ON Parts.ID = Order_Items.Part_ID
WHEERE Order_Items.Order_ID = 12345
If in your Orders table you had a field for the ID of the supplier Supplier_ID, pointing to a Suppliers table, and an ID of the person placing the order Customer_ID, pointing to a Customers table, then the first query would look like this:
SELECT *
FROM Orders
JOIN Suppliers ON Suppliers.ID = Orders.Supplier_ID
JOIN Customers ON Customers.ID = Orders.Customer_ID
WHERE Orders.Order_ID = 12345
Hope this is enough for you to learn further on your own :).
I am creating a query and I used the LEFT JOIN to join two tables. But I'm having trouble in getting the fb_id value from the table, it gives me an empty result. Here is my code:
$sql = "SELECT * FROM tblfeedback a LEFT JOIN tblreply b ON a.fb_id = b.fb_id WHERE a.fb_status = 1 ORDER BY a.fb_id DESC";
$res = $con->query($sql);
....
....
The query above would give me a result like this :
I think that's why I don't get the fb_id value is because the last column is null. How do I get the value of fb_id from the first column? Thanks. I am really having trouble with this. I hope someone can enlightened my mind.
You should give an alias to the column in the parent table, because the column names are the same in both tables. When fetch_assoc() fills in $row['fb_id'], it gets the last one in the result row, which can be NULL because it comes from the second table.
SELECT a.fb_id AS a_id, a.*, b.*
FROM tblfeedback a
LEFT JOIN tblreply b ON a.fb_id = b.fb_id
WHERE a.fb_status = 1
ORDER BY a_id DESC
Then you can access $row['a_id'] to get this column.
More generally, I recommend against using SELECT *. Just select the columns you actually need. So you can select a.fb_id without selecting b.fb_id, and it will always be filled in.
Because you are using a left join, the 2 rows in your result set image are the rows from tblfeedback whose fb_id were not found in tblreply. We know this is true because all the tblreply columns in the result set are null.
With that said, its not real clear what you are asking for. If you are asking how you access the tblfeedback.fd_id column from your query via php, you can use the fetch_array method and use the 0 index.
$sql = "SELECT * FROM tblfeedback a LEFT JOIN tblreply b ON a.fb_id = b.fb_id WHERE a.fb_status = 1 ORDER BY a.fb_id DESC";
$res = $con->query($sql);
while($row = $res->fetch_array()) {
echo "fb_id: " . $row[0] . "<br>";
}
Alright, so I have a table outputting data from a MySQL table in a while loop. Well one of the columns it outputs isn't stored statically in the table, instead it's the sum of how many times it appears in a different MySQL table.
Sorry I'm not sure this is easy to understand. Here's my code:
$query="SELECT * FROM list WHERE added='$addedby' ORDER BY time DESC";
$result=mysql_query($query);
while($row=mysql_fetch_array($result, MYSQL_ASSOC)){
$loghwid = $row['hwid'];
$sql="SELECT * FROM logs WHERE hwid='$loghwid' AND time < now() + interval 1 hour";
$query = mysql_query($sql) OR DIE(mysql_error());
$boots = mysql_num_rows($query);
//Display the table
}
The above is the code displaying the table.
As you can see it's grabbing data from two different MySQL tables. However I want to be able to ORDER BY $boots DESC. But as its a counting of a completely different table, I have no idea of how to go about doing that.
There is a JOIN operation that is intended to... well... join two different table together.
SELECT list.hwid, COUNT(log.hwid) AS boots
FROM list WHERE added='$addedby'
LEFT JOIN log ON list.hwid=log.hwid
GROUP BY list.hwid
ORDER BY boots
I'm not sure if ORDER BY boots in the last line will work like this in MySQL. If it doesn't, just put all but the last line in a subquery.
But the result of the query into an array indexed by $boots.
AKA:
while(..){
$boot = mysql_num_rows($query);
$results[$boot][] = $result_array;
}
ksort($results);
foreach($results as $array)
{
foreach($array as ....)
{
// display table
}
}
You can switch between ksort and krsort to switch the orders, but basically you are making an array that is keyed by the number in $boot, sorting that array by that number, and then traversing each group of records that have a specific $boot value.
I have a MySQL database called "bookfeather." It contains 56 tables. Each table has the following structure:
id site votes_up votes_down
The value for "site" is a book title. The value for "votes_up" is an integer. Sometimes a unique value for "site" appears in more than one table.
For each unique value "site" in the entire database, I would like to sum "votes_up" from all 56 tables. Then I would like to print the top 25 values for "site" ranked by total "votes_up".
How can I do this in PHP?
Thanks in advance,
John
You can do something like this (warning: Extremely poor SQL ahead)
select site, sum(votes_up) votes_up
from (
select site, votes_up from table_1
UNION
select site, votes_up from table_2
UNION
...
UNION
select site, votes_up from table_56
) group by site order by sum(votes_up) desc limit 25
But, as Dav asked, does your data have to be like this? There are much more efficient ways of storing this kind of data.
Edit: You just mentioned in a comment that you expect there to be more than 56 tables in the future -- I would look into MySQL limits on how many tables you can UNION before going forward with this kind of SQL.
Here's a PHP code snip that should get it done.
I have not tested it so it might have some typos and stuff, make sure you replace DB_NAME
$result = mysql_query("SHOW TABLES");
$tables = array();
while ($row = mysql_fetch_assoc($result)) {
$tables[] = '`'.$row["Tables_in_DB_NAME"].'`';
}
$subQuery = "SELECT site, votes_up FROM ".implode(" UNION ALL SELECT site, votes_up FROM ",$tables);
// Create one query that gets the data you need
$sqlStr = "SELECT site, sum(votes_up) sumVotesUp
FROM (
".$subQuery." ) subQuery
GROUP BY site ORDER BY sum(votes_up) DESC LIMIT 25";
$result = mysql_query($sqlStr);
$arr = array();
while ($row = mysql_fetch_assoc($result)) {
$arr[] = $row["site"]." - ".$row["sumVotesUp"];
}
print_r($arr)
The UNION part of Ian Clelland answer can be generated using a statement like the following. The table INFORMATION_SCHEMA.COLUMNS has a column TABLE_NAME to get all tables.
select * from information_schema.columns
where table_schema not like 'informat%'
and column_name like 'VOTES_UP'
Join all inner SELECT with UNION ALL instead of UNION. UNION is doing an implicit DISTINCT (on oracle).
The basic idea would be to iterate over all your tables (using a SQL SHOW TABLES statement or similar) in PHP, then for every table, iterate over the rows (SELECT site,votes_up FROM $table). Then, for every row, check the site against an array that you're building with sites as keys and votes up as values. If the site is already in the array, increment its votes appropriately; otherwise, add it.
Vaguely PHP-like pseudocode:
// Build an empty array for use later
$votes_array = empty_array();
// Get all the tables and iterate over them
$tables = query("SHOW TABLES");
for($table in $tables) {
$rows = query("SELECT site,votes_up FROM $table");
// Iterate over the rows in each table
for($row in $rows) {
$site = $row['site'];
$votes = $row['votes_up'];
// If the site is already in the array, increment votes; otherwise, add it
if(exists_in_array($site, $votes_array)) {
$votes_array[$site] += $votes;
} else {
insert_into_array($site => $votes);
}
}
}
// Get the sites and votes as lists, and print out the top 25
$sorted_sites = array_keys($votes_array);
$sorted_votes = array_values($votes_array);
for($i = 0; $i < 25; $i++) {
print "Site " . $sorted_sites[$i] . " has " . $sorted_votes[$i] . " votes";
}
"I allow users to add tables to the database." - I hope all your users are benevolent and trustworthy and capable. Do you worry about people dropping or truncating tables, creating incorrect new tables that break your code, or other things like that? What kind of security do you have when users can log right into your database and change the schema?
Here's a tutorial on relational database normalization. Maybe it'll help.
Just in case someone else that comes after you wants to find what this could have looked like, here's a single table that could do what you want:
create database bookfeather;
create user bookfeather identified by 'bookfeather';
grant all on bookfeather.* to 'bookfeather'#'%';
use bookfeather;
create table if not exists book
(
id int not null auto_increment,
title varchar(255) not null default '',
upvotes integer not null default 0,
downvotes integer not null default 0,
primary key(id),
unique(title)
);
You'd vote a title up or down with an UPDATE:
update book set upvotes = upvotes + 1 where id = ?
Adding a new book is as easy as adding another row:
insert into book(title) values('grails in action')
I'd strongly urge that you reconsider.