Is there any possibility to duplicate specific entries by auto updating context sensitive relations?
Given a table 'table1' like:
My goal is to duplicate all entries with categoryId 42 while updating parentId if neccessary:
id is an auto incremented column and parentId is used to identify relations between the entries.
Currently I'm inserting them one by one, selecting the old data and managing the logic for the parentId in PHP.
//Thats simplified what I do ($conn is a Zend_Db_Adapter_Abstract)
$rows = $conn->fetchAll("SELECT * FROM table1 WHERE categoryId = 42 ORDER BY parentId ASC");
$newIds = [];
foreach($rows as $row){
if(array_key_exists($row['parentId'],$newIds))
$row['parentId'] = $newIds[$row['parentId']];
else
$row['parentId'] = null;
$conn->query("INSERT INTO table1 (parentId,info,categoryId) VALUES (?,?,?)",[$row['parentId'],$row['info'],$row['categoryId']]);
$newId = $conn->lastInsertId('table1');
$newIds[$row['id']] = $newId;
}
I'm stuck with this because I need the lastInsertedId of the new element to set the new parentId for the next one.
But I'm experiencing this to be pretty slow (in relation to one single query which contains the logic).
Is there any possibility to give a query some kind of incremental element sensitive logic? Or have you any suggestions on how to fasten this up?
Any help appreciated! :)
You are not showing any code. I guess you use mysqli_insert_id() to get the last inserted ID. Maybe you can do something with MAX(ID)+1.
Something like:
INSERT INTO table1
SELECT MAX(ID)+1,
info,
categoryId
FROM
table1
WHERE .............. -- the conditions to retreive the parent
Related
How would I go about deleting a row from the table 'subjects' that has a primary id 'subject_id' based on the number of rows in another table named 'replies' that uses a 'subject_id' column as a reference.
Example in pseudo code:
If ('subject' has less than 1 reply){
delete 'subject'}
I don't know much about SQL triggers so I have no clue if I would be able to incorporate this directly in the database or if I'd have to write some PHP code to handle this...
To delete any subjects that have had no replies, this query should do the trick:
DELETE s.* FROM subjects AS s
WHERE NOT EXISTS
(
SELECT r.subject_id
FROM replies AS r
WHERE r.subject_id = s.subject_id
);
Demo: DB Fiddle Example
One of the MySQL gurus will need to weigh in on whether or not you can do this directly, but in PHP you could...
$query = "SELECT subject_id FROM subjects WHERE subject='test'";
$return = mysqli_query($mysqli, $query);
$id = mysqli_fetch_assoc($return);
$query = "SELECT reply_id FROM replies WHERE subject_id='".$id[0]."'";
$return = mysqli_query($mysqli, $query);
if(mysqli_num_rows($return) < 1){
$query = "DELETE FROM subjects WHERE subject_id='1'";
$return = mysqli_query($mysqli, $query);
}
This example assumes the "subject" is unique. In other words, SELECTing WHERE subject='test' will only ever return one subject_id. If you were doing this as a periodic cleaning, you would grab all the subject_id values (no WHERE clause) and loop through them to remove them if no replies.
You can achieve this in one query by selecting all (unique) subject-ids from the replies table, and delete all subjects that doesn't have a reply in there. Using SELECT DISTINCT, you don't get the IDs more than once (if a subject has more than one reply), so you don't get unnecessary data.
DELETE FROM subjects
WHERE subject_id NOT IN (SELECT DISTINCT subject_id FROM replies)
Any subject that doesn't have a reply should be deleted!
So you want to delete all subjects with no replies:
DELETE FROM subjects WHERE subject_id NOT IN
(SELECT subject_id FROM replies);
I think this is what you want...
i want to display record related to a specific primary key based on the foreign keys in other tables. How to display records for that primary key in other tables using php??
for example:
table1
primary key 1: plate#1
primary key 2: plate#2
primary key 3: plate#3
table2
primary key 1: destination|route|revenue|plate# 1
primary key 2: destination|route|revenue|plate# 3
table3
primary key 1: diesel price|number of liters|plate# 1
primary key 2: diesel price|number of liters|plate# 3
I already created a page that will display all the data in table1. I want to display the data in table1 and table2 that are related to the data in table1 when I made the database they already had relationship with each other. My problem is just displaying the record related to table1. I want to display records for just plate#1, another for plate#2 and so on.
Here's a crude example:
<?php
$truck_id = mysql_real_escape_string($_GET['truck_id']);
$sql_PK = "SELECT * FROM table1 WHERE id = '{$truck_id}'";
$res_PK = mysql_query($sql_PK);
$row_PK = mysql_fetch_assoc($res_PK);
$truck_plate = $row_PK['plate'];
$truck_plate = mysql_real_escape_string($truck_plate);
$sql = "SELECT table2.plate, table2.destination, table2.route, table3.plate, table3.diesel_price, table3.num_of_liters FROM table2, table3 WHERE table2.plate = table3.plate AND table2.plate = '{$truck_plate}'";
$res = mysql_query($sql) or die(mysql_error());
$row = mysql_fetch_array($res);
// this will give you the details from the following
//TABLE2
// plate
// destination
// route
//TABLE3
// plate
// diesel_price
// num_of_liters
?>
There's a couple of important things to remember. First is get in the habit of naming your fields without spaces in the database and use _ instead. so diesel price is diesel_price, etc. Secondly is to make sure that you are protecting yourself from any injections like I have shown here using the mysql_real_escape_string
So when someone clicks on truckdetails.php?plate=mrtchi it's going to query the database based on the plate number: mrtchi
You are probably looking to join tables to get the final results. Join allows you to essentially merge tables together and get a single result based on the selections. Google php mysql join table and look at some of the examples. Here's one for you: Join Tables
I have a table called tblActivities. There are two fields ID and Attendees.
ID Attendees
1 Jon Jhonson
2 Ive Iveson
Which PHP function or MySQL statement do I need to use to get to this result:
ID Attendees
1 Jon Jhonson, Ive Iveson, Adam Adamer
2 Ive Iveson
In other words, how can I add new data to existing data in my database?
You need something like:
UPDATE tblActivities
SET Attendees = CONCAT(Attendees, "Ive Iveson, Adam Adamer")
WHERE id = 1;
But maybe you should change the database layout. It is preferable to have only one value in one field. You could find more information under http://en.wikipedia.org/wiki/Database_normalization.
use mysql's update statement. If you want to exeute through php then
first get the existing value using select statement,
SELECT Attendees from <table-name> WHERE ID = 1
you will get attendees existing value, put it in a php variable, now concatenate your value..
and then update,
UPDATE <table_name>
SET Attendees=<new-value>
WHERE ID=1
you would have to use the php's mysql functions to run the above queries
I think you're better off restructuring. Make a table:
ID (primary key)
ACTIVITY_ID (foreign key to activity table)
ATTENDEE (foreign key to USERS table)
Then select everything from that event and concat in PHP.
$q = mysql_query( 'SELECT ATTENDEE FROM tblActivities '.
'WHERE ACTIVITY_ID = $activity_id' );
$attendees = array();
while( $row = mysql_fetch_assoc( $q ) )
{
$attendees[] = $row['attendee'];
}
$attendees =implode( ' ', $attendees );
I need to select category ids from my sql database.
I have a variable $product_id and for each product id there are three rows in a table that i need to select using PHP.
If I do "SELECT * FROM table_name WHERE product_id='$prodid'"; I only get the one on the top.
How can I select all three category_ids which contain the same product_id?
I suppose you are using PHP's mysql functions, is this correct? I am figuring that your query is actually returning all three rows but you aren't fetching all of them.
$sql = "SELECT * FROM table_name WHERE product_id='$prodid'";
$r = mysql_query($sql, $conn); //where $conn is your connection
$x = mysql_fetch_SOMETHING($r); //where something is array, assoc, object, etc.
The fetch function gives only one row at a time. You say you need three so it needs to be executed three times.
$x[0] = mysql_fetch_assoc($r);
$x[1] = mysql_fetch_assoc($r);
$x[2] = mysql_fetch_assoc($r);
OR this would be better
while($curRow = mysql_fetch_assoc($r)) //this returns false when its out of rows, returns false
{
$categoryIds[] = $curRow['category_id'];
}
If this doesn't do it then your query is actually returning only one row and we need to see your tables/fields and maybe sample data.
SQL seems to be correct, but Why do you store product_id in categories table? if it's one-to-many relation it would be better to store only category_id in products table.
The SQL query is correct for what you want to do. It will select all the records in table_name with the field product_id = $prodid (not only 1 or 3 but any that matches the variable)
To select a few records you should use the LIMIT keyword
You should look inside your table structure and the variable $prodid to find problems.
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.