I just started with sqlite3 and was stuck on using (union) and combining multiple queries, is there a proper way of doing this
I am dealing with the following data
sql table =>
account |ledger_num | date_from | interest | name | loan | loan_type
____________|___________|_______________|_______________|___________|_________|____________________
Book_1 | HP2269 | 2017-03-31 | 15 | Samuel | 120 | type_a
Book_3 | HP2236 | 2020-03-31 | -20 | bob | 190 | type_a
Book_1 | HP2269 | 2019-04-01 | 250 | Samuel | 1500 | type_b
Book_3 | HP2236 | 2019-04-01 | -3 | bob | 36 | type_b
Book_2 | L01 | 2020-03-31 | 16 | josh | 95 | type_a
Book_1 | HP2269 | 2022-04-01 | 400 | Samuel | 4050 | type_c
and this is what I am trying to do
var sql_1 = `SELECT SUM(loan) ,SUM(interest) FROM allYearAllAcc WHERE (account = 'Book_1' AND date_from > '2018-03-31')`
var sql_2 = `SELECT SUM(interest) FROM allYearAllAcc WHERE (account = 'Book_2' AND date_from > '2017-04-01' AND interest < 0)`
var sql_3 = `SELECT SUM(loan) FROM allYearAllAcc WHERE (account = 'Book_3' AND date_from < '2022-04-01')`
temp_db.all(``+sql_1+`UNION`+sql_2+`UNION`+sql_3+``,(err,data)=>{
console.log(data)
})
this clearly isn't working I want this to appear as an json array which I have to use it in my code
Use conditional aggregation with a single query on your table:
SELECT
SUM(CASE WHEN account = 'Book_1' AND date_from > '2018-03-31'
THEN loan ELSE 0 END) AS loan1,
SUM(CASE WHEN account = 'Book_1' AND date_from > '2018-03-31'
THEN interest ELSE 0 END) AS interest1,
SUM(CASE WHEN account = 'Book_2' AND date_from > '2017-04-01' AND interest < 0
THEN interest ELSE 0 END) AS interest2,
SUM(CASE WHEN account = 'Book_3' AND date_from < '2022-04-01'
THEN loan ELSE 0 END) AS loan2
FROM allYearAllAcc;
I want to concatenate the bits of two number values like this:
+---------+---------+---------+-----------------+
| | val1 | val2 | concatenate val |
+---------+---------+---------+-----------------+
| decimal | 18 | 16 | 592 |
| hexa | 0x12 | 0x10 | 0x250 |
| binary | 0b10010 | 0b10000 | 0b1001010000 |
+---------+---------+---------+-----------------+
I have try to just concatenate with + like that :
const test = 0b10010, test1 = 0b10000
console.log(test+test1)//return 34
It does not concatenate the values but adds them together.
You could shift the first value by the bitwise length of the second value before adding.
const
add = (a, b) => (a << Math.ceil(Math.log2(b)) + 1) + b;
test = 0b10010,
test1 = 0b10000,
console.log(add(test, test1)); //
I watched a youtube video where this basic permutation tree was shown. If you look at this bit of code:
function recursion(input, set = [], result = []) {
if (!input.length) {
result.push([...set].join(''));
}
for (let i = 0; i < input.length; i++) {
const newArr = input.filter((n, index) => index !== i);
set.push(input[i]);
recursion(newArr, set, result);
set.pop();
}
return result.join(', ');
}
you can see that the base case (if statement) is at the top before the parameter nums is filtered. So, my whole question is how the tree and the code makes sense because for me the code would remove one too many digits from the set array. Becuase it pops an item of when returning and doesn't it return more than two times?
Does this log add clarity?
/ entering recursion with input = [1,2,3], set = [], result = []
| looping, i = 0
| adding 1 to set
| / entering recursion with input = [2,3], set = [1], result = []
| | looping, i = 0
| | adding 2 to set
| | / entering recursion with input = [3], set = [1,2], result = []
| | | looping, i = 0
| | | adding 3 to set
| | | / entering recursion with input = [], set = [1,2,3], result = []
| | | | adding 123 to result
| | | \ returning [123]
| | | removing 3 from set
| | \ returning [123]
| | removing 2 from set
| | looping, i = 1
| | adding 3 to set
| | / entering recursion with input = [2], set = [1,3], result = [123]
| | | looping, i = 0
| | | adding 2 to set
| | | / entering recursion with input = [], set = [1,3,2], result = [123]
| | | | adding 132 to result
| | | \ returning [123,132]
| | | removing 2 from set
| | \ returning [123,132]
| | removing 3 from set
| \ returning [123,132]
| removing 1 from set
| looping, i = 1
| adding 2 to set
| / entering recursion with input = [1,3], set = [2], result = [123,132]
| | looping, i = 0
| | adding 1 to set
| | / entering recursion with input = [3], set = [2,1], result = [123,132]
| | | looping, i = 0
| | | adding 3 to set
| | | / entering recursion with input = [], set = [2,1,3], result = [123,132]
| | | | adding 213 to result
| | | \ returning [123,132,213]
| | | removing 3 from set
| | \ returning [123,132,213]
| | removing 1 from set
| | looping, i = 1
| | adding 3 to set
| | / entering recursion with input = [1], set = [2,3], result = [123,132,213]
| | | looping, i = 0
| | | adding 1 to set
| | | / entering recursion with input = [], set = [2,3,1], result = [123,132,213]
| | | | adding 231 to result
| | | \ returning [123,132,213,231]
| | | removing 1 from set
| | \ returning [123,132,213,231]
| | removing 3 from set
| \ returning [123,132,213,231]
| removing 2 from set
| looping, i = 2
| adding 3 to set
| / entering recursion with input = [1,2], set = [3], result = [123,132,213,231]
| | looping, i = 0
| | adding 1 to set
| | / entering recursion with input = [2], set = [3,1], result = [123,132,213,231]
| | | looping, i = 0
| | | adding 2 to set
| | | / entering recursion with input = [], set = [3,1,2], result = [123,132,213,231]
| | | | adding 312 to result
| | | \ returning [123,132,213,231,312]
| | | removing 2 from set
| | \ returning [123,132,213,231,312]
| | removing 1 from set
| | looping, i = 1
| | adding 2 to set
| | / entering recursion with input = [1], set = [3,2], result = [123,132,213,231,312]
| | | looping, i = 0
| | | adding 1 to set
| | | / entering recursion with input = [], set = [3,2,1], result = [123,132,213,231,312]
| | | | adding 321 to result
| | | \ returning [123,132,213,231,312,321]
| | | removing 1 from set
| | \ returning [123,132,213,231,312,321]
| | removing 2 from set
| \ returning [123,132,213,231,312,321]
| removing 3 from set
\ returning [123,132,213,231,312,321]
You can see how I added the logging to your code in this snippet:
const log = (depth, message) =>
console .log ('| '.repeat (depth) + message)
function recursion(input, set = [], result = [], depth = 0) {
log (depth, `/ entering recursion with input = [${input}], set = [${set}], result = [${result}]`)
if (!input.length) {
log (depth, `| adding ${[...set].join('')} to result`)
result.push([...set].join(''));
}
for (let i = 0; i < input.length; i++) {
log (depth, `| looping, i = ${i}`)
const newArr = input.filter((n, index) => index !== i);
log (depth, `| adding ${input[i]} to set` )
set.push(input[i]);
recursion(newArr, set, result, depth + 1);
log (depth, `| removing ${input[i]} from set` )
set.pop();
}
log (depth, `\\ returning [${result}]`)
return result.join(', ');
}
console .log (recursion([1, 2, 3]))
.as-console-wrapper {min-height: 100% !important; top: 0}
(but the console output there is limited to the last 50 lines.)
I am running this query
SELECT distinct make, model FROM `used`order by make
in order to get all the available make / models in the database.
My end goal though is to edit the date using PHP and have this result:
var alfa-romeo= '<select name="stage_type">
<option value="Mito">Mito</option>
</select>';
var audi= '<select name="stage_type">
<option value="A4">A4</option>
<option value="A5">A5</option>
<option value="Allroad">Allroad</option>
</select>';
for all the results.
How can I do this?
Consider the following...
<?php
/*
DROP TABLE IF EXISTS cars;
CREATE TABLE cars
(car_id SERIAL PRIMARY KEY
,make VARCHAR(20) NOT NULL
,model VARCHAR(20) NOT NULL
,UNIQUE(make,model)
);
INSERT INTO cars (make,model) VALUES
('Alfa Romeo','Mito'),
('Audi','A4'),
('Audi','A5'),
('Audi','Allroad'),
('Audi','Q5'),
('Audi','S3'),
('Audi','SQ5'),
('Audi','TT'),
('BMW','1.14'),
('BMW','116'),
('BMW','320'),
('BMW','525'),
('BMW','X1'),
('BMW','X3'),
('BMW','X5'),
('Chevrolet','Aveo'),
('Chevrolet','Orlando'),
('Chevrolet','Spark'),
('Citroen','Berlingo'),
('Citroen','C-ELYSEE'),
('Citroen','C1'),
('Citroen','C3'),
('Citroen','C4'),
('Citroen','C4 Grand Picasso'),
('Citroen','C4 Picasso');
SELECT * FROM cars ORDER BY make,model;
+--------+------------+------------------+
| car_id | make | model |
+--------+------------+------------------+
| 1 | Alfa Romeo | Mito |
| 2 | Audi | A4 |
| 3 | Audi | A5 |
| 4 | Audi | Allroad |
| 5 | Audi | Q5 |
| 6 | Audi | S3 |
| 7 | Audi | SQ5 |
| 8 | Audi | TT |
| 9 | BMW | 1.14 |
| 10 | BMW | 116 |
| 11 | BMW | 320 |
| 12 | BMW | 525 |
| 13 | BMW | X1 |
| 14 | BMW | X3 |
| 15 | BMW | X5 |
| 16 | Chevrolet | Aveo |
| 17 | Chevrolet | Orlando |
| 18 | Chevrolet | Spark |
| 19 | Citroen | Berlingo |
| 20 | Citroen | C-ELYSEE |
| 21 | Citroen | C1 |
| 22 | Citroen | C3 |
| 23 | Citroen | C4 |
| 24 | Citroen | C4 Grand Picasso |
| 25 | Citroen | C4 Picasso |
+--------+------------+------------------+
25 rows in set (0.00 sec)
*/
require('path/to/connection/stateme.nts');
$query = "
SELECT * FROM cars ORDER BY make,model;
";
$result = mysqli_query($conn,$query);
$array = array();
while($row = mysqli_fetch_assoc($result)){
$array[] = $row;
}
$new_array = array();
foreach($array as $v){
$new_array[$v['make']][] = $v['model'];
}
print_r($new_array);
?>
Outputs:
Array
(
[Alfa Romeo] => Array
(
[0] => Mito
)
[Audi] => Array
(
[0] => A4
[1] => A5
[2] => Allroad
[3] => Q5
[4] => S3
[5] => SQ5
[6] => TT
)
[BMW] => Array
(
[0] => 1.14
[1] => 116
[2] => 320
[3] => 525
[4] => X1
[5] => X3
[6] => X5
)
[Chevrolet] => Array
(
[0] => Aveo
[1] => Orlando
[2] => Spark
)
[Citroen] => Array
(
[0] => Berlingo
[1] => C-ELYSEE
[2] => C1
[3] => C3
[4] => C4
[5] => C4 Grand Picasso
[6] => C4 Picasso
)
)