How to select data from columns that match a criteria? - php

Question

Hi i have a table in mysql database which has columns like UI_12-Apr-2016,DA_12-Apr-2016.
----------------------------------------------------------------------
| DA_12-Apr-2016 | UI_13-Apr-2016 | UI_12-Apr-2016 | DA_13-Apr-2016 |
|---------------------------------------------------------------------
| |
----------------------------------------------------------------------
How do i fetch data from the table whose columns have 12-Apr-2016 in it. Is there any way to select data according to the criteria. I know i can run this query:-
SELECT UI_12-Apr-2016,DA_12-Apr-2016 from table;
but the date and the code before it can be anything. I want to create a dynamic query to fetch data from the columns that match the date criteria.
I would be highly grateful if anyone can please provide a solution to do it.

Answer 1

This may help:
select COLUMN_name
from INFORMATION_SCHEMA.COLUMNS
where TABLE_NAME='Table1' and column_name like "%12-Apr-2016%";
in the above query, we can search for column names from INFORMATION_SCHEMA.COLUMNS table(which holds data related to table specified).
Output of above query should:
UI_12-Apr-2016
DA_12-Apr-2016
The result of above query is a list of columns with the pattern given(%12-Apr-2016%). You may save this result in a list and use it to fetch data accordingly from the table.
There is no direct method or query to do the same.

Answer 2

maybe you can create a new table tbl_column with one column column_name include all your columns~
SELECT * FROM tbl_column WHERE column_name LIKE '%12-Apr-2016%';
then you can use the query result to generate a new dynamic sql~

Answer 3

Did you mean:
SELECT * FROM table WHERE column LIKE '%13-Apr-2016%';
?
Update: you should look into information_schema database:
SELECT COLUMN_NAME from `COLUMNS` where `TABLE_NAME` = 'table_name' AND `COLUMN_NAME` LIKE '%12-Apr-2016%';
Simple PHP example script could be:
$table_name = 'table_name';
$col_pattern = '12-Apr-2016';
$mysqli = new mysqli($config['host'], $config['user'], $config['password'], $config['dbname']);
$sql1 = "SELECT COLUMN_NAME from `COLUMNS` where `TABLE_NAME` = '{$table_name}' AND `COLUMN_NAME` LIKE '%{$col_pattern}%'";
$res1 = $mysqli->query($sql1);
$acol = array();
while ($r1 = $res1->fetch_assoc()) {
$acol[] = $r1['COLUMN_NAME'];
}
if (!empty($acol)) {
$sql2 = 'SELECT ' . implode(', ', $acol) . ' FROM ' . $table_name;
$res2 = $mysqli->query($sql2);
while ($r2 = $res2->fetch_assoc()) {
echo var_export($r2, 1) . PHP_EOL;
}
}

Answer 4

You must put together your SQL statement dynamically. E.g. the following statement uses the Oracle view ALL_TAB_COLUMNS to produce a select column_name from table_name statement for every column that contains 12-Apr-2016:
SELECT 'SELECT ' || COLUMN_NAME || ' FROM ' || TABLE_NAME || ';'
FROM ALL_TAB_COLUMNS
WHERE TABLE_NAME = 'MY_TABLE'
AND COLUMN_NAME LIKE '%12-Apr-2016%'
This may not be exactly what you want, however, this illustrates the general idea.

Answer 5

It is not possible to dynamically come up with names of columns since MySQL does not have an eval() function. Since you are using PHP, it is preferable to construct the query from your application in order to achieve this.
If you really insist for a solution using MySQL, there is one using prepared statements that is not too hard to come up with as follows:
set #table='your_table_name';
select group_concat(column_name)
from information_schema.columns
where table_name='your_table_name'
and column_name like '%12_Apr_2016'
into #colnames;
set #construct= CONCAT('SELECT ', #colnames, ' FROM ', #table);
prepare query from #construct;
execute query;
SQL Fiddle for reference.
But overall, having columns named using dynamically generated DDL is not the best schema modeling practice.