Substring reurns a number value instead of varchar - mysql

$query = mysql_query(SELECT * FROM test WHERE SUBSTRING(date,0,4)="1392") //for example 1392
echo query['name'];
i want to get 4 first character and check its equal with 1392 or not .
but its return 1054 my column name type is varchar and i have no idea about this.

MySQL count char from 1, not from 0 as php.
try
SUBSTRING(date,1,4)="1392"
or if you want the year
YEAR(date)="1392"
http://dev.mysql.com/doc/refman/5.5/en/string-functions.html#function_substring
EDIT test exemple
CREATE TABLE babak (name VARCHAR(20) NOT NULL, date VARCHAR(10) NOT NULL);
INSERT INTO babak SELECT 'test', '1391/11/11';
INSERT INTO babak SELECT 'correct', '1932/11/11';
SELECT * FROM babak WHERE YEAR(date) = 1932;
+---------+------------+
| name | date |
+---------+------------+
| correct | 1932/11/11 |
+---------+------------+
SELECT * FROM babak WHERE SUBSTRING(date, 1, 4) = 1932;
+---------+------------+
| name | date |
+---------+------------+
| correct | 1932/11/11 |
+---------+------------+

So your date is in varchar with the format yyyy/MM/dd, try this one:
SELECT *
FROM test
WHERE SUBSTRING(`date`,1,4) = '1932'
CLICK HERE FOR DEMO

mysql_query only returns a resource, you need to use that resource to fetch the result
$resource = mysql_query('SELECT * FROM test WHERE SUBSTRING(date,1,4)="1392"')
$result = mysql_fetch_assoc($resource);
echo $result['name'];

Related

MYSQL select function from field value?

I have a table and its data are mentioned below :
id | function
1 | current_date
2 | UUID()
3 | RAND()
Structure of the table is
id int, function varchar(50)
Query : select * from func_table;
My excepted result is
id | function
1 | 2020-08-24
2 | 70d6cffc-ae01-11ea-80ca-c11529136ae3630
3 | 0.982584554752
Thanks in advance.
You can use a giant case expression:
select (case when function = 'current_date' then cast(current_date as char)
when function = 'uuid()' then cast(uuid as char)
when function = 'rand()' then cast(rand as char)
end) as value
If you actually want to evaluate the function directly, then you probably have a problem with your data model. SQL does not directly support such functionality.

Using LIKE with prepared statement

I have a statement I wish to execute to find if a column containing a string contains a certain value.
+----+------+
| id | st |
+----+------+
| 0 | 2183 |
| 1 | 5820 |
| 2 | 2984 |
| ...| ... |
+----+------+
Say I wish to find all rows where st contains a 1, I would use these where conditions:
WHERE st LIKE "%1%"
OR st LIKE "1%"
OR st LIKE "1"
OR st LIKE "%1"
But how do I do this in a prepared statement?
$ps = $db->prepare("
SELECT id
FROM table
WHERE st LIKE "%:a%"
OR st LIKE ":a%"
OR st LIKE ":a"
OR st LIKE "%:a"
");
$ps->execute(array(
':a' => $var
));
This doesn't work evidently.
The % sign must be part of $var not part of the prepared Statement. also you only need %:a% it include all other parts of your where clause
$ps = $db->prepare("
SELECT id
FROM table
WHERE st LIKE :a
");
$ps->execute(array(
':a' => "%".$var."%"
));
ps = $db->prepare("
SELECT id
FROM table
WHERE st LIKE :a");
$ps->execute(array(
':a' => "%".$var."%"
));
Try above code.
And one more thing if you require rows which contains 1 in it,then there is no requirement of 4 like condition you only can achieve using '%1%'.
Hope this will help.

Choosing from multiple columns by column name stored in a column

I want to select from a table like this
|id|col_name|col1|col2|col3|col4|col5|...|col100|
| 1|col1 | 142| 241| 333| 417| 713|...| 125|
| 2|col5 | 927| 72| 403| 104| 136|...| 739|
| 3|col100 | 358| 842| 150| 125| 174|...| 103|
Select from column specified by col_name field. Something like
SELECT id,valueof(col_name) val FROM table1
which returns
|id|val|
| 1|142|
| 2|136|
| 3|103|
If you are using PHP (or modify the logic accordingly), you can do something like-
$colNames = array(col1, col5, col100); // or SELECT col_name FROM
table_name and store it in $colNames
foreach ($colNames as $val) {
$query = "SELECT $val FROM table_name WHERE col_name={$val}";
//execute this and store its result one-by-one into another array.
}
as I'm not sure if it could be done with single query.

mysql update table with values from another table?

I am trying to update my table 'supplier_stats' with the values from my other table 'supplier_change_request'.
My two tables look like the following:
Supplier_change_request
id | user_id | company_name | supplier_number
1 123 hewden V0001
Supplier_stats
Id | user_id | company_name | address | reference | supplier_number
1 123 pie n/a 12345 V0001
2 145 gates n/a 12345 V0002
Here is my MySQL:
$reference = '12345'
$query = "UPDATE supplier_stats
SET supplier_stats.company_name = (
SELECT supplier_change_request.company_name
FROM supplier_change_request
WHERE supplier_change_request.reference = '$reference' AND supplier_change_request.supplier_number = supplier_stats.supplier_number";
mysql_select_db('hewden1');
$retval = mysql_query( $query, $conn )
by my calculation this should be setting the value of company_name where supplier_number is 'V0001' in my table 'supplier_stats' to 'hewden'. However the company_name is not being updated.
Can someone please show me where I am going wrong? Thank you in advance
I think the syntax is a bit off in your query and that it should look like this (just the SQL, adapt to PHP as needed):
UPDATE supplier_stats ss
JOIN supplier_change_request scr ON scr.supplier_number = ss.supplier_number
SET ss.company_name = scr.company_name
WHERE ss.reference = '$reference'
The column reference pointed to the supplier_change_request in your sample query, but to supplier_stats in your sample data - I assumed the sample data was correct; change if not.
This query should change the company_name in supplier_stats from pie to hewden.

How to achieve default value if column value is NULL?

I want to retrieve some column values from table with these conditions.
If value is NULL (or) Empty String , return some user defined value
If not above condition , return it's value.
How can I figure it out ?
Here is my Table query..
CREATE TABLE AUCTION_CAR_BID(
bid_seq bigint NOT NULL AUTO_INCREMENT,
auction_car_seq bigint NOT NULL,
bid_group_seq bigint NOT NULL,
bid_price int DEFAULT 0 NOT NULL,
over_bid_price int DEFAULT -1 NOT NULL,
result_id int DEFAULT 0 NOT NULL,
remark varchar(500),
PRIMARY KEY (bid_seq))
ENGINE = InnoDB DEFAULT CHARACTER SET utf8;
Here is my efforted codes to get it..
SELECT
COALESCE(OVER_BID_PRICE, -1)
FROM
AUCTION_CAR_BID
WHERE
BID_SEQ = 2354435345;
Another :
SELECT
CASE
WHEN OVER_BID_PRICE IS NULL
OR TRIM(OVER_BID_PRICE) = '' THEN -1
ELSE OVER_BID_PRICE
END OVER_BID_PRICE
FROM
AUCTION_CAR_BID
WHERE
BID_SEQ = 2354435345;
But I always get empty String value(not -1) if given id is not in my table.
Any suggestions would be really appreciated !
If you write this:
SELECT
COALESCE(OVER_BID_PRICE, -1)
FROM
AUCTION_CAR_BID
WHERE
BID_SEQ = 2354435345;
The results can be two types.
First result: Your query no returns rows! Your WHERE condition is unsatisfact so you'll read NULL
Second result: Your query returns rows but the value of your field is NULL, your COALESCE works fine in this case
To resolve you can try this:
SELECT COALESCE(
(SELECT
COALESCE(OVER_BID_PRICE, -1)
FROM AUCTION_CAR_BID
WHERE BID_SEQ = 2354435345)
,-1);
Tell me if it's OK
How about this:
select
case when price is null or id <> 1
then -1
else price
end price
from mytable
DROP TABLE prices;
CREATE TABLE prices (price_id INT NOT NULL AUTO_INCREMENT PRIMARY KEY,price INT NULL);
INSERT INTO prices (price) VALUES (' '),(''),(NULL);
SELECT * FROM prices;
+----------+-------+
| price_id | price |
+----------+-------+
| 1 | 0 |
| 2 | 0 |
| 3 | NULL |
+----------+-------+
SELECT price_id,COALESCE(price,-1) price FROM prices;
+----------+-------+
| price_id | price |
+----------+-------+
| 1 | 0 |
| 2 | 0 |
| 3 | -1 |
+----------+-------+
If there's no row for USER_SEQ = 2354435345 in your table there's no row returned. But aggregate functions always return a row even if the result is empty :-)
SELECT
COALESCE(MIN(OVER_BID_PRICE), -1)
FROM
USER_PARAM
WHERE
USER_SEQ = 2354435345;