Am I blind, or what is wrong with my query?
select
STRCMP( message, 'LogMessage') = 1
from
LogEntries;
works fine. However
select
IF STRCMP( message, 'LogMessage') = 1 THEN 'bla' END IF
from
LogEntries;
returns:
You have an error in your SQL syntax; check the manual that
corresponds to your MySQL server version for the right syntax to use
near 'STRCMP( message, 'LogMessage') = 1 THEN 'bla' END IF from
LogEntries' at line 2
What is wrong with this statement?
You might want to use CASE WHEN:
SELECT
CASE WHEN STRCMP( message, 'LogMessage') = 1 THEN 'bla' END AS your_column
FROM
LogEntries;
when the condition is true it will return 'bla' otherwise, since there's no else part, it will return NULL.
Related
UPDATE table_one
SET user_accessible = 1
WHERE volume == 2
AND lesson_order == '04'
LIMIT 1
It's giving the following error:
You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near '= '2' AND lesson_order == '04') LIMIT 1' at line 1
The query looks correct to me :(
You made a mistake on your comparison. Use single = not ==
UPDATE table_one
SET user_accessible = 1
WHERE volume = 2
AND lesson_order = '04'
LIMIT 1
Good Evning My Brothers/Sisters.
I want to check whether a column is exist from select result.
I have tried some effort like below but instead it gives me an error.
SHOW COLUMNS
FROM (SELECT
hasil.skpd, hasil.kode_skpd, SUM(hasil.sudah_isi) AS sudah_isi, SUM(hasil.belum_isi) AS belum_isi
FROM
(
SELECT
a.id_pegawai, c.skpd, c.kode_skpd, a.id_satuan_organisasi, a.nama_pegawai, a.nip,
CASE
WHEN i.id_pegawai IS NULL THEN 0
ELSE 1
END AS sudah_isi,
CASE
WHEN i.id_pegawai IS NULL THEN 1
ELSE 0
END AS belum_isi
FROM
tbl_pegawai a
LEFT JOIN ref_skpd c ON a.id_satuan_organisasi = c.id_skpd
LEFT JOIN tbl_bapertarum i ON (a.id_pegawai = i.id_pegawai AND YEAR(i.tgl_lapor)='2015')
GROUP BY
a.id_pegawai
) hasil
WHERE
1 AND hasil.kode_skpd LIKE 'Badan Kepegawaian Daerah%'
GROUP BY
hasil.id_satuan_organisasi
ORDER BY
hasil.kode_skpd) WHERE Field = "kode_skpd"
This is the full error result:
#1064 - You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near '(SELECT
hasil.skpd, hasil.kode_skpd, SUM(hasil.sudah_isi) AS s' at line 2
How do i to find the existence of a column or columns from SELECT result ?
Thanks in advance Brothers/Sisters.
here's the generated query:
UPDATE namelist
SET 'submitterName' = 'Jim'
,'actorName' = 'dingle'
,'setYear' = '1103'
,'country' = 'tanata'
,'blink' = 'on'
,'crush' = 'on'
,'initialize' = 'on'
,'entered' = 'on'
,'stuck' = 'on'
,'catapult' = 'on'
,'ruck' = 'on'
WHERE id = 31
it generates this (less than helpful) error:
You have an error in your SQL syntax; check the manual that
corresponds to your MySQL server version for the right syntax to use
near (query snippet) at line 1
for the life of me, i can't spot where the problem is. yes, column names match, yes
TIA for helping out.
WR!
You have used single quotes before and after columns in the query, replace those single quotes with backquotes.
So the query like
UPDATE namelist
SET `submitterName`='Jim',
`actorName`='dingle',
`setYear`='1103',
`country`='tanata',
`blink`='on',
`crush`='on',
`initialize`='on',
`entered`='on',
`stuck`='on',
`catapult`='on',
`ruck`='on'
WHERE id=31;
user ` instead of '
like this
UPDATE namelist SET `submitterName`='Jim',`actorName`='dingle',`setYear`='1103',`country`='tanata',`blink`='on',`crush`='on',`initialize`='on',`entered`='on',`stuck`='on',`catapult`='on',`ruck`='on' WHERE id=31
Why Can't I use or why does it encounters an error. I'm trying to use a script as such as the one below. I'm a newbie when it comes to mysql and I tried creating scripts like this but I get the errr like the below
#1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '1 THEN dogs.dogcount = dogcount - dogdetails.total, dogdetails.statu' at line 4
Here's my script
UPDATE dogs
LEFT JOIN dogdetails ON dogs.username = dogdetails.username
SET dogs.dogcount =
CASE WHEN dogdetails.dogcount IS NOT 1 THEN dogs.dogcount = dogcount - dogdetails.total, dogdetails.status = 'Checked'
WHEN dogdetails.dogcount = 1 THEN dogs.pto = pto - dogdetails.total, dogdetails.status = 'Checked'
WHERE dogdetails.id=4
Is there somethng I'm missing or overlooked?
remember that you have to use the keyword 'end' after your case statements and to always return a value. The IS NOT keyword is also incorrect when comparing numbers. Here's a version of your query that should work:
UPDATE dogs
LEFT JOIN dogdetails ON dogs.username = dogdetails.username
SET dogs.dogcount = CASE WHEN dogdetails.dogcount != 1 THEN dogs.dogcount = dogcount-dogdetails.total else dogs.dogcount end,
dogs.pto= case WHEN dogdetails.dogcount = 1 THEN pto - dogdetails.total else pto end,
dogdetails.status = 'Checked'
WHERE dogdetails.id=4
I have this query:
$this->db->select('COUNT(tran_ID) AS count, CASE WHEN MONTH(days)>=4 THEN concat(YEAR(days), "-", YEAR(days)+1) ELSE concat(YEAR(days)-1, "-", YEAR(days)) END AS format_date,product_class AS saleCol');
I get syntax error on
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'WHEN MONTH(days)>=4 THEN concat(YEAR(days), `"-"`, YEAR(days)+1) ELSE concat(YEA' at line 1
SELECT COUNT(tran_ID) AS count, `CASE` WHEN MONTH(days)>=4 THEN concat(YEAR(days), `"-"`, YEAR(days)+1) ELSE concat(YEAR(days)-1, `"-"`, YEAR(days)) END AS format_date, `product_class` AS saleCol FROM (`transactions`) WHERE `trans_type` = 'E' AND `product_class` != '0' GROUP BY `format_date`, `product_class`
Try this query, added FALSE as the second parameter in select statement
$this->db->select(
'COUNT(tran_ID) AS count,
CASE WHEN MONTH(days)>=4 THEN concat(YEAR(days), "-", YEAR(days)+1)
ELSE concat(YEAR(days)-1, "-", YEAR(days))
END AS format_date,
product_class AS saleCol',FALSE);