i need to update a lot of db values, so i guess it's better to use a sql statement, maybe creating and uploading a php file and running it from time to time.
in my db i have 3 related tables, let's say
tableA_label
tableB_image
tableC_text
the relations are as follows:
tableaA_label.ImageID refers to tableB_image.ID
tableB_image.TextID refers to tableC_text.ID
my goal is:
update tableA_label.Name
tableA_label.Name = tableC_text.title
where
tableC_text.ID = tableB_image.TextID
and
tableB_image.ID = tableA_label.ImageID
.....
how can accomplish this using an sql statement?
thank you for supporting
Try this query:
UPDATE tableA_label SET
tableA_label.Name = (SELECT TableC_text.title FROM TableC_text INNER
JOIN TableB_image ON TableB_image.TextID = TableC_text.ID
WHERE TableB_image.ID = tableA_label.imageID)
Related
I have almost thousands of data to display for my reports and it makes my browser lags due to the heavy data. I think that my query is the real problem. How can I optimized my query? is there something that I should add in my query?
I am using Xampp which supports PHP7.
SELECT
`payroll_billed_units`.`allotment_code`,
`payroll_billed_units`.`category_name`,
`payroll_billed_units`.`ntp_number`,
`payroll_billed_units`.`activity`,
`payroll_billed_units`.`regular_labor`,
`payroll_sub`.`block_number`,
(SELECT
GROUP_CONCAT(DISTINCT `lot_number` SEPARATOR ', ')
FROM
`payroll_billed_units` `lot_numbers`
WHERE
`lot_numbers`.`allotment_code` = `payroll_billed_units`.`allotment_code`
AND `lot_numbers`.`category_name` = `payroll_billed_units`.`category_name`
AND `lot_numbers`.`ntp_number` = `payroll_billed_units`.`ntp_number`
AND `lot_numbers`.`activity` = `payroll_billed_units`.`activity`) AS `lot_numbers`,
(SELECT
COUNT(`billed`.`ntp_id`)
FROM
`regular_ntp` `billed`
WHERE
`billed`.`allotment_code` = `payroll_billed_units`.`allotment_code`
AND `billed`.`category_name` = `payroll_billed_units`.`category_name`
AND `billed`.`ntp_number` = `payroll_billed_units`.`ntp_number`
AND `billed`.`activity` = `payroll_billed_units`.`activity`) AS `billed`,
(SELECT
COUNT(`approved`.`id`)
FROM
`payroll_billed_units` `approved`
WHERE
`approved`.`allotment_code` = `payroll_billed_units`.`allotment_code`
AND `approved`.`category_name` = `payroll_billed_units`.`category_name`
AND `approved`.`ntp_number` = `payroll_billed_units`.`ntp_number`
AND `approved`.`activity` = `payroll_billed_units`.`activity`) AS `approved`
FROM
`payroll_billed_units`
JOIN payroll_transaction ON payroll_billed_units.billing_number =
payroll_transaction.billing_number
JOIN payroll_sub ON payroll_transaction.billing_number =
payroll_sub.billing_number
WHERE payroll_billed_units.billing_date = '2019-02-13'
AND payroll_transaction.contractor_name = 'Roy Codal' GROUP BY allotment_code, category_name, activity
I was expecting that it will load or display all my data.
The biggest problem are the dependendt sub-selects, they are responsible for a bad performance. A sub-select will be executed for EVERY ROW of the outer query. And if you cascade subs-selects, you'll quickly have a query run forever.
If any of the parts would yield only 5 resultsets, 3 sub-select would mean that the database has to run 625 queries (5^4)!
Use JOINs.
Several of your tables need this 'composite' index:
INDEX(allotment_code, category_name, ntp_number, activity) -- in any order
payroll_transaction needs INDEX(contractor_name), though it may not get used.
payroll_billed_units needs INDEX(billing_date), though it may not get used.
For further discussion, please provide SHOW CREATE TABLE for each table and EXPLAIN SELECT ...
Use simply COUNT(*) instead of COUNT(foo). The latter checks the column for being not-NULL before including it. This is usually not needed. The reader is confused by thinking that there might be NULLs.
Your GROUP BY is improper because it is missing ntp_number. Read about the sql_mode of ONLY_FULL_GROUP_BY. I bring this up because you can almost get rid of some of those subqueries.
Another issue... Because of the "inflate-deflate" nature of JOIN with GROUP BY, the numbers may be inflated. I recommend you manually check the values of the COUNTs.
Heyy i think i need to do a query inside a query.
Here is what i tried to do(it seems to work but it takes forever to load) :
$query="SELECT * FROM Table1";
$results=mysql_query($query);
while($row=mysql_fetch_array($results)){
$bnr=$row['COL 1'];
$usg=$row['COL 19'];
echo"$bnr hat $usg <br /> ";
$sqlupdate="UPDATE TABLE_1
SET
Usg='$usg'
WHERE
bnr = '$bnr'";
mysql_query($sqlupdate);
}
Yes, updating many rows individually will take a long time regardless of whether you're doing it inside another query.
But certainly running two queries on the same database, at the same time will not help the situation at all!
That said, the good news is that in your example you're able to perform the update entirely on the server using an update with a join.
UPDATE TABLE_1 as TDest
INNER JOIN TABLE1 as TSrc
ON TDest.bnr = TSrc.`COL 1`
SET TDest.Usg = TSrc.`COL 19`
I think the MySQL query that you want is:
update tdest t1 join
tdest t2
on t1.bnr = t2.`Col 1`
set t1.usg = t2.`Col 19`;
Hi there i am working on PHP code that is selecting columns from two tables.
Here is my code:
$result2 = mysql_query("SELECT *
FROM `videos`, `m_subedvids`
WHERE `videos.approved`='yes' AND
`videos.user_id`='$subedFOR'
ORDER BY `videos.indexer`
DESC LIMIT $newVID");
while($row2 = mysql_fetch_array($result2))
{
$indexer = addslashes($row2['videos.indexer']);
$title_seo = addslashes($row2['videos.title_seo']);
$video_id = addslashes($row2['videos.video_id']);
$title = addslashes($row2['videos.title']);
$number_of_views = addslashes($row2['videos.number_of_views']);
$video_length = addslashes($row2['videos.video_length']);
}
When i try to print $indexer with echo $indexer; it's not giving me any results.
Where is my mistake in this code?
It seems to me like the key 'indexer' isn't in your results. It's hard to tell, since you haven't listed a definition for your table and you're using SELECT * so we can't see the names.
It makes the program easier to read later, if instead of SELECT *..., you use SELECT col1, col2, .... Yes, SELECT * will save you some typing right now, but you'll lose that time later when you or anyone else who works on your code has to check the table definition every time they work with that line of code.
So, try changing your query to explicitly select the columns you use. If it's an invalid column you'll get an error right away rather than this silent failure you're getting now, and you'll thank yourself later as well.
So long as videos.indexer is a unique field name among all tables used in the query you can change
$indexer = addslashes($row2['videos.indexer']);
to
$indexer = addslashes($row2['indexer']);
You don't need to (or can not) use the table name when referring to the result.
Problem
I have a table tbl_student_courses which join 2 tables student and courses now when data is inserted it is the combination of 2 ids course_id and student_id. I just want there would be no duplication of this combination in tbl_student_courses.
Code
foreach($_POST['sel_course'] as $val) {
$query_std_course = "
INSERT INTO
`tbl_student_courses`
SET
`course_id` = '".$val."',
`std_id` = '".$_POST['std']."',
WHERE NOT EXISTS (
SELECT * FROM `tbl_student_courses` WHERE course_id=$val AND std_id=$std
)";
}
Help
This query giving SQL syntax error.
Can any body help me?
Thanks in advance.
Probably you are missing quotes one inner query values.
You SQL query should look like this
$sql = "
INSERT INTO
`tbl_student_courses`
SET
`course_id` = '".$val."',
`std_id` = '".$_POST['std']."',
WHERE NOT EXISTS (
SELECT * FROM `tbl_student_courses` WHERE course_id='".$val."' AND std_id='".$std."'
)";
NOTE: Inserting in database not prepared statements like std_id = '".$_POST['std']."' is not of a good manner. Consider using PDO or filter data yourself, bec. this can be easily used for SQL Iinjection therefore it is potential security breach.
UPDATE: Try to use ON DUPLICATE KEY UPDATE or INSERT IGNORE INTO table.
You can find more information regarding your implementation - http://bogdan.org.ua/2007/10/18/mysql-insert-if-not-exists-syntax.html
And read about proposed implementation - http://dev.mysql.com/doc/refman/5.1/en/insert-on-duplicate.html
The SQL syntax you seek is the MERGE statement, or its equivalent on your platform
http://en.wikipedia.org/wiki/Merge_(SQL)
I have a tables like this
Results
-------
id - autoincrement value
TestCase - varchar
Verdict - varchar
AppID - varchar
TestCases
---------
id - autoincrementr value
TestCase - varchar
TestCase_container - varchar
Basically I am displaying the results in php code. while displaying the testcase, I am storing the testcase in a variable. in the while loop of mysql_query, I am creating another connection to DB and passing this variable to TestCases table to get the TestCase_Container assiciated with it.
This is a long way of doing this but I am unable to figure out proper direct SQL query using join or any other thing. Can someone point me in right direction please?
Thanks,
LIke this?
select r.id,r.TestCase,r.Verdict,r.AppId,tc.TestCase_container
from Results r,TestCases tc
where Results.TestCase=TestCases.TestCase
For DB normalization, results table must have testcase_id field instead of TestCase
Kind of hard to say, but I think what you're trying to do is a query like this maybe?
SELECT b.id AS result_id, a.TestCase, b.Verdict, b.AppID, a.id AS testcase_id, a.TestCase_container
FROM TestCases a
LEFT JOIN Results b
ON b.TestCase = a.TestCase;
Would probably be better to be joining on an indexed integer/id field, but that would work fine.
You could easily select all data from your tables by this SQL query:
SELECT Results.TestCase AS TestCase, Results.Verdict AS Verdict, Results.AppID AS AppID, TestCases.TestCase_container AS Container FROM Results JOIN TestCases ON Results.TestCase = TestCases.TestCase
After you should iterate getted array of values in any loop (for example while) like that:
$query = "SELECT Results.TestCase, Results.Verdict, Results.AppID, TestCases.TestCase_container FROM Results JOIN TestCases ON Results.TestCase = TestCases.TestCase";
$res = mysql_query($query) or die(mysql_error());
while ($row=mysql_fetch_array($res)) {
echo $row['TestCase'], ":", $row['Verdict'], ":", $row['AppID'], ":", $row['Container'], "\n";
}