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How should I make a sql update query only IF a value of the table is equal to "something"?
I would not want to use case because I don't have any "else" statement and it is regulated by another simple value of the table, so there are no more cases.
EDIT: Since there is so much need to see one single line of code because certainly my question has no answer this way, I'll leave it here:
$query = "IF seen=1 UPDATE something SET other_thing = 100 WHERE yet_another_thing= 'outro' ";
You use a where statement:
update t
set foo = bar
where value = 'something';
Looking at everyone's answers, here is the code for YOU.
$query = "UPDATE something SET other_thing = 100 WHERE yet_another_thing= 'outro' AND seen = 1";
This is where use use a WHERE clause:
UPDATE
SomeTable
SET
field = 1234
WHERE
anotherField = 5678
UPDATE tablename SET updatevalue = 'value' WHERE avalue = 'something'
Related
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I am retrieving multiple records from a table in MySQL. Is there a way that I can retrieve one row which totals each of the columns from this query? I was hoping that there might be an SQL solution, as I'll be repeating the process for many tables as it's being used for a dynamic report.
If there is no SQL solution, is there a way to easily add together all the nested arrays inside of an array? This would achieve the same result, but in PHP.
I can't seem to find any information on either approach.
You can do this using using aggregation. You don't give much guidance on what your query looks like, but something like:
select sum(col1) as sumcol1, sum(col2) as sumcol2, . . .
from t;
Without the group by clause, this returns one row, which is a summary of everything in t.
I am not sure what you exactly want to achieve by reading your question as you did not provide any codes or data. Try the following:
SELECT id, col_1, col_2, col_3, col_4, sum(t.Col)
FROM
(SELECT id, col_1, col_2, col_3, col_4,
SUM(col_1+col_2+col_3+col_4)
AS Col
FROM datas
WHERE id = '1'
GROUP BY id)t
You can see the fiddle here
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I'm trying to replace the first instance of a NULl or ' ' value in a column with another value. But only the first instance and nothing else.
So far I've put this together:
UPDATE table_name SET column = CONCAT(REPLACE(LEFT(column , INSTR(column , '')), '', 'new_value'), SUBSTRING(column , INSTR(column , '') +1))
I could replace all the values but I don't want that:
UPDATE table_name SET column = REPLACE (column , 'old_value', 'new_value')
Just try using limit 1
UPDATE table_name SET column='new_value' WHERE column='' limit 1
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I have two queries in my PHP file:
$q1 = " INSERT INTO team_password ( team_name, password ) VALUES ( '". $team_name."', '". $password."' )";
$result = mysqli_query($con,$q1);
it is working fine.
$q2 = " INSERT INTO all_users ( user_name, team_name ) VALUES ('". $user_name."', '". $team_name."')";
$all_users_var = $mysqli_query($con,$q2);
But this query is not responding.
all_users have four columns user_name, team_name, longitude and latitude.
longitude and latitude have default have values NULL.
Any help will be appreciated.
I think the guilty here is the $ sign before your mysqli_query().
When you have a doubt, you should check your php_error.log :)
It is the $ sign before mysqli_query() which is creating problems, remove $ sign and execute the code
Try this:
$q2 = " INSERT INTO all_users ( user_name, team_name ) VALUES ('$user_name','$team_name')";
$all_users_var = mysqli_query($con,$q2);
or this one
$q2 = mysqli_query("INSERT INTO all_users(user_name, team_name) VALUES('$user_name','$team_name')") or die(mysqli_error());
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I'm trying to let users insert a subject to an image. I know I can't use a WHERE clause in combination with an INSERT INTO statement, and I know that I should use SELECT. I am very new to mysql and I didn't understand the results I tried to search on google. So I needed a more specific answer;)
$classed = mysql_query("INSERT INTO images (subject) VALUES ('$_POST[subject]') WHERE image_id='$_POST[id]'");
You can't use WHERE in your INSERT statement. Consider trying UPDATE if you want to update existing rows, or INSERT without the WHERE to insert new rows.
Don't use the mysql_query function, as it deprecated. Try using PDO or mysqli_query instead.
Don't ever use unfiltered input in your query.
if you already have image_id value then you should UPDATE your table not INSERT.
you should escape your variables before inserting them in your table. by
mysql_real_escape_string()
try this:
$subject = mysql_real_escape_string($_POST['subject']) ;
$id =mysql_real_escape_string($_POST['id'] ) ;
$classed = mysql_query("UPDATE images
SET subject = '".$subject."'
WHERE image_id='".$id."' ");
You cannot use INSERT in existing rows, try UPDATE the SQL will be something like
UPDATE images SET subject = '...' WHERE id=x;
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mysql_query("
INSERT INTO `LMS`.`Presentation`
('Pre_Name' ,'Path' ,'PLec_ID' ,'pdatein' ,'pdesc','PSems_ID')
values
('$fname','$newname','$com',NOW(),'$filedesc','$semes')"
) or die("failed");
Dear All,
I have a table named presentation and I am going to enter value to it, it is mentionable that $com and $sems are comboboxs value, but the query show failed, anyone could help please,
thanks in advance
You're using quotes when you should be using backticks:
mysql_query("INSERT INTO `LMS`.`Presentation` (`Pre_Name`, `Path`, ...
Or simply don't use any special character. The backtick is only necessary if you do something silly like use a reserved word as a column name and I would hope people would choose their column names to be more readable.
In other words, date and in and select are silly names for columns, you should be using expiry_date, isInLocation and selectionStatus.
change or die("failed") into or die(mysql_error()) and you'll know why.
btw, consider changing from mysql functions to mysqli functions. And use parameterized queries. Otherwise you will be open to SQL injection.
mysql_query("
INSERT INTO table
(column1, column2, column3, column4 .... columnX)
VALUES(column1Data,column2Data, column3Data, column4Data ... columnXdata)
") or die(mysql_error());
> rove on this example and if there isnt a alias for a table then could not use a alias.
everything hiddends on details..