I have a set of joined tables that I am querying in the following way:
$query = $this->mysqli->query("
SELECT DISTINCT name, website, email
FROM sfe
INNER JOIN ef ON sfe.ID_SFE = ef.ID_SFE
INNER JOIN f ON f.ID_F = ef.ID_F
INNER JOIN ad ON ad.ID_SFE = ef.ID_SFE
WHERE name LIKE '%{$sanitized}%' OR
website LIKE '%{$sanitized}%' OR
business_name LIKE '%{$sanitized}%' OR
email LIKE '%{$sanitized}%'
");
where ID_SFE is the primary key of table sfe and also the foreign key of ef.
When I make this query, I then echo the list of results with the following:
while ($result = $query->fetch_object()) {
$query_result = $result->"name";
echo "$query_result"
}
Because now I would like to also find the value of ID_SFE inside the same while loop, I tried to add ID_SFE in the list of SELECT DISTINCT together with name, website, email, however, I get ERROR: There was a problem with the query.
How can I get the value of ID_SFE and store it to another variable inside the while loop?
Thanks
You can not simply add ID_SFE to the list of fields to retrieve, because such field exists in the ad and ef tables.
You will be able to add ad.ID_SFE and/or ad.ID_SFE fields -note you need to specify the table name when specifying the field, as the fields needs to be referenced unequivocally.
Related
I have a query where i have "TEST"."TABLE" LEFT JOINED to PUBLIC."SchemaKey". Now in my final select statement i have a case statement where i check if c."Type" = 'FOREIGN' then i want to grab a value from another table but the table name value i am using in that select statement is coming from the left joined table column value. I've tried multiple ways to get to work but i keep getting an error, although if i hard code the table name it seems to work. i need the table name to come from c."FullParentTableName". Is what i am trying to achieve possible in snowflake and is there a way to make this work ? any help would be appreciated !
SELECT
c."ParentColumn",
c."FullParentTableName",
a."new_value",
a."column_name"
CASE WHEN c."Type" = 'FOREIGN' THEN (SELECT "Name" FROM TABLE(c."FullParentTableName") WHERE "Id" = 'SOME_ID') ELSE null END "TestColumn" -- Need assistance on this line...
FROM "TEST"."TABLE" a
LEFT JOIN (
select s."Type", s."ParentSchema", s."ParentTable", s."ParentColumn", concat(s."ParentSchema",'.','"',s."ParentTable",'"') "FullParentTableName",s."ChildSchema", s."ChildTable", trim(s."ChildColumn",'"') "ChildColumn"
from PUBLIC."SchemaKey" as s
where s."Type" = 'FOREIGN'
and s."ChildTable" = 'SOMETABLENAME'
and "ChildSchema" = 'SOMESCHEMANAME'
) c
on a."column_name" = c."ChildColumn"
Thanks !
In Snowflake you cannot dynamically use the partial results as tables.
You can use a single bound value via identifier to bind a value to table name
But you could write a Snowflake Scripting but it would need to explicitly join the N tables. Thus if you N is fixed, you should just join those.
I am writing a SQL query and I have an array in one table and in that array I stores IDs and want to compare that array of IDs with another table to show data against that ids. when I run it gives me following error.
No database selected Select the default DB to be used by double-clicking.
Here is my query
select TagId
, Name
from ctrData2.Tag
Left
outer join ctrData2.CallDetail
On Tag.TagId = array(CallDetail.Tag)
where CallDetail.ContactId = 'f9d4787a-f1ac-41af-97d8-ea324daad018'
this is how I store IDs in array in Tag Column
here is the other table from where I want to show data against these ids
you can set the default schema by below query first
use ctrData2;
And run below query
select TagId, Name
from Tag, CallDetail
where CallDetail.ContactId = 'f9d4787a-f1ac-41af-97d8-ea324daad018' and LOCATE(Tag.TagId, CallDetail.Tag) > 0;
Hey I tried this thread but it doesn't work and i can't figure out why...
here's my SQL:
SELECT * FROM gone_items
LEFT JOIN items
ON gone_items.item_ID=items.ID
WHERE
gone_items.aus_ID='$ID'
ORDER BY items.name ASC
Now, I fetch that via PHP and have a $row and try another mysql to get the individual ID's of the gone_items table. But if i use $row['ID'] I get the ID of the items.ID not the one from gone_items.ID.
I tried setting the variable manually in the first query but it doesn't work.
I also tried this: MYSQL Left join A.table and b.table while retaining a.table id
Also didn't help me...
All I want is to retain the ID (Primary key) from the gone_items table..
Can anyone please tell me what I'm doing wrong ?
Love
Gram
EDIT
//Query for Joined infos
$sqlx="SELECT foto_res_ausgeliehene_geg.ID, foto_res_ausgeliehene_geg.aus_ID, foto_res_ausgeliehene_geg.geg_ID, foto_res_ausgeliehene_geg.zusaetzliches, foto_res_gegenstaende.ID, foto_res_gegenstaende.bezeichnung, foto_res_gegenstaende.seriennummer, foto_res_gegenstaende.interne_seriennummer, foto_res_gegenstaende.zusaetzliches FROM foto_res_ausgeliehene_geg
LEFT JOIN foto_res_gegenstaende
ON foto_res_ausgeliehene_geg.geg_ID=foto_res_gegenstaende.ID
WHERE
foto_res_ausgeliehene_geg.aus_ID='$ID'
ORDER BY foto_res_gegenstaende.bezeichnung ASC
";
$ergebnisx = mysqli_query($db,$sqlx);
while ($zeilex = mysqli_fetch_assoc($ergebnisx))
{
//Query for individual infos
$sqly="SELECT * FROM foto_res_ausgeliehene_geg
WHERE `geg_ID`='".$zeilex['ID']."'
AND `aus_ID`='$ID'
GROUP BY `geg_ID`
";
$ergebnisy = mysqli_query($db,$sqly);
while ($zeiley = mysqli_fetch_assoc($ergebnisy))
{};
Now I did select all items individually. The foto_res_ausgeliehene_geg.ID still merges with the foto_res_gegenstanede.ID due to the LEFT JOIN.
So if i access $zeilex['ID'] im getting the ID of foto_res_gegenstaende.ID.
Would it help if I rename the ID field in one of the tables into lets say item_ID ?
Thanks alot.
Love
Gram.
Instead of using select *, you should explicitly state what items you want to select. Else you can get conflicts with multiple id fields. In your case something like:
select gone_items.id, gone_items.column1, gone_items.column2, items.column1, items.column2
It is also considered good practice, to limit the amount of data there is being selected. But is meanwhile also a highly debateable what is the right way. Performance issue in using SELECT *?
WORKS!
I simply renamed one of the Primary ID keys to something else, in this case, one of them got ID -> item_ID. The other one still is ID that way the left join won't merge them.
yolo
EDIT
WORKING CODE
$sqlx="SELECT foto_res_ausgeliehene_geg.item_ID, foto_res_ausgeliehene_geg.aus_ID, foto_res_ausgeliehene_geg.geg_ID, foto_res_ausgeliehene_geg.zusaetzliches, foto_res_gegenstaende.ID, foto_res_gegenstaende.bezeichnung, foto_res_gegenstaende.seriennummer, foto_res_gegenstaende.interne_seriennummer, foto_res_gegenstaende.zusaetzliches FROM foto_res_ausgeliehene_geg
LEFT JOIN foto_res_gegenstaende
ON foto_res_ausgeliehene_geg.geg_ID=foto_res_gegenstaende.ID
WHERE
foto_res_ausgeliehene_geg.aus_ID='$ID'
ORDER BY foto_res_gegenstaende.bezeichnung ASC
";
$ergebnisx = mysqli_query($db,$sqlx);
while ($zeilex = mysqli_fetch_assoc($ergebnisx))
{
//Query for individual infos
$sqly="SELECT * FROM foto_res_ausgeliehene_geg
WHERE `item_ID`='".$zeilex['item_ID']."'
";
$ergebnisy = mysqli_query($db,$sqly);
while ($zeiley = mysqli_fetch_assoc($ergebnisy))
{
Let's say I have two tables, and both their primary identifiers use the name 'id'. If I want to perform a join with these two tables, how would I alias the id of the table that I want to join with the former table?
For example:
SELECT * FROM `sites_indexed` LEFT JOIN `individual_data` ON `sites_indexed`.`id` = `individual_data`.`site_id` WHERE `url` LIKE :url
Now, site_id is supposed to link up with sites_indexed.id. The actual id which represents the row for individual_data however has the same title as sites_indexed.
Personally, I like to just use the name id for everything, as it keeps things consistent. When scripting server-side however, it can make things confusing.
e.g.
$var = $result['id'];
Given the aforementioned query, wouldn't this confuse the interpreter?
Anyway, how is this accomplished?
Instead of selecting all fields with "SELECT *" you should explicitly name each field you need, aliasing them with AS as required. For example:
SELECT si.field1 as si_field1,
si.field2 as si_field2,
ind_data.field1 as ind_data_field1
FROM sites_indexed as si
LEFT JOIN individual_data as ind_data
ON si.id = ind_data.site_id
WHERE `url` LIKE :url
And then you can reference the aliased names in your result set.
This thread is old and i found because i had the same problem. Now i have a better solution.
The answer given by Paul McNett and antun forces you to list all fields but in some cases this is impossible (too much fields to list), so you can keep the * and alias only the fields you want (typically the fields that have the same name and will override each other).
Here's how :
SELECT *, t.myfield as myNewName
FROM table t ... continue your query
you can add as much aliases as you want by adding comas.
Using this expression you will get results with columns id (from table sites_indexed) and id2 (alias for column id from table individual_data)
SELECT t1 . *, t2 . * FROM sites_indexed t1
LEFT JOIN (select id as id2, other_field1, other_field2 FROM individual_data) t2 ON t1.id = t2.site_id WHERE your_statement
The problem is that you're using the * wildcard. If you explicitly list the column names in your query, you can give them aliases:
SELECT `sites_indexed`.`id` AS `sites_indexed_id`,
`individual_data`.`id` AS `individual_data_id`
FROM `sites_indexed`
LEFT JOIN `individual_data` ON `sites_indexed`.`id` = `individual_data`.`site_id`
WHERE `url` LIKE :url
Then you can reference them via the alias:
$var = $result['sites_indexed_id'];
$var_b = $result['individual_data_id'];
How I can select name of creator and editor from users table, creator and editor are different ids in same table, but user table is different table
Is it what you mean?
$sql = "
SELECT id,name
FROM users
WHERE users.id = editors_table.$editor_id
OR users.id = creators_table.$creator_id";
from what i understand, are you saying you have 3 tables - 1 with creator data, 1 with editor data, and a third that references a record in each of the tables using an id?
if so, you'll have to use JOINs to achieve what you want - something like:
SELECT id, name, editors_table.editor_id, creators_table.creator_id
FROM users
LEFT JOIN editors_table ON user.editor_id = editor_table.editor_id
LEFT JOIN creators_table ON user.creator_id = creator_table.creator_id
WHERE editor_table.editor_id = $editor_id_var
OR creator_table.creator_id = $creator_id_var
(you'll want to go through the query as I'm guessing here)