mysql distinct statement not working - php

$query = "SELECT DISTINCT t.task_id, e.url,e.error_id, p.page_rank, e.scan_status, e.url_is_route,e.url_is_route,e.forbidden_word,e.google_host_fail,e.google_cache_fail,e.google_title_fail,e.robots_noindex_nofollow,e.xrobots_noindex_nofollow,e.title_fetch_warn,e.h1_fail,e.h2_fail,e.h3_fail,e.h1_warn,e.h2_warn,e.h3_warn
FROM `error_report` AS e
INNER JOIN task_pages AS t ON t.task_id=e.task_id
INNER JOIN `pages` AS p ON p.page_id=t.page_id
WHERE t.task_id=" . $this->task_id ;
I want the query to be distinct only by t.task_id. The problem is that when I add more fields..the query isnt distinct anymore. Is there still a way to select distinct by t.task_id only?


instead of distinct use group by
$query = "SELECT t.task_id, e.url,e.error_id, p.page_rank, e.scan_status, e.url_is_route,e.url_is_route,e.forbidden_word,e.google_host_fail,e.google_cache_fail,e.google_title_fail,e.robots_noindex_nofollow,e.xrobots_noindex_nofollow,e.title_fetch_warn,e.h1_fail,e.h2_fail,e.h3_fail,e.h1_warn,e.h2_warn,e.h3_warn
FROM `error_report` AS e
INNER JOIN task_pages AS t ON t.task_id=e.task_id
INNER JOIN `pages` AS p ON p.page_id=t.page_id
WHERE t.task_id=" . $this->task_id
."group by t.task_id";


If you add more fields, you should use GROUP BY iirc.
Note that your extra fields should be aggregates, e.g. MIN, MAX, etc. MySQL is more forgiving and shows the first value of each field (though not reliably apparently).

Related

Joining the same table twice with different values

I'm trying to replace the value of "transfers.pickup_areas_group_id" and "transfers.drop_areas_group_id" with the values from the table "areas_group", using IDs
I'm using this query:
SELECT
transfers.id AS transfer_id,
transfers.name AS transfer_name,
transfers.pickup_areas_group_id AS transfer_pickup_areas_group_id,
transfers.drop_areas_group_id AS transfer_drop_areas_group_id,
transfers_pricing.vehicle_id AS vehicle_id,
transfers_pricing.date_start AS date_start,
transfers_pricing.date_end AS date_end,
transfers_pricing.price AS price
FROM transfers
INNER JOIN transfers_pricing ON transfers_pricing.transfer_id = transfers.id
I tried an extra INNER JOIN to replace the first value "transfers.pickup_areas_group_id", but I couldn't find a way to replace the second one "transfers.drop_areas_group_id"
I tried this query:
SELECT
transfers.id AS transfer_id,
transfers.name AS transfer_name,
transfers.pickup_areas_group_id AS transfer_pickup_areas_group_id,
areas_group.area_id AS pickup_area_ids,
transfers.drop_areas_group_id AS transfer_drop_areas_group_id,
transfers_pricing.vehicle_id AS vehicle_id,
transfers_pricing.date_start AS date_start,
transfers_pricing.date_end AS date_end,
transfers_pricing.price AS price
FROM transfers
INNER JOIN transfers_pricing ON transfers_pricing.transfer_id = transfers.id
INNER JOIN areas_group ON areas_group.id = transfers.pickup_areas_group_id
Thank you,

Basically, you need another join on areas_group; to disambiguate the two references to the same table, you need to use a table alias.
Actually, it is a good practice to use table aliases for all tables that come into play in the query: this makes the query shorter to read and write.
SELECT
t.id AS transfer_id,
t.name AS transfer_name,
t.pickup_areas_group_id AS transfer_pickup_areas_group_id,
ag1.area_id AS pickup_area_ids,
t.drop_areas_group_id AS transfer_drop_areas_group_id,
ag2.area_id AS drop_area_ids
tp.vehicle_id AS vehicle_id,
tp.date_start AS date_start,
tp.date_end AS date_end,
tp.price AS price
FROM transfers t
INNER JOIN transfers_pricing tp ON tp.transfer_id = t.id
INNER JOIN areas_group ag1 ON ag1.id = t.pickup_areas_group_id
INNER JOIN areas_group ag2 ON ag2.id = t.drop_areas_group_id

Select DISTINCT with multiple OR in sql query

I am trying to write a query with multiple OR and using DISTINICT for multiple column.But the result is NULL what i am missing in my query. Anyone can help me here please.
SELECT DISTINCT
P.user_p_id,
P.surl,
M.user_id,
M.user_p_id
FROM users P, page M
WHERE (P.user_p_id = '$urlid' OR P.surl = '$urlid' OR M.user_id = '$urlid' OR M.user_p_id = '$urlid')

You should do a proper join with an ON clause:
SELECT DISTINCT
P.user_p_id,
P.surl,
M.user_id,
M.user_p_id
FROM users P INNER JOIN page M
ON M.user_id = P.user_p_id
WHERE '$urlid' IN (P.user_p_id, P.surl, M.user_id, M.user_p_id)
I use ON M.user_id = P.user_p_id although it's not clear if you want these columns to be matched.
You may change the type of the JOIN to LEFT if this is what you need.

MYSQL/PHP: Concat returning to many fields on LEFT JOIN

I had a SELECT query with a LEFT JOIN working as desired. I then added one more table via a smilar LEFT JOIN and now I am getting a wierd result. Basically, for a group_concat where I was getting one item for every record, I am getting eight records. I don't see why this is happening because the join to the new table is analagous to several other joins that do not have this problem (that I have omitted from the example for clarity).
Here is the query that is fine:
$sql = "SELECT t.*,
group_concat(tf.todoid) as `tftodoid`,
group_concat(tf.id) as `tfid`,
group_concat(tf.filedescript) as `tffiledescript`,
group_concat(tf.filename) as `tffilename`,
group_concat(tf.founderid) as `tffounderid`,
group_concat(tf.ext) as `tfext`,
group_concat(tf.lasttouched) as `tilt`
FROM titles `t`
LEFT JOIN titlefiles `tf`
ON (tf.todoid = t.id AND tf.founderid = '$userid')
WHERE t.userid='$userid'
GROUP BY t.id";
And here is the query with the extra join that is now spilling out the multiple copies of the items:
$sql = "SELECT t.*,
group_concat(tf.todoid) as `tftodoid`,
group_concat(tf.id) as `tfid`,
group_concat(tf.filedescript) as `tffiledescript`,
group_concat(tf.filename) as `tffilename`,
group_concat(tf.founderid) as `tffounderid`,
group_concat(tf.ext) as `tfext`,
group_concat(tf.lasttouched) as `tilt`,
group_concat(s.id) as `stepid`,
group_concat(s.step) as `steps`
FROM titles `t`
LEFT JOIN titlefiles `tf`
ON (tf.titleid = t.id AND tf.founderid = '$userid')
LEFT JOIN steps `s`
ON s.titleid = t.id
WHERE t.userid='$userid'
GROUP BY t.id";
Here is an example of output in JSON showing the difference:
First query:
"tfid":"56,57,58,59,60,61,62,63,64,65,66,67,68,75,76,81"
Second query:
"tfid":"56,57,58,59,60,61,62,63,64,65,66,67,68,75,76,81,56,57,58,59,60,61,62,63,64,65,66,67,68,75,76,81,56,57,58,59,60,61,62,63,64,65,66,67,68,75,76,81,56,57,58,59,60,61,62,63,64,65,66,67,68,75,76,81,56,57,58,59,60,61,62,63,64,65,66,67,68,75,76,81,56,57,58,59,60,61,62,63,64,65,66,67,68,75,76,81,56,57,58,59,60,61,62,63,64,65,66,67,68,75,76,81,56,57,58,59,60,61,62,63,64,65,66,67,68,75,76,81,56,57,58,59,60,61,62,63,64,65,66,67,68,75,76,81",
I suspect the problem has something to do with the JOIN or with the Group By statements but I can't see how to fix.
How can I ensure that I get only one fileid per file as opposed to eight?

Alter the line as follows:
group_concat(DISTINCT tf.id) as `tfid`,
This then only gets you the unique ids.
If you want them ordered add:
group_concat(DISTINCT tf.id ORDER BY tf.id ASC) as `tfid`,

UNION SELECT tables with inner join and GET values with "ALIAS"

So I have 2 tables, Matches and Teams, what I want to do is get some values from the match and Inner join "Teams" to get the names of both teams and add them to a php array later on (getting it all in one sql)
Matches
- IDMatch
- IDLocalTeam
- IDVisitorTeam
- Time
- Half
- Stopped
Teams
- IDTeam
- name
What I have by now is
$query = "SELECT * FROM `Matches`
INNER JOIN `Teams` ON `Matches`.IDLocalTeam = `Teams`.IDTeam
UNION SELECT * FROM `Matches` INNER JOIN `Teams`
ON `Matches`.IDVisitorTeam = `Teams`.IDTeam
ORDER BY IDMatch DESC;";
If someone could help me it would be great! Thanks alot

Wouldn't it be easier to use INNER JOIN twice? It's definitely quicker way of doing joins anyway.
SELECT
*
FROM
`Matches` m
INNER JOIN `Teams` t1
ON m.IDLocalTeam = t1.IDTeam
INNER JOIN `Teams` t2
ON m.IDVisitorTeam = t2.IDTeam
ORDER BY
m.IDMatch DESC;
Also start using aliases instead of table names to identify fields in query, it will SQL much smaller.

PHP SQL Adding Multiple Selects

I am using the Sql query below:
$query = mysql_query("SELECT * FROM `venues`
LEFT JOIN `follows` USING (venue_id)
LEFT JOIN `stats` USING (venue_id, user_id)
WHERE follows.user_id = $userid");
The problem is that it's not showing some fields in the stats table.
So what I would thinking the problem might be (might be wrong), is that I need to select all the fields of that table?
If this is the case, is there a way of telling it to select * the fields for the 3 tables?
For example:
SELECT * FROM `venues`, SELECT * FROM `follows`, SELECT * FROM `stats` LEFT JOIN ....

use alias for each table then
select A., B., C.* from venues as A left join Follows as B.....

You can use alias for tables. Here's your request using those aliases:
$query = mysql_query("SELECT a.*, b.*, c.* FROM `venues` as b
LEFT JOIN `follows` as b USING (venue_id)
LEFT JOIN `stats` as c USING (venue_id, user_id)
WHERE follows.user_id = $userid");

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