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Help with SQL query

Thread began 12/07/2015 2:16 pm by iain331081 | Last modified 12/11/2015 7:03 pm by Ray Borduin | 565 views | 9 replies |

iain331081

Help with SQL query

I have a page on a site using the following query:

SELECT DISTINCT LodgeID, Lodge, GSG_URL, banner_image_new, Country, banner_image_new_title, photo_one, photo_one_title, Summary_2013, lodges.CountryID, UserIDIT, LodgeIDPL, UserIDPL
FROM lodges
INNER JOIN countries ON lodges.CountryID = countries.CountryID
INNER JOIN product_users ON lodges.LodgeID = product_users.LodgeIDIT
LEFT JOIN product_likes ON lodges.LodgeID = product_likes.LodgeIDPL
WHERE Product_Type = 'Property'
GROUP BY LodgeID

which lists all lodges in the lodges table which have a corresponding country in the countries table, and also a user in a corresponding product_users table. (lodges and products are the same thing) And also lists them if whether there is or is not a record in the product_likes table.

The product_likes table is a simple table that records when the user currently logged in 'likes' a lodge.

However, what I am trying to have is that page list all the lodges that the user currently logged in has not yet liked. i.e. when they like a lodge, that lodge no longer appears on that list, but moves to a list of liked lodges. (which I have working).

I have been trying things like:

$ParamUserID_WADAlodges= "-1";
if (isset($_SESSION["SecurityAssist_UserID"])) {
$ParamUserID_WADAlodges = $_SESSION["SecurityAssist_UserID"];
}
mysql_select_db($database_connSafari, $connSafari);
$query_WADAlodges = sprintf("SELECT DISTINCT LodgeID, Lodge, GSG_URL, banner_image_new, Country, banner_image_new_title, photo_one, photo_one_title, Summary_2013, lodges.CountryID, UserIDIT, LodgeIDPL, UserIDPL
FROM lodges
INNER JOIN countries ON lodges.CountryID = countries.CountryID
INNER JOIN product_users ON lodges.LodgeID = product_users.LodgeIDIT
LEFT JOIN product_likes ON lodges.LodgeID = product_likes.LodgeIDPL
WHERE Product_Type = 'Property' AND UserIDPL <> %s
GROUP BY LodgeID
", GetSQLValueString($ParamUserID2_WADAlodgelikes, "int"));

i.e. basically just adding the condition UserIDPL <> UserID of the currently logged in User - so it should only list lodges that the user hasn't yet liked.

But I can't see to get it to work.

Not sure I have explained that too well, but hopefully you get what I mean.

Any pointers much appreciated.

Let me know if anything isn't clear. I have attached a copy of the page as it currently is.

Thanks again.

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Ray BorduinWebAssist

SELECT DISTINCT LodgeID, Lodge, GSG_URL, banner_image_new, Country, banner_image_new_title, photo_one, photo_one_title, Summary_2013, lodges.CountryID, UserIDIT, LodgeIDPL, UserIDPL
FROM lodges
INNER JOIN countries ON lodges.CountryID = countries.CountryID
INNER JOIN product_users ON lodges.LodgeID = product_users.LodgeIDIT
LEFT JOIN product_likes ON lodges.LodgeID = product_likes.LodgeIDPL
WHERE Product_Type = 'Property' AND product_likes.UserIDPL = %s AND product_likes.LodgeIDPL IS NULL
GROUP BY LodgeID

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iain331081

Thanks Ray - that returns no results though.

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Ray BorduinWebAssist

How about:

SELECT DISTINCT LodgeID, Lodge, GSG_URL, banner_image_new, Country, banner_image_new_title, photo_one, photo_one_title, Summary_2013, lodges.CountryID, UserIDIT, LodgeIDPL, UserIDPL
FROM lodges
INNER JOIN countries ON lodges.CountryID = countries.CountryID
INNER JOIN product_users ON lodges.LodgeID = product_users.LodgeIDIT
LEFT JOIN product_likes ON lodges.LodgeID = product_likes.LodgeIDPL AND product_likes.UserIDPL = %s
WHERE Product_Type = 'Property' AND product_likes.LodgeIDPL IS NULL
GROUP BY LodgeID

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iain331081

That's returning all the records:

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Ray BorduinWebAssist

Your parameter name was wrong on line 111... you had:
GetSQLValueString($ParamUserID2_WADAlodgelikes, "int")

and it should be:
GetSQLValueString($ParamUserID2_WADAlodges, "int")

The query was correct

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iain331081

Thanks Ray - paying too much attention to the SQL bit and missed that.

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iain331081

Just a quick follow up to this one - I realised there were five pages where this is an issue. I have it correct on four, but can't seem to get it correct on the search results page. So the same idea - if someone connects to a product, that should no longer appear on the search results.

Details below...

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Ray BorduinWebAssist

Did you do this query by hand? You forgot the sprintf() function to replace the parameters, the code should be:

$query_WADAlodges = sprintf("SELECT DISTINCT LodgeID, Lodge, GSG_URL, banner_image_new, Country, banner_image_new_title, photo_one, photo_one_title, Summary_2013, lodges.CountryID, UserIDIT, LodgeIDPL, UserIDPL
FROM lodges
INNER JOIN countries ON lodges.CountryID = countries.CountryID
INNER JOIN product_users ON lodges.LodgeID = product_users.LodgeIDIT
LEFT JOIN product_likes ON lodges.LodgeID = product_likes.LodgeIDPL AND product_likes.UserIDPL = %s
WHERE Product_Type = 'Property' AND product_likes.LodgeIDPL IS NULL
", GetSQLValueString($ParamUserID_WADAlodges, "int"));

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iain331081

Thanks Ray - I did edit it by hand and missed that.

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