CODEWITHSUNDEEP
mysql
Richard
Richard

Reputation: 89

Is there a way to identify those records not found within a where IN() statement?

From PHP Code $Lines is defined as a list of accessions e.g. 123,146,165,1546,455,155

plant table has sequential records with the highest idPlant (unique identifier) of say 1000.

My simple SQL Query:

SELECT * FROM plant WHERE `plant`.idPlant IN($Lines) order by plant.idPlant;

This brings back row data for '123,146,165' etc.

Is there away to be told that '1546' was not found? (and thus the user probably entered a typo, I can not use a 'confirm all numbers are below X' because in the real data the idPlant may not be sequential and the upper bound will increase during use).

Update:

Looking to get an output that will tell me what Numbers were not found.

Upvotes: 5

Views: 81

Answers (2)

Hristo Valkanov
Hristo Valkanov

Reputation: 1687

You can create a temporary table and compare it to the original table. It goes something like this:

CREATE TEMPORARY TABLE IF NOT EXISTS plantIDs  (
    ID INT(11) NOT NULL UNIQUE,
    found INT(11) NOT NULL);

INSERT INTO plantIDs(ID) VALUES (123),(146),(165),(1546),(455),(155);

SELECT plantIDs.ID, COALESCE(plant.name, "Not Found") as PlantName, plant.* FROM plant RIGHT JOIN plantIDs ON plant.idPlant=plantIDs.ID ORDER BY plantIDs.ID;

Assuming you have a field named name inside the table plant, this code will produce a row for each plant and the column named PlantName will contain the name of hte plant or the text "Not Found", ofc you can change the coalesce value to anything that fits your needs.

Upvotes: 0

Kickstart
Kickstart

Reputation: 21513

You can build up a sub query using unions that returns a list of all your values, then LEFT JOIN against that, checking for NULL in the WHERE clause to find the non matching values.

Basic php for this would be something like this:-

<?php

$sub_array = explode(',', $Lines);

$sub = '(SELECT '.implode(' AS i UNION SELECT ', $sub_array).' AS i) sub0';

$sql = "SELECT sub0.i 
    FROM $sub
    LEFT OUTER JOIN plant 
    ON  plant.idPlant = sub0.i 
    WHERE plant.idPlant IS NULL";

?>

Upvotes: 1

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