Query che non restituisce risultati

Ciao a tutti

qualcuno mi spiega questo mistero,
questa query non restituisce niente nonostante il valore Linda Blair (I), Linda Denise Blair e' presente nella tabella actor.performer_aka :roll::roll:


https://www.upload.ee/image/13419231/sshot-1.png


:shock::shock:

Codice PHP:

---

SET @variable = 'Linda Denise Blair';
SELECT
actor.actor_id,
actor.nome,
actor.performer_aka,
ptgweb.alias,
ptgweb2.websites
FROM
ptgweb
INNER JOIN actor ON ptgweb.actor_id = actor.actor_id
INNER JOIN ptgweb2 ON actor.actor_id = ptgweb2.actor_id
WHERE
actor.performer_aka LIKE concat ('%',@variable,'%')
OR
ptgweb.alias LIKE concat ('%',@variable,'%')
OR
ptgweb2.websites LIKE concat ('%',@variable,'%')


---

https://www.upload.ee/image/13419250/sshot-2.png

risolto :=):

la parola magica e' LEFT JOIN :wink:

Codice PHP:

---


SET @variable = 'Linda Denise Blair';
SELECT
actor.actor_id,
actor.nome,
actor.performer_aka,
ptgweb.alias,
ptgweb2.websites
FROM
actor
LEFT JOIN ptgweb ON actor.actor_id = ptgweb.actor_id
LEFT JOIN ptgweb2 ON actor.actor_id = ptgweb2.actor_id
WHERE
actor.performer_aka LIKE concat ('%',@variable,'%')
OR ptgweb.alias LIKE concat ('%',@variable,'%')
OR ptgweb2.websites LIKE concat ('%',@variable,'%')


---

https://www.upload.ee/image/13420516/sshot-1.png

caspita, pensavo di avere risolto ed invece altri problemi:incavolat

neella ricerca il secondo risultato appare senza id (actor_id)

https://www.upload.ee/image/13420559/sshot-2.png

non capisco che succede :roll:

ecco la query

Codice PHP:

---

<?php
$_POST['input_nome']=trim($_POST['input_nome']); // funzione per eliminare spazi vuoti
$value = $_POST["input_nome"];
//$query = mysql_query ("SELECT nome FROM attrici WHERE nome = '".$value."'", $db);
$query = mysql_query ("
SELECT
actor.*,
ptgweb.*,
ptgweb2.*
FROM actor
LEFT JOIN ptgweb ON actor.actor_id = ptgweb.actor_id
LEFT JOIN ptgweb2 ON actor.actor_id = ptgweb2.actor_id
WHERE
(nome LIKE '%" . $value . "%')
OR (performer_aka LIKE '%" . $value . "%')
OR (career_status LIKE '%" . $value . "%')
OR (start LIKE '%" . $value . "%')
OR (end LIKE '%" . $value . "%')
OR (ptgweb.alias LIKE '%" . $value . "%')
OR (ptgweb2.websites LIKE '%" . $value . "%')
", $db);

?>

---

risolto :=): l'errore era in questo punto

Codice PHP:

---

ptgweb.*,
ptgweb2.*


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siccome anche in queste 2 tabelle era presente il campo actor_id la query entrava in conflitto

risolto cosi :wink:

Codice PHP:

---

ptgweb.alias,
ptgweb2.websites


---

Citazione:

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NOTA: La ricerca FULLTEXT e' abilitata in certi campi

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query completa funzionate,

Codice PHP:

---

SELECT
actor.*,
ptgweb.alias,
ptgweb2.websites
FROM actor
LEFT JOIN ptgweb ON actor.actor_id = ptgweb.actor_id
LEFT JOIN ptgweb2 ON actor.actor_id = ptgweb2.actor_id
WHERE
actor.nome LIKE ('%" . $value . "%')
OR actor.performer_aka LIKE ('%" . $value . "%')
OR ptgweb.alias LIKE ('%" . $value . "%')
OR ptgweb2.websites LIKE ('%" . $value . "%')
OR actor.career_status LIKE ('%" . $value . "%')
OR actor.start LIKE ('%" . $value . "%')
OR actor.end LIKE ('%" . $value . "%')


---