Dein Code 1:1 kopiert funktioniert aber bei mir:
PHP-Code:
$Patientennummer = 'p0815';
$abfrage ="SELECT DISTINCT
PATIENT.p_number,
INSOLE.insole_1,
PATIENT.p_name,
PATIENT.p_vorname,
PATIENT.p_street,
PATIENT.p_postcode,
PATIENT.p_city,
PATIENT.p_country,
PATIENT.p_picture
FROM
PATIENT
CROSS JOIN INSOLE
WHERE
INSOLE.insole_1 LIKE '$Patientennummer'
AND
INSOLE.insole_1 IS NOT NULL
ORDER BY
PATIENT.p_number,
PATIENT.p_vorname,
PATIENT.p_name,
PATIENT.p_street,
PATIENT.p_postcode,
PATIENT.p_city,
PATIENT.p_country,
INSOLE.insole_1";
echo $abfrage;
Ausgabe:
Code:
SELECT DISTINCT PATIENT.p_number, INSOLE.insole_1, PATIENT.p_name, PATIENT.p_vorname, PATIENT.p_street, PATIENT.p_postcode, PATIENT.p_city, PATIENT.p_country, PATIENT.p_picture FROM PATIENT CROSS JOIN INSOLE WHERE INSOLE.insole_1 LIKE 'p0815' AND INSOLE.insole_1 IS NOT NULL ORDER BY PATIENT.p_number, PATIENT.p_vorname, PATIENT.p_name, PATIENT.p_street, PATIENT.p_postcode, PATIENT.p_city, PATIENT.p_country, INSOLE.insole_1
Ohne Fehlermeldung!