有条件加入MySQL的

我有这样的SQL语句有条件加入MySQL的

SELECT f.form_id 
    , f.tba_id 
    , f.patient_id 
    , f.location_code 
    , f.date_created 
    , p.lmp 
    , p.first_name 
    , p.last_name 
    , p.phone 
    , u.phone 
    FROM forms f 
    JOIN patients p 
    ON f.patient_id = p.id 
    JOIN users u 
    ON f.tba_id = u.id 
WHERE f.patient_id = 20

返回以下JSON

[ 
{ 
    "form_id": "11", 
    "tba_id": "10", 
    "patient_id": "20", 
    "location_code": "", 
    "date_created": "2014-08-30 16:46:16", 
    "lmp": "", 
    "first_name": "Mariam", 
    "last_name": "Oti", 
    "phone": "2348061356894" 
}, 
{ 
    "form_id": "12", 
    "tba_id": "10", 
    "patient_id": "20", 
    "location_code": "", 
    "date_created": "2014-08-30 17:02:26", 
    "lmp": "", 
    "first_name": "Mariam", 
    "last_name": "Oti", 
    "phone": "2348061356894" 
} 
]

我也想从f.form_id另一个表加入，但也并不总是对的记录对于每form_id即警报表，如果我执行语句

SELECT status 
    , cug_id 
    , alert_id 
    FROM alert 
WHERE form_id = 11

和

SELECT status 
    , cug_id 
    , alert_id 
    FROM alert 
WHERE form_id = 12

我可能会得到第一条语句的一条记录和第二条语句的0条记录。

我的目标是得到这样

[ 
{ 
    "form_id": "11", 
    "tba_id": "10", 
    "patient_id": "20", 
    "location_code": "", 
    "date_created": "2014-08-30 16:46:16", 
    "lmp": "", 
    "first_name": "Mariam", 
    "last_name": "Oti", 
    "phone": "2348061356894", 
    "status": "0", 
    "cug_id": "17", 
    "alert_id": "20" 
}, 
{ 
    "form_id": "12", 
    "tba_id": "10", 
    "patient_id": "20", 
    "location_code": "", 
    "date_created": "2014-08-30 17:02:26", 
    "lmp": "", 
    "first_name": "Mariam", 
    "last_name": "Oti", 
    "phone": "2348061356894", 
    "status": "", 
    "cug_id": "", 
    "alert_id": "" 
} 
]

结果所以，当我尝试做一个加入我最终得到一个记录，而不是

来源

2014-09-02 Amanni

见左[OUTER]加入 – Strawberry 2014-09-02 11:24:25

使用左连接的警报表

SELECT f.form_id 
    , f.tba_id 
    , f.patient_id 
    , f.location_code 
    , f.date_created 
    , p.lmp 
    , p.first_name 
    , p.last_name 
    , p.phone 
    , u.phone 
    , a.status 
    , a.cug_id 
    , a.alert_id 
    FROM forms f 
    JOIN patients p 
    ON f.patient_id = p.id 
    JOIN users u 
    ON f.tba_id = u.id 
    left join alert a 
    on a.form_id = f.form_id 
WHERE f.patient_id = 20

来源

2014-09-02 11:24:48 senK

有条件加入MySQL的

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