表字段值和外鍵 - LOAD DATA INFILE

Question

我目前正在使用mysql LOAD DATA INFILE將csv值插入名爲test的表中。當事情變得更加複雜時，到目前爲止所有的東西都非常好。我有另一個表occupations其中包含occupation_id我正在使用test表中的外鍵。原始csv文件僅帶有以下字段First Name,Last Name,Age,Date Of Birth和Occupation（請參閱下面的示例值）。我想根據csv文本字段Occupation來計算occupation_id。這怎麼可能？在csv file

列標題各自的價值

+------------+-----------+-----+---------------+------------+ 
| First Name | Last Name | Age | Date of Birth | Occupation | 
+------------+-----------+-----+---------------+------------+ 
| Lionel  | Messi  | 27 | 6/24/1987  | Soccer  | 
| Michael | Jordan | 51 | 2/17/1963  | Basketball | 
| Lebron  | James  | 30 | 12/30/1984 | Actor  | 
+------------+-----------+-----+---------------+------------+ 

表occupation

+---------------+-----------------+ 
| occupation_id | occupation_name | 
+---------------+-----------------+ 
|    1 | Basketball  | 
|    2 | Soccer   | 
|    3 | Actor   | 
+---------------+-----------------+ 

結果CSV後插入到表test

+------------+-----------+-----+-------------+---------------+-----------------+ 
| first_name | last_name | age | dob  | occupation_id | occupation_name | 
+------------+-----------+-----+-------------+---------------+-----------------+ 
| Lionel  | Messi  | 27 | 1987-06-24 |    2 | Soccer   | 
| Michael | Jordan | 51 | 1963-02-17 |    1 | Basketball  | 
| Lebron  | James  | 30 | 1984-30-12 |    3 | Actor   | 
+------------+-----------+-----+-------------+---------------+-----------------+ 

PHP/SQL - 我的查詢到目前爲止

$db_insert = $db_con->prepare("LOAD DATA LOCAL INFILE '".$filename."' 
    INTO TABLE test FIELDS TERMINATED BY ',' 
    OPTIONALLY ENCLOSED BY '\"' 
    LINES TERMINATED BY '\r\n' 
    IGNORE 1 LINES 
    (@column1, @column2, @column3, @column4, @column5) 
    SET first_name=@column1, last_name=@column2, age=@column3, dob = STR_TO_DATE(@column4, '%m/%d/%Y'), occupation=@column5 
"); 
$db_insert->execute(); 

Answer 1

我不會在LOAD DATA聲明中試圖做到這一點。從理論上講，您可以在LOAD DATA聲明中執行子查詢來查找相應的occupation_id，但即使可以，也會損害批量加載的性能。

下面是它會怎麼看，但我希望表現是可怕的，如果你不是行的一個微不足道的數目裝載更多：

LOAD DATA LOCAL INFILE 't.csv' 
INTO TABLE test FIELDS TERMINATED BY ',' 
OPTIONALLY ENCLOSED BY '\"' 
LINES TERMINATED BY '\r\n' 
IGNORE 1 LINES 
(@column1, @column2, @column3, @column4, @column5) 
SET first_name=@column1, last_name=@column2, age=@column3, 
    dob = STR_TO_DATE(@column4, '%m/%d/%Y'), occupation=@column5, 
    occupation_id=(SELECT occupation_id FROM occupation WHERE occupation_name=@column5 LIMIT 1); 

相反，我會做LOAD DATA和離開occupation_id空。然後LOAD DATA完成後，運行UPDATE加入到其他表：

UPDATE test JOIN occupation ON test.occupation = occupation.occupation_name 
SET test.occupation_id = occupation.occupation_id; 

Answer 2

首先，我會幹掉領域的test.occupation_name

然後，您可以在兩個步驟來做到：

1. 按原樣將LOAD DATA加載到類似csv結構的表中。讓我們把它叫做test_csv，並使用與那些test
2. 兼容的字段名執行以下操作：

。

INSERT INTO test 
SELECT tc.first_name, tc.last_name, tc.age, tc.dob, o.occupation_id 
FROM test_csv tc 
JOIN occupation o ON (tc.occupation_name=o.occupation_name) 

您將結束表test引用職業表occupations

希望這有助於。