这不是在数据库中存储信息的好方法。但我马上就会谈到这一点。要直接回答您的问题,您可以将其用作 SQL 查询:
UPDATE TEST SET File = CONCAT(File, '$dbfilename')
WHERE id='$p_id'
AND File NOT LIKE '%$dbfilename%'
AND Lingua='$linguadilavoro'
pineapple.jpg
但是,当您尝试添加一个文件时,这可能会导致一些问题another-pineapple.jpg
真的,我认为您应该考虑一下这是一种非常糟糕的数据库方法。考虑将文件分成第二个表。例如:
# a table for the fruit names
CREATE TABLE fruits (
id INT UNSIGNED NOT NULL PRIMARY KEY AUTO_INCREMENT,
name VARCHAR(250) NOT NULL,
UNIQUE INDEX(name)
);
# a table for file names
CREATE TABLE files (
fileid INT UNSIGNED NOT NULL DEFAULT PRIMARY KEY AUTO_INCREMENT,
fruitid INT UNSIGNED,
filename VARCHAR(250),
UNIQUE INDEX(fruitid, filename)
);
# find all of the fruits with their associated files
SELECT fruits.id, fruits.name, files.filename
FROM fruits LEFT JOIN files ON fruits.id=files.fruitid
# add a file to a fruit
INSERT INTO files (fruitid, filename)
VALUES ('$fruitID', '$filename')
ON DUPLICATE KEY UPDATE fruitid=LAST_INSERT_ID(id)