php 如何更新字段以将值添加到现有值？
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How to update field to add value to existing value?
提问by Jam Ville
How to update field to add value to existing value?
For example I have
如何更新字段以将值添加到现有值？
例如我有
Table name: table
表名： table
id credit
1 4
2 5
3 3
Is there a way to simply add value to the credit?
like
有没有办法简单地为信用增加价值？
喜欢
UPDATE table SET credit = '+7' WHERE id='1'
I want to add 7 to 4 so that the credit=11 where id='1'
How to do this?
我想将 7 添加到 4 以便 credit=11 where id='1'
如何做到这一点？
回答by MatthewMcGovern
UPDATE table SET credit = credit + 7 WHERE id = 1
回答by John Woo
This is just a simple UPDATE. Try the following.
这只是一个简单的UPDATE. 请尝试以下操作。
UPDATE tableName
SET Credit = Credit + 7
WHERE ID = 1
note that ID = 1and ID = '1'is the same as the server automatically parses it.
注意ID = 1和ID = '1'服务器自动解析它是一样的。
回答by hims056
Simply use credit = credit + 7instead of credit = '+7'in UPDATEstatement
简单地使用credit = credit + 7而不是credit = '+7'在UPDATE语句中
UPDATE tablename SET credit = credit + 7 WHERE id = 1
See this SQLFiddle
看到这个 SQLFiddle
回答by iLaYa ツ
Try this code
试试这个代码
UPDATE table SET credit = credit + 7 WHERE id='1'
回答by Gopesh Sharma
Just try this...
试试这个...
UPDATE table SET credit = credit + 7 WHERE id = 1
回答by Hussain Akhtar Wahid 'Ghouri'
well '+' is an operator so u need to provide the parameter it requires. '+' operator is an binary operator thus we need to provide two parameters to it with the syntax
'+' 是一个运算符，所以你需要提供它需要的参数。'+' 运算符是一个二元运算符，因此我们需要使用语法为其提供两个参数
value1+value2
though it may take parameters of many data types by writing '+7' you are only sending a String value "+7" replacing your previous value
虽然它可以通过写入 '+7' 来获取多种数据类型的参数，但您只会发送一个字符串值“+7”来替换您之前的值
so u better use
所以你最好用
UPDATE table SET credit = '+7' WHERE id='1'
don't confuse '+' operator with other increment operators
不要将“+”运算符与其他增量运算符混淆
回答by KisnardOnline
I wanted to add to this with an 'ON DUPLICATE KEY UPDATE' example(based on the answer by @hims056). I found this answer but needed 'ON DUP...' so I figured may as well post it here.
我想添加一个“ON DUPLICATE KEY UPDATE”示例（基于@hims056的答案）。我找到了这个答案，但需要“ON DUP...”所以我想不妨把它贴在这里。
INSERT INTO table1
(`id`, `credit`)
VALUES (1, 4)
ON DUPLICATE KEY UPDATE
`credit` = `credit` + 7;
