问题描述
您好，如果数据库中的数据为NULL，我正在与mysqli一起替换表上的默认值.我已经在PHPmyAdmin上尝试过了，并且可以正常工作，但是在我的代码上却不行:(
Hi I am working with mysqli to replace a default value on the table if the data from the database is NULL. I already tried it on PHPmyAdmin and it's working but not on my code :(
这是我的SELECT查询:
Here's my SELECT query:
$query="SELECT pro_id, pro_name, unit_name, cat_name, IFNULL(quantity,'empty') AS quantity FROM products, unit, categories WHERE products.unit=unit.unit_id AND products.pro_cat=categories.cat_id";
推荐答案
如您的评论之一所示，如果得到的错误是:
If, as one of your comments seems to indicate, the error you're getting is:
Incorrect parameter count in the call to native function 'ISNULL'
那么这是一个简单的错字. ISNULL 与 IFNULL .
then it's a simple typo. ISNULL is not the same as IFNULL.
如果前者的 one 参数为null，则前者将返回真值.
The former returns a truth value if its one argument is null.
如果第一个为null，则后者返回第二个参数，否则返回第一个参数.
The latter returns the second argument if the first is null, otherwise it returns the first argument.
如果将以下代码放入SqlFiddle中，您将看到此消息:
You can see this if you put the following code into SqlFiddle:
-- DDL
create table xyzzy (plugh int);
insert into xyzzy (plugh) values (null);
insert into xyzzy (plugh) values (42);
select plugh, isnull(plugh) from xyzzy;
select plugh, ifnull(plugh,-1) from xyzzy;
select plugh, isnull(plugh,-1) from xyzzy;
前两个选择语句的输出与预期的一样，而第三个选择语句则产生您所描述的错误:
The output is as expected for the first two select statements while the third generates the error you describe:
plugh isnull(plugh)
------ -------------
(null) 1
42 0
plugh ifnull(plugh,-1)
------ ----------------
(null) -1
42 42
Incorrect parameter count in the call to native function 'isnull'
这篇关于IFNULL不起作用的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！