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警告:SQLite3 :: query()[sqlite3.query]:2列的1个值

[英]Warning: SQLite3::query() [sqlite3.query]: 1 values for 2 columns

I am using an sqlite3 db in php wich i want to fill, i need to fill in a foreign key and the name of the subcategory, here's the code that i use. 我在PHP中使用sqlite3 db,我要填写,我需要填写外键和子类别的名称,这是我使用的代码。

var_dump('"'.$categoryRow.','.$subcategory.'"');
            $db->query('INSERT INTO Subcategories (fk_cat, subcat) VALUES ("'.$categoryRow.','.$subcategory.'")');
            $subCategoryRow = $db->lastInsertRowID();
            die();

But when i run this code, it returns: 但是,当我运行此代码时,它返回:

string(11) ""1,subcategory"" 
Warning: SQLite3::query() [sqlite3.query]: 1 values for 2 columns in /mysite/product.php on line 94

but when i log them i get 1 and ''subcategory' back. 但是当我登录它们时,我得到1和“子类别”。

Does anyone know what i am doing wrong? 有人知道我在做什么错吗?

your variables should be both surrounded by double quotes sql should be insert into (field1, field2) values ("value1","value2") 您的变量都应该用双引号括起来sql应该插入到(field1,field2)值(“ value1”,“ value2”)中

var_dump('"'.$categoryRow.','.$subcategory.'"');
            $db->query('INSERT INTO Subcategories (fk_cat, subcat) VALUES ("'.$categoryRow.'","'.$subcategory.'")');
            $subCategoryRow = $db->lastInsertRowID();
            die();

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