正如我在本文中所读到的那样Sequelize increment function returning error
我已经完成了下一个代码:
exports.create = function (req, res) {
models.sparepart_request.create({
codWO: req.body.codWO,
codSparePart: req.body.codSparePart,
quantity: req.body.quantity,
date_request: req.body.date_request,
codUser: req.body.codUser,
request_return: req.body.request_return,
received: req.body.received
}).then(function (item) {
return models.sparepart.decrement(
'stock', {
by: req.body.quantity,
where: {
codSparePart: req.body.codSparePart
}
}
);
}).then(function (item) {
res.status(200);
res.json(({
success: true,
message: 'Query successful',
data: item //This is null, I need the values from creating
}));
}).catch(function (error) {
logger.error(JSON.stringify(error));
res.json({
success: false,
message: 'Query not successful and error has occured cretaing',
error: error,
stackError: error.stack
});
return res.status(500);
});
问题如下:
我的JSON答案如下:
{
"success": true,
"message": "Query successful",
"data": [
[
null,
1
]
]
}
如何获取创建项的值?
注意:我已尝试进行交易但总是返回错误或停止并给我一个超时错误。我不知道如何与实例进行交易。
答案 0 :(得分:1)
您可以使用:
returning: true, // to get updated data back
要从更新,销毁和其他函数返回结果,它只返回受影响的行数,例如
return models.sparepart.decrement(
'stock', {
by: req.body.quantity,
where: {
codSparePart: req.body.codSparePart
}
returning: true, // to get updated data back
plain: true, // for plain object and not raw data
}
);