Mis*_*r_L 4 mongoose mongodb node.js
考虑这个命令:
WorkPlan.findOneAndUpdate({ _id: req.params.id }, updateObj, function(err) {
...
})
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与此相对:
WorkPlan.findOneAndUpdate({ _id: req.params.id }, { '$set': updateObj }, function(err) {
...
})
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在开发我的项目时,我惊讶地发现第一个命令的结果与第二个命令的结果相同:它updateObj被合并到数据库中的现有记录中,即使在第一个命令应该是代替它.这是mongoose/mongodb中的错误还是我做错了什么?如何在更新时替换对象而不是合并它?我正在使用mongoose 4.0.7.
谢谢.
==========
更新:
这是实际的WorkPlan架构定义:
workPlanSchema = mongoose.Schema({
planId: { type: String, required: true },
projectName: { type: String, required: true },
projectNumber: { type: String, required: false },
projectManagerName: { type: String, required: true },
clientPhoneNumber: { type: String, required: false },
clientEmail: { type: String, required: true },
projectEndShowDate: { type: Date, required: true },
segmentationsToDisplay: { type: [String], required: false },
areas: [
{
fatherArea: { type: mongoose.Schema.ObjectId, ref: 'Area' },
childAreas: [{ childId : { type: mongoose.Schema.ObjectId, ref: 'Area' }, status: { type: String, default: 'none' } }]
}
],
logoPositions: [
{
lat: { type: Number, required: true },
lng: { type: Number, required: true }
}
],
logoPath: { type: String, required: false },
}, { collection: 'workPlans' });
WorkPlan = mongoose.model('WorkPlan', workPlanSchema);
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这是一个例子updateObj:
var updateObj = {
projectManagerName: projectManagerName,
clientEmail: clientEmail,
clientPhoneNumber: clientPhoneNumber,
segmentationsToDisplay: segmentationsToDisplay ? segmentationsToDisplay.split(',') : []
}
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因此,当我不使用$ set标志时,我会期望该字段projectNumber,例如,不存在于新记录中,但我发现它仍然存在.
Mongoose update将所有顶级密钥视为$set操作(在较旧的文档中更明确:Mongoose 2.7.x更新文档).
为了获得所需的行为,您需要将该overwrite选项设置为true:
WorkPlan.findOneAndUpdate({ _id: req.params.id }, updateObj, { overwrite: true }, function(err) {
...
})
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