Question

给出一些背景- 我正在制作一个错误跟踪器应用程序，并使用MongoDB进行数据存储。我需要更新文档中的单个值-错误的状态。我的“ bug”对象存储在mongoDB中。目前，我正在通过其ID值检索文档/对象/错误。然后，我将通过JVM更新对象，并重新保存/覆盖整个原始文档/错误，以便它现在包含更新的状态。虽然这样做有效，但我不确定这是否是在MongoDB中更新数据的“专业”方式。预先感谢您的建议/意见。

@PutMapping("/{bugID}/{status}")
public Bug setBugStatus(@PathVariable String projectID, @PathVariable String bugID, @PathVariable String status) {
    return bugService.setBugStatus(projectID, bugID, status);
}

public Bug setBugStatus(String projectID, String bugID, String status) {
    Project project = getProjectByID(projectID);
    Bug bug = project.getBugQueue().findBugByID(bugID);
    
    if (status.equalsIgnoreCase("solved")) {
        bug.setStatusSolved();
        
    } else if (status.equalsIgnoreCase("inprogress")) {
        bug.setStatusInProgress();
        
    } else if (status.equalsIgnoreCase("unsolved")) {
        bug.setStatusUnsolved();
    }

    mongoTemplate.save(project);
    return bug;
}


private Project getProjectByID(String projectID) {
    return mongoTemplate.findById(projectID, Project.class);
}

这是文档的样子：

{
"projectID": "1",
"title": "Amazon",
"description": "An online ECommerce store.",
"dateCreated": "Mon Sep 14 01:18:40 IST 2020",
"timestamp": "14-09-2020 01:18:40",
"bugQueue": {
    "priorityQueue": [
        {
            "priority": 10,
            "bug": {
                "bugID": "1",
                "dateCreated": "Mon Sep 14 01:18:40 IST 2020",
                "dateResolved": null,
                "title": "Order processing failed.",
                "bugDescription": "Orders are not being processed after 9.32pm.",
                "status": "Unsolved",
                "discoveredBy": null,
                "solutionDescription": null,
                "bugProgressReportList": {
                    "reportsList": []
                },
                "bugAssignedToList": {
                    "employeeList": []
                },
                "priority": "Highest"
            }
        }
    ],
    "empty": false
},
"employeeList": {
    "employeeList": [
        {
            "employeeID": "2",
            "firstName": "Alex",
            "secondName": "Wands"
        },
        {
            "employeeID": "3",
            "firstName": "Alice",
            "secondName": "Black"
        }
    ]
}

}

这是访问，更新和保存mongoDB的正确方法吗？

0 个答案: