如何检查数组中是否存在值以及值的计数是否大于1？

Question

我有一个阵列

[
    {"name":"Ticket1","releases":[{"needToBeDeliver":true,"name":"release1","delivered":false},{"needToBeDeliver":true,"name":"release2","delivered":false}]},
    {"name":"Ticket2","releases":[{"needToBeDeliver":true,"name":"release1","delivered":false},{"needToBeDeliver":true,"name":"release2","delivered":false},{"needToBeDeliver":false,"name":"unplanned","delivered":false}]},
    {"name":"Ticket3","releases":[{"needToBeDeliver":false,"name":"release1","delivered":false},{"needToBeDeliver":false,"name":"unplanned","delivered":false}]},
    {"name":"Ticket4","releases":[{"needToBeDeliver":false,"name":"unplanned","delivered":false}]}
]

在上面的数组中，我必须检查releases数组是否包含“unplanned”条目和count（releases.needToBeDeliver == true）> 0，然后取消发布数组中的“unplanned”条目。

例如

• 在第一个索引中，它将保持原样，因为它不包含版本数组中的任何未计划条目
• 在第二个索引中，它包含未计划的条目，并且needToBeDeliver值多次为true，删除未计划的条目
• 在第三个索引中，它包含未计划的条目，但needToBeDeliver不等于true，不删除未计划的条目
• 在第四个索引中，它包含未计划的条目，但needToBeDeliver不正确，不删除未计划的条目

O / p应该遵循

[
    {"name":"Ticket1","releases":[{"needToBeDeliver":true,"name":"release1","delivered":false},{"needToBeDeliver":true,"name":"release2","delivered":false}]},
    {"name":"Ticket2","releases":[{"needToBeDeliver":true,"name":"release1","delivered":false},{"needToBeDeliver":true,"name":"release2","delivered":false}]},
    {"name":"Ticket3","releases":[{"needToBeDeliver":false,"name":"release1","delivered":false},{"needToBeDeliver":false,"name":"unplanned","delivered":false}]},
    {"name":"Ticket4","releases":[{"needToBeDeliver":false,"name":"unplanned","delivered":false}]}
]

到目前为止我尝试过的：

tickets.forEach(ticketsData => {
    var i = 0;
    ticketsData.releases.forEach(release => {
        if(release.needToBeDeliver === true){
            i++;
        }       
    });
});

但我没有得到如何在循环中添加第二个条件来检查每个索引的版本数组中是否存在计划外条目。 请帮我继续这个。

Answer 1

使用逻辑运算符可以非常轻松地检查任何语句中的第二个条件。

在您的示例中，只需检查

if(release.needToBeDeliver == true && release.name == "unplanned"){
            i++;
        }

将允许您“过滤掉”需要传递的元素并将其命名为“unplanned”。 希望，这就是你要找的东西。

Answer 2

你先得到计数，然后过滤releases 。

 var array = [{ name: "Ticket1", releases: [{ needToBeDeliver: true, name: "release1", delivered: false }, { needToBeDeliver: true, name: "release2", delivered: false }] }, { name: "Ticket2", releases: [{ needToBeDeliver: true, name: "release1", delivered: false }, { needToBeDeliver: true, name: "release2", delivered: false }, { needToBeDeliver: false, name: "unplanned", delivered: false }] }, { name: "Ticket3", releases: [{ needToBeDeliver: false, name: "release1", delivered: false }, { needToBeDeliver: false, name: "unplanned", delivered: false }] }, { name: "Ticket4", releases: [{ needToBeDeliver: false, name: "unplanned", delivered: false }] }]; array.forEach(function (o) { var count = o.releases.reduce((s, { needToBeDeliver }) => s + needToBeDeliver, 0); o.releases = o.releases.filter(a => !(a.name === 'unplanned' && count)); }); console.log(array); 

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Answer 3

要求“count（releases.needToBeDeliver == true）> 0”可以改为“有一个.needToBeDeliver == true的版本”，您可以使用.some ：

data.forEach(d => {
    if (d.releases.some(r => r.needToBeDeliver))
        d.releases = d.releases.filter(r => r.name !== 'unplanned')
});

Answer 4

我会做以下事情：

var result = tickets.map(ticket => {
  if (!ticket.releases.some(r => r.needToBeDeliver)) return ticket;
  return Object.assign(ticket, { releases: ticket.releases.filter(r => r.name !== 'unplanned') });
});

您检查是否有包含needToBeDeliver返回原始版本（如果该版本为false）。 如果为true，则过滤掉任何unplanned版本。