javascript - 如何从具有值的 JSON 对象中获取索引？

Question

这是我的 JSON 字符串。

[{
    "name": "placeHolder",
    "section": "right"
}, {
    "name": "Overview",
    "section": "left"
}, {
    "name": "ByFunction",
    "section": "left"
}, {
    "name": "Time",
    "section": "left"
}, {
    "name": "allFit",
    "section": "left"
}, {
    "name": "allbMatches",
    "section": "left"
}, {
    "name": "allOffers",
    "section": "left"
}, {
    "name": "allInterests",
    "section": "left"
}, {
    "name": "allResponses",
    "section": "left"
}, {
    "name": "divChanged",
    "section": "right"
}]

现在，我有了值 allInterests，我想找出上述字符串中该对象的索引(本例中为“7”)。我尝试了以下代码，但它总是返回 -1。

var q = MY_JSON_STRING
console.log(q.indexOf( 'allInterests' ) );

Answer 1

您将不得不使用 Array.find 或 Array.filter 或 Array.forEach。

由于您的值是数组并且您需要元素的位置，因此您将不得不对其进行迭代。

数组查找

var data = [{"name":"placeHolder","section":"right"},{"name":"Overview","section":"left"},{"name":"ByFunction","section":"left"},{"name":"Time","section":"left"},{"name":"allFit","section":"left"},{"name":"allbMatches","section":"left"},{"name":"allOffers","section":"left"},{"name":"allInterests","section":"left"},{"name":"allResponses","section":"left"},{"name":"divChanged","section":"right"}];
var index = -1;
var val = "allInterests"
var filteredObj = data.find(function(item, i){
  if(item.name === val){
    index = i;
    return i;
  }
});

console.log(index, filteredObj);

Array.findIndex() @ Ted Hopp的建议

var data = [{"name":"placeHolder","section":"right"},{"name":"Overview","section":"left"},{"name":"ByFunction","section":"left"},{"name":"Time","section":"left"},{"name":"allFit","section":"left"},{"name":"allbMatches","section":"left"},{"name":"allOffers","section":"left"},{"name":"allInterests","section":"left"},{"name":"allResponses","section":"left"},{"name":"divChanged","section":"right"}];

var val = "allInterests"
var index = data.findIndex(function(item, i){
  return item.name === val
});

console.log(index);

默认 Array.indexOf() 会将 searchValue 与当前元素匹配，而不是其属性。可以引用Array.indexOf - polyfill在 MDN 上