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如何在json数组中搜索值,然后在找到值时删除索引

uppy javascript

2

calyxofheld · 技术社区 · 4 年前

我得到了这个json字符串,我需要从中解析和删除数据,但我不确定如何处理它。假设我有以下json:

<input 
 name="uppyResult"
 type="hidden"
 value="[{
  &quot;successful&quot;:[
   {&quot;id&quot;:&quot;uppy-a8geo/fw/400x400/jpg-1d-2v-1e-image/jpeg-10097-1626922525568&quot;},
   {&quot;id&quot;:&quot;uppy-maxresdefault/jpg-1e-image/jpeg-81700-1626406845772&quot;}
  ],
  &quot;failed&quot;:[],
  &quot;uploadID&quot;:&quot;ckss6e4xv00023h63uov94n5e&quot;
 }]"
>

我得到的元素与 document.getElementsByName("uppyResult")[0].value; 然后我用 const obj = JSON.parse(json)

然后我如何删除只有索引在哪里 id: uppy-maxresdefault/jpg-1e-image/jpeg-81700-1626406845772 然后将其作为字符串重新插入DOM?

编辑:以前的版本有 " 而不是 " 在…内 value

2 回复 | 直到 4 年前

1

2

Yosvel Quintero 4 年前

你可以做:

const data = [{
  "successful":[
   {"id":"uppy-a8geo/fw/400x400/jpg-1d-2v-1e-image/jpeg-10097-1626922525568"},
   {"id":"uppy-maxresdefault/jpg-1e-image/jpeg-81700-1626406845772"}
  ],
  "failed":[],
  "uploadID":"ckss6e4xv00023h63uov94n5e"
 }]

const idToRemove = "uppy-maxresdefault/jpg-1e-image/jpeg-81700-1626406845772"
const result = data.map(obj => Object.entries(obj).reduce((a, [k, v]) => {
  a[k] = Array.isArray(v) 
    ? v.filter(item => item.id !== idToRemove) 
    : v
  return a
}, {}))

console.log(result)

2

1

Amir MB 4 年前

const data = [{
  "successful":[
   {"id":"uppy-a8geo/fw/400x400/jpg-1d-2v-1e-image/jpeg-10097-1626922525568"},
   {"id":"uppy-maxresdefault/jpg-1e-image/jpeg-81700-1626406845772"}
  ],
  "failed":[],
  "uploadID":"ckss6e4xv00023h63uov94n5e"
 }];
 
const toRemove = "uppy-maxresdefault/jpg-1e-image/jpeg-81700-1626406845772"; 
data.forEach(item => {
  Object.values(item).forEach(array => {
    if (!Array.isArray(array))
        return;
    const index = array.findIndex(elm => elm.id === toRemove);
    if (index > -1)
      array.splice(index, 1);
  });
});

console.log(data);

3

1

Tibrogargan 4 年前

您对要删除的内容不是特别清楚,但以下是两种可能性:

第一个过滤器使用分解并查找id匹配的任何元素。
第二个筛选器将删除成功数组中与搜索id匹配的所有条目。请注意,注意不要更改原始数据-复制数组中的每个对象并更改它们(不确定这是否重要)。

为测试目的向值数组添加了一个额外的元素。

let value='[{ \
  "successful":[ \
   {"id":"uppy-a8geo/fw/400x400/jpg-1d-2v-1e-image/jpeg-10097-1626922525568"}, \
   {"id":"uppy-maxresdefault/jpg-1e-image/jpeg-81700-1626406845772"} \
  ], \
  "failed":[], \
  "uploadID":"ckss6e4xv00023h63uov94n5e" \
 },{ \
  "successful":[ \
   {"id":"uppy-a8geo/fw/400x400/jpg-1d-2v-1e-image/jpeg-10097-1626922525568"}, \
   {"id":"not-uppy-maxresdefault/jpg-1e-image/jpeg-81700-1626406845772"} \
  ], \
  "failed":[], \
  "uploadID":"some_other_id" \
 }]'
 
let data = JSON.parse(value)
var filter_data = value => data.filter( ({successful}) => ! successful.find( ({id}) => id == value ))
var filter_successful = value => data.map( elem => {
  elem = Object.assign( {}, elem ) // copy element to avoid mutation
  elem.successful = elem.successful.filter( ({id}) => id != value )
  return elem
})
 
console.log('Remove any object from the values array where the search id is in the successful array')
console.log(filter_data('uppy-maxresdefault/jpg-1e-image/jpeg-81700-1626406845772'))
console.log('Remove successful entries that match the search id from all values')
console.log(filter_successful('uppy-a8geo/fw/400x400/jpg-1d-2v-1e-image/jpeg-10097-1626922525568'))

.reduce() 做到这一点,并创建副本,避免变异。同样,不清楚这是否是一项要求。