我有 10000 行的結果 json 文件。在一個數組對象中，有一些我需要刪除的不必要的 json 對象。我嘗試了很多方法，但它對我不起作用。附上json文件的那一行[ { "product_id": "easybridge", "errors": [] }, { "product_id": "learningstudio", "errors": [] }, { "product_id": "pearsontestprep", "errors": [] }, { "product_id": "productization", "errors": [] }, { "product_id": "equella", "errors": [ { "property": "instance.test_ids[1]", "message": "requires property \"maintenance\"", "schema": { "$id": "#/properties/test_ids/items", ], "properties": { "trend": { "$id": "#/properties/test_ids/items/properties/trend", "examples": [ true ] }, "display": { "$id": "#/properties/test_ids/items/properties/display", "type": "boolean", "examples": [ true ] }, "test_id": { "$id": "#/properties/test_ids/items/properties/test_id", "type": "string", }, "test_name": { "$id": "#/properties/test_ids/items/properties/test_name", "type": "string", }, "maintenance": { "$id": "#/properties/test_ids/items/properties/maintenance", "type": "boolean", ] }, "instance": { "trend": false, "display": false, "test_id": "8597ae3c-e2a9-45c7-b279-bde1710681be", "test_name": "Equella Pearsonresearch Ping Test", "nrAlertStatus": "enabled", "test_locations": [ { "alert_state": false, "location_name": "AWS_US_WEST_2", "location_label": "Portland, OR, USA", "included_to_health": false }如果錯誤 json 數組不為空。我什至不需要這個 json 我如何使用 java 腳本或 java 刪除“模式”json 對象和其他不必要的 json 對象和數組，特別是“模式”json 對象。請幫忙
3 回答
吃雞游戲
TA貢獻1829條經驗 獲得超7個贊
遍歷數組，查看每個對象，然后通過復制所需的數據來創建一個新數組。
例如，我認為如果對象的錯誤數組為空，則您不關心對象，并且您schema永遠不關心：
let newJSON = [];
//Assume the json variable is the parsed JSON file you posted.
for (let element of json) {
//Must have at least one error
if (element.errors.length > 0) {
//Create a new object
let newObj = {
"product_id" : element.product_id,
"errors" : []
};
//Add each errror
for (let error of element.errors) {
//Only copy across what we need
newObj.errors.push({
"property" : error.property,
"message" : error.message
});
}
//Add object to our new array of JSON
newJSON.push(newObj);
}
}
//newJSON is your processed JSON output
浮云間
TA貢獻1829條經驗 獲得超4個贊
最簡單的解決方案可以是：
const records = [{
"product_id": "learningstudio",
"errors": []
},
{
"product_id": "pearsontestprep",
"errors": []
},
{
"product_id": "equella",
"errors": [{
"property": "instance.test_ids[1]",
"message": "requires property \"maintenance\"",
"schema": {
"$id": "#/properties/test_ids/items",
}
}]
}];
const filteredRecords = records.map((record) => {
record.errors = record.errors.map((error) => {
return {property: error. property, message: error.message};
});
return record;
});
console.log(filteredRecords);
函數式編程
TA貢獻1807條經驗 獲得超9個贊
您可以使用映射和解構分配來僅捕獲所需的屬性
let json = [{"product_id": "equella", "errors": [{"property": "instance.test_ids[1]","message": "requires property \"maintenance\"",'xyz': 'not needed','useless': 'not needed',},{'xyz': 'not needed',}]},]
let op = json.map(({product_id,errors}) =>{
let { property, message } = errors[0]
return { product_id, errors: {property,message}}
})
console.log(op)
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