英文:
Add values from Array to every Object in Array
问题
以下是代码部分的翻译：
closed_status = [{"project_no": 5,"priority": 3,"s_status": "S8","project_Status: closed": "closed"},{"project_no": 8,"priority": 1,"s_status": "S5","project_Status: closed": "closed"},{"project_no": 12,"priority": 2,"s_status": "S2","project_Status: closed": "closed"}]str = ["Value 1","Value 2","Value 3",]let result = []closed_status.forEach((obj) => {str.forEach((newValue) => {result.push({ ...obj,newValue})})})
请注意，这只是代码的翻译部分，不包括问题和结果。如果您需要问题的翻译或结果的翻译，请提出明确的请求。
英文:
For a project i need to push the values of an array into every object in another array.
What i have so far:
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closed_status = [{"project_no": 5,"priority": 3,"s_status": "S8","project_Status: closed"},{"project_no": 8,"priority": 1,"s_status": "S5","project_Status: closed"},{"project_no": 12,"priority": 2,"s_status": "S2","project_Status: closed"}]str = ["Value 1","Value 2","Value 3",]let result = []closed_status.forEach((obj) => {str.forEach((newValue) => {result.push({ ...obj,newValue})})})
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...and this is the result i get
result = [{"project_no" : 5,"priority" : 3,"s_status": "S8","project_Status: closed","newValue": "Value 1"},{"project_no" : 5,"priority" : 3,"s_status": "S8","project_Status: closed","newValue": "Value 2"},{"project_no" : 5,"priority" : 3,"s_status": "S8","project_Status: closed","newValue": "Value 3"},{"project_no" : 8,"priority" : 1,"s_status": "S5","project_Status: closed","newValue": "Value 1",},{"project_no" : 8,"priority" : 1,"s_status": "S5","project_Status: closed","newValue": "Value 2",}]
...and so on...
But what i am trying to get is:
{"project_no" : 5,"priority" : 3,"s_status": "S8","project_Status: closed","newValue": "Value 1",},{"project_no" : 8,"priority" : 1,"s_status": "S5","project_Status: closed","newValue": "Value 2",},{"project_no" : 12,"priority" : 2,"s_status": "S2","project_Status: closed","newValue": "Value 3",},
Anyone knows where my mistake is?
Thanks in advance
答案1
得分: 1
以下是翻译好的部分：
假设closed_status和str的长度始终相同，从您的示例和期望结果来看，您可以执行以下操作。
for (let i = 0; i < closed_status.length; i++) {closed_status[i].newValue = str[i]}
英文:
Assuming, that closed_status and str always have the same length, which it looks like from your example and wanted result, then you could do the following.
for (let i = 0; i < closed_status.length; i++) {closed_status[i].newValue = str[i]}
答案2
得分: 0
我猜你想用字符串的值填充对象数组。
你可以通过跟踪索引来实现这个目标：
closed_status = [{"project_no": 5,"priority": 3,"s_status": "S8","project_Status": "closed",},{"project_no": 8,"priority": 1,"s_status": "S5","project_Status": 'closed'},{"project_no": 12,"priority": 2,"s_status": "S2","project_Status": "closed"}]str = ["Value 1","Value 2","Value 3",]closed_status.forEach((obj,index) => {obj.newValue = str[index]})console.log(closed_status)
希望这对你有所帮助。
英文:
I guess you want to fill the array of objects with the values of str.
You could do this by keeping track of the index:
<!-- begin snippet: js hide: false console: true babel: false -->
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closed_status = [{"project_no": 5,"priority": 3,"s_status": "S8","project_Status": "closed"},{"project_no": 8,"priority": 1,"s_status": "S5","project_Status": 'closed'},{"project_no": 12,"priority": 2,"s_status": "S2","project_Status": "closed"}]str = ["Value 1","Value 2","Value 3",]closed_status.forEach((obj,index) => {obj.newValue = str[index]})console.log(closed_status)
<!-- end snippet -->
答案3
得分: 0
如果你确定 str 和 closed_status 的项数相同，你可以使用 str.shift() 将单个值移出，然后将其扩展到原始对象中。
closed_status = closed_status.map(e => ({ ...e, newValue: str.shift() }) );console.log(closed_status);
英文:
If you're sure str and closed_status have the same amount of items, you could use str.shift() to shift out a single value that you spread into the original object.
<!-- begin snippet: js hide: false console: true babel: false -->
<!-- language: lang-js -->
closed_status = [{"project_no": 5,"priority": 3,"s_status": "S8","project_Status": "closed"},{"project_no": 8,"priority": 1,"s_status": "S5","project_Status": "closed"},{"project_no": 12,"priority": 2,"s_status": "S2","project_Status": "closed"}]str = ["Value 1","Value 2","Value 3",]closed_status = closed_status.map(e => ({ ...e, newValue: str.shift() }) );console.log(closed_status);
<!-- end snippet -->
答案4
得分: 0
以下解决方案假设第一个数组比第二个数组长。但如果长度始终相同，那么可以将 i < str.length ? i : i % str.length 替换为 i。
constclosed_status = [{"project_no": 5,"priority": 3,"s_status": "S8","project_Status": "closed"},{"project_no": 8,"priority": 1,"s_status": "S5","project_Status": "closed"},{"project_no": 12,"priority": 2,"s_status": "S2","project_Status": "closed"}],str = ["Value 1","Value 2","Value 3",],result = closed_status.map((status,i) => ({...status, newValue:str[i < str.length ? i : i % str.length]}));console.log( result );//alternatively, (EQUAL LENGTHS)console.log( str.map((newValue,i) => ({...closed_status[i],newValue})) );
英文:
The following solution assumes the first array is longer than the second. However, the lengths are always the same then i < str.length ? i : i % str.length can be replaced by i.
<!-- begin snippet: js hide: false console: true babel: false -->
<!-- language: lang-js -->
constclosed_status = [{"project_no": 5,"priority": 3,"s_status": "S8","project_Status": "closed"},{"project_no": 8,"priority": 1,"s_status": "S5","project_Status": "closed"},{"project_no": 12,"priority": 2,"s_status": "S2","project_Status": "closed"}],str = ["Value 1","Value 2","Value 3",],result = closed_status.map((status,i) => ({...status, newValue:str[i < str.length ? i : i % str.length]}));console.log( result );//alternatively, (EQUAL LENGTHS)console.log( str.map((newValue,i) => ({...closed_status[i],newValue})) );
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