如何使用 java 將漂亮格式的 json 文件轉換為簡單的單行 json 文件？

Question

我需要將 Pretty 格式的 json 文件轉換為一行 json 文件。 但不能這樣做。 誰可以幫我這個事？

例子：下面需要轉換

[
 {
  "Employee ID": 1,
  "Name": "Abhishek",
  "Designation": "Software Engineer"
 },
 {
  "Employee ID": 2,
  "Name": "Garima",
  "Designation": "Email Marketing Specialist"
 }
]
[
   {
      "Employee ID": 1,
      "Name": "Abhishek",
      "Designation": "Software Engineer"
   },
   {
      "Employee ID": 2,
      "Name": "Garima",
      "Designation": "Email Marketing Specialist"
   }
]

對此：

 '[{"Employee ID":1,"Name":"Abhishek","Designation":"Software Engineer"},' \ 
            '{"Employee ID":2,"Name":"Garima","Designation":"Email Marketing Specialist"}]'

Answer 1

這個答案的一個更好、更完整的版本，帶有正確的錯誤處理和刪除額外的空格。

提到的答案的一個問題是它使用.concat()而沒有.trim() - 嵌套非常深的 pretty-JSON 的縮進將在最后的 output 中作為額外的空格可見。

String unprettyJSON = null;

try {
    unprettyJSON = Files.readAllLines(Paths.get("pretty.json"))
                        .stream()
                        .map(String::trim)
                        .reduce(String::concat)
                        .orElseThrow(FileNotFoundException::new);
} catch (IOException e) {
    e.printStackTrace();
}

Output 帶裝飾：

[{"Employee ID": 1,"Name": "Abhishek","Designation": "Software Engineer"},{"Employee ID": 2,"Name": "Garima","Designation": "Email Marketing Specialist"}][{"Employee ID": 1,"Name": "Abhishek","Designation": "Software Engineer"},{"Employee ID": 2,"Name": "Garima","Designation": "Email Marketing Specialist"}]

Output 無修剪：

[ {  "Employee ID": 1,  "Name": "Abhishek",  "Designation": "Software Engineer" }, {  "Employee ID": 2,  "Name": "Garima",  "Designation": "Email Marketing Specialist" }][   {      "Employee ID": 1,      "Name": "Abhishek",      "Designation": "Software Engineer"   },   {      "Employee ID": 2,      "Name": "Garima",      "Designation": "Email Marketing Specialist"   }]

Answer 2

如果您通過Files.readAllLines(Path)讀取文件內容，那么您已經將每一行作為列表中的不同項目獲取，您可以依次將其合並為一行。

例如：

String oneLine = Files.readAllLines(Paths.get("input.json"))
.stream()
.reduce(String::concat)
.orElseThrow();

System.out.println(oneLine);

Answer 3

try {
    unprettyJSON = Files.readAllLines(Paths.get("pretty.json"))
                        .stream()
                        .map(String::trim)
                        .reduce(String::concat)
                        .orElseThrow(FileNotFoundException::new);
} catch (IOException e) {
    e.printStackTrace();
}

上述解決方案效果很好。 感謝您發布它。 您可以從 a.json 文件中讀取，該文件轉換為單行字符串，也可用於 POST https 請求。