TypeScript中如何将JSON对象数组里的空字符串替换为null

问题场景 #

给定以下JSON数组（注：原JSON存在语法错误，第二个对象与第三个对象之间缺少逗号，需修正后再处理）：

[{"test":"a","test1":"1","test2":""},{"test":"b","test1":"","test2":"hi"},{"test":"","test1":"3","test2":""}]

需要将所有值为空字符串""的字段替换为null，得到目标结果：

[{"test":"a","test1":"1","test2":null},{"test":"b","test1":null,"test2":"hi"},{"test":null,"test1":"3","test2":null}]

解决方案示例 #

1. JavaScript实现

遍历数组中的每个对象，逐一检查键值对，将空字符串替换为null：

const originalJson = '[{"test":"a","test1":"1","test2":""},{"test":"b","test1":"","test2":"hi"},{"test":"","test1":"3","test2":""}]';
// 解析JSON为JS对象
const data = JSON.parse(originalJson);

// 处理数据
const processedData = data.map(obj => {
  const newObj = {...obj};
  for (const key in newObj) {
    if (newObj[key] === "") {
      newObj[key] = null;
    }
  }
  return newObj;
});

// 转换回JSON字符串
const resultJson = JSON.stringify(processedData);
console.log(resultJson);

2. Python实现

通过嵌套遍历处理每个对象的键值对：

import json

original_json = '[{"test":"a","test1":"1","test2":""},{"test":"b","test1":"","test2":"hi"},{"test":"","test1":"3","test2":""}]'
data = json.loads(original_json)

processed_data = []
for obj in data:
    new_obj = {}
    for key, value in obj.items():
        new_obj[key] = None if value == "" else value
    processed_data.append(new_obj)

result_json = json.dumps(processed_data)
print(result_json)

注意事项 #

• 处理前必须确保原始JSON语法正确，否则解析阶段会直接报错（比如原示例中缺失的逗号需先补全）。
• 上述代码针对单层对象结构，若JSON存在多层嵌套，可扩展为递归遍历处理。

内容的提问来源于stack exchange，提问作者Sirth