Php 更新现有的JSON
假设我有以下JSON结构:Php 更新现有的JSON,php,json,Php,Json,假设我有以下JSON结构: {"employees":[ {"firstName":"John", "lastName":"Doe"}, {"firstName":"Anna", "lastName":"Smith"}, {"firstName":"Peter", "lastName":"Jones"} ]} 如何使用PHP更新它并添加另一个属性,例如年龄? 这是我的预期结果: {"employees":[ {"firstName":"John", "lastN
{"employees":[
{"firstName":"John", "lastName":"Doe"},
{"firstName":"Anna", "lastName":"Smith"},
{"firstName":"Peter", "lastName":"Jones"}
]}
如何使用PHP更新它并添加另一个属性,例如年龄?
这是我的预期结果:
{"employees":[
{"firstName":"John", "lastName":"Doe", "age":"20"},
{"firstName":"Anna", "lastName":"Smith", "age":"30"},
{"firstName":"Peter", "lastName":"Jones", "age":"40"}
]}
更新:
这是我的方法:
$json1 = '{"employees":[
{"firstName":"John", "lastName":"Doe"},
{"firstName":"Anna", "lastName":"Smith"},
{"firstName":"Peter", "lastName":"Jones"}
]}';
$json2 = '{"employees":[
{"age":"20"},
{"age":"30"},
{"age":"40"}
]}';
$data1 = json_decode($json1, true);
$data2 = json_decode($json2, true);
$i = 0;
foreach ($data1 as $key => $entry) {
$data1[$key][$i] += $data2["employees"][$i];
$i++;
}
$json1 = json_encode($data1);
echo ($json1)
但不幸的是,我只能得到这样的结果:
{
"employees":[
{
"firstName":"John",
"lastName":"Doe",
"age":"20"
},
{
"firstName":"Anna",
"lastName":"Smith"
},
{
"firstName":"Peter",
"lastName":"Jones"
}
]
}
因此,它只将其添加到第一个条目中。这取决于您使用的json文件的语言,但基本上是一个
for employe in json['employees']:
employe['age'] = '20'
但对于这种方法,所有员工的年龄都是20岁,我会做如下事情:
ages ={'John':'20','Anna':'30','Peter':'40'}
for employe in json['employess']:
name = employe['firstname']
age= ages[name]
employe[name] = age
对于javascript:
var ages ={'John':'20','Anna':'30','Peter':'40'}
for(var i=0;i<json['employees'].length;i++){
var employe = json['employees'][i];
var name = employe[name];
var age = ages[name];
employe[name] = age;
};
您需要在阵列上循环。此时,您正在对象上循环,因此第一个项目是employees,然后就没有其他项目了
foreach ($data1['employees'] as $key => $entry) {
$data1['employees'][$key] += $data2["employees"][$key];
}
解析它。修改生成的数组。序列化它。有什么问题吗?@close投票者该问题的编辑质量尚可。这取决于您使用的语言-该问题被标记为PHP