JCa*_*Cam 19 javascript jquery json
我有一个JSON string(?)我已经回来$.ajax()并命名的data.有些值为空,我需要为某些键添加值并将其发送回我的PHP脚本.
我通过data.keyName访问现有值.如何在"数据"中添加或更改某些键的值?
这就是data看起来像.
{
"ID":"48",
"userID":"0",
"address":"750 North High Street",
"city":"Columbus",
"state":"OH",
"zip":"43215",
"lat":"39.977673",
"lng":"-83.003357",
"busNumber":"55",
"isClaimed":"N",
"whereFound":"",
"busNum":"",
"email":"",
"fname":"",
"lname":"",
"comments":""
}
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cdh*_*wie 37
解码完JSON后,结果就是一个JavaScript对象.像对待任何其他对象一样操纵它.例如:
data.busNum = 12345;
...
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sje*_*397 20
var temp = data.oldKey; // or data['oldKey']
data.newKey = temp;
delete data.oldKey;
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看来您的密钥是否保存在变量中。data.key = value不会工作。
你应该使用 data[key] = value
例子:
data = {key1:'v1', key2:'v2'};
var mykey = 'key1';
data.mykey = 'newv1';
data[mykey] = 'newV2';
console.log(data);
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结果:
{
"key1": "newV2",
"key2": "v2",
"mykey": "newv1"
}
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