Question

我在$ result中有$result = json_encode($json)的JSON输出。我如何访问Mickey，Maddu，mickey @ gmail等值。我需要将这些值存储在变量中。

JSON输出

{"jsonFirstName":"Mickey","jsonLastName":"Maddu","jsonEmail":"mickey@gmail.com","jsonPassword":"QWERqwer!@#$","jsonDob":"2014-01-01","jsonDobTime":"dobtime","jsonLocaldob":"2014-01-01T01:00","jsonSsn":"123-12-1234","jsonPhonenumber":"123-123-1234","jsonCreditcardnumber":"123412341234"}

Answer 1

您只需使用extract()即可。（^{只要数组保持一维}）

$json = '{"jsonFirstName":"Mickey","jsonLastName":"Maddu","jsonEmail":"mickey@gmail.com","jsonPassword":"QWERqwer!@#$","jsonDob":"2014-01-01","jsonDobTime":"dobtime","jsonLocaldob":"2014-01-01T01:00","jsonSsn":"123-12-1234","jsonPhonenumber":"123-123-1234","jsonCreditcardnumber":"123412341234"}';

extract(json_decode($json, true));

echo $jsonLastName;

Example

Answer 2

使用json_decode，您可以将JSON作为对象或关联数组进行访问，看起来：

作为对象：

$string = '{"foo": "bar", "foo2": "attr"}';
$the_json = json_decode($string);

echo $the_json->foo;
//output: "bar"

作为关联数组：

$string = '{"foo": "bar", "foo2": "attr"}';
$the_json = json_decode($string, true);

echo $the_json['foo'];
//output: "bar"

请参阅官方文档：PHP JSON

PHP - 如何在$ firstName，$ lastName等变量中访问和存储JSON值

2 个答案: