php CodeIgniter
string(112) "{"code":"100","user":{"name":"张三"},"msg":"获取成功"}"
PHP 怎么获取 数组里面的 user 括号里面name 值呢?
回复讨论(解决方案)
$s = '{"code":"100","user":{"name":"张三"},"msg":"获取成功"}';echo json_decode($s)->user->name; //张三 json_decode 后就是对象/数组了,之后就容易了吧。
$s = '{"code":"100","user":{"name":"张三"},"msg":"获取成功"}';echo json_decode($s)->user->name; //张三 版主太牛X
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