Question

我在我的php函数中使用json_encode将数据库中的多个结果发送到ajax调用。

我以下列格式获得结果..

[
    {
        "id": 24,
        "title": "Mr",
        "first_name": "Patrick",
        "last_name": "Vinc",
        "gender": "male",
        "email": "nupur@gmail.com",
        "password": "$2y$10$yCyGBOtX6kF3ghy/k8YuXe4wR9W5hYtTGDkl5trTEd7.s5LntOQ.u",
        "phone_type": null,
        "phone_number": "0000000000",
        "pager_number": "00000000000000",
        "address_line_1": "",
        "address_line_2": "",
        "city": "",
        "postal_code": "",
        "province": "BC",
        "country": null,
        "emc_contact": "",
        "emc_phone": "000000000000000000",
        "emc_relation": "",
        "passcode": "",
        "locker": "999999",
        "combination": "abc567",
        "its_username": null,
        "its_password": null,
        "dictation_number": null,
        "emailed": 1,
        "signed": 0,
        "student_num": "12345634",
        "level": "Default",
        "persist_code": "",
        "activated_at": "2014-08-23 16:04:18",
        "program": null,
        "school": "",
        "service": "",
        "undergrad_year": null,
        "undergrad_level": null,
        "activated": 1,
        "activation_code": "",
        "undergrad_text": null,
        "cpso_num": 0,
        "start_date": "2014-08-01",
        "end_date": "2014-08-31",
        "learner_start_date": "0000-00-00",
        "learner_end_date": "0000-00-00",
        "vacation_start_date": "0000-00-00",
        "vacation_end_date": "0000-00-00",
        "physician": "1",
        "affiliates": null,
        "mask": "",
        "mask_fit_month_year": "",
        "learner_type": null,
        "status": 1,
        "last_login": "0000-00-00 00:00:00",
        "reset_password_code": "",
        "permissions": "",
        "created_at": "-0001-11-30 00:00:00",
        "updated_at": "2014-08-22 21:27:17"
    }
]

如何访问此数据？我想得到名字和姓氏。

Answer 1

希望这有帮助

$json = json_decode($jsondata);

foreach ($json as $item)
{
    echo $item->first_name;
    echo $item->last_name;
}

在Javascript中执行相同操作：

for(var i=0;i<data.length;i++){
   var item=data[i];
   console.log(item['first_name']);//Or item.first_name also works
   console.log(item['last_name']);// 
}

Answer 2

我认为您正试图通过AJAX请求将JSON检索到PHP应用程序吗？

您需要向PHP应用程序发出一个AJAX请求，该请求应该适当地响应（使用JSON数据）

我相信你需要这样的东西：

$.post( "ajax/test.html", function( data ) {
  for (item in data){
    console.log(item.first_name);
    console.log(item.last_name);
  }
});

确保您实际上回应了PHP应用程序的响应。

http://api.jquery.com/jquery.post/

http://api.jquery.com/jquery.ajax/

访问从php传递的json数据（json_encode（databaseresults））

2 个答案: