我正在进行一个项目,必须使用 json 来实现我的目标。我正在使用 mysql 和使用 json_encode() 的 php。我怎么得到这个:
{"contacts": [ { "id": "c200","name": "Ravi Tamada","email": "ravi@gmail.com","address": "xx-xx-xxxx,x - street, x - country","gender" : "male","phone": { "mobile": "+91 0000000000","home": "00 000000","office": "00 000000"}},{ "id": "c201","name": "Johnny Depp","email": "johnny_depp@gmail.com","address": "xx-xx-xxxx,x - street, x - country","gender" : "male","phone": {"mobile": "+91 0000000000","home": "00 000000","office": "00 000000"}},
长成这样?
{
"contacts": [
{
"id": "c200",
"name": "Ravi Tamada",
"email": "ravi@gmail.com",
"address": "xx-xx-xxxx,x - street, x - country",
"gender" : "male",
"phone": {
"mobile": "+91 0000000000",
"home": "00 000000",
"office": "00 000000"
}
},
{
"id": "c201",
"name": "Johnny Depp",
"email": "johnny_depp@gmail.com",
"address": "xx-xx-xxxx,x - street, x - country",
"gender" : "male",
"phone": {
"mobile": "+91 0000000000",
"home": "00 000000",
"office": "00 000000"
}
}
]
};
我知道这是可以做到的,因为当多个教程“回显”对象时,json 对象看起来像这样,就像这个http://api.androidhive.info/contacts/。我使用了错误的功能吗?这是我的代码。
$employee = array();
while($employee = mysql_fetch_array($result, MYSQL_ASSOC)) {
$employee[] = array('employee'=>$employee);
}
$output = json_encode(array('employee' => $employee));
}