微信小程序獲取用戶信息簡單,可是在存入本身服務器數據庫的過程當中研究了一天多的時間,而且網上搜索不到該資源,故發出來供你們參考。php
index.jshtml
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Page({
data: {
nickName:
"微信帳號登陸"
,
avatarUrl:
"./user-unlogin.png"
,
},
onLoad:
function
() {
var
that =
this
;
var
nickName = that.data.nickName ;
var
avatarUrl = that.data.avatarUrl;
var
db=
"no"
;
wx.getUserInfo({
success:
function
(res) {
that.data.nickName = res.userInfo.nickName,
that.data.avatarUrl = res.userInfo.avatarUrl,
that.setData({
nickName: that.data.nickName,
avatarUrl: that.data.avatarUrl,
}),
that.setData({
db:
"ok"
// 設置變量db,只有成功獲取用戶信息後才寫入數據庫
})
if
(db =
"ok"
) {
var
name, url;
wx.request({
header: {
"Content-Type"
:
"application/x-www-form-urlencoded"
},
method:
"POST"
,
data: {
name: res.userInfo.nickName,
url: res.userInfo.avatarUrl,
},
success:
function
() {
console.log(
"success"
)
},
fail:
function
() {
console.log(
"fail"
)
}
})
}
},
fail:
function
(res) {
that.data.nickName =
"未受權沒法獲取用戶信息"
,
that.setData({
nickName: that.data.nickName
})
}
})
},
onShow:
function
() {
}
})
|
getuserinfo.php前端
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<?php
$name
=
$_POST
[
'name'
];
$url
=
$_POST
[
'url'
];
$servername
=
"localhost"
;
$username
=
"root"
;
$password
=
""
;
//服務器中鏈接數據庫的密碼
$dbname
=
"mydb"
;
//使用的數據庫名
// 建立鏈接
$conn
=
new
mysqli(
$servername
,
$username
,
$password
,
$dbname
);
// 檢測鏈接
if
(
$conn
->connect_error) {
die
(
"connect server fail: "
.
$conn
->connect_error);
}
$sql
= "INSERT INTO userinfo
VALUES (
'$name'
,
'$url'
)";
if
(
$conn
->query(
$sql
) === TRUE) {
echo
"insert success"
;
}
else
{
echo
"Error: "
.
$sql
.
"<br>"
.
$conn
->error;
}
$conn
->close();
?>
|
測試經過,在服務器中查詢有用戶信息vue