• 需求的场景描述（希望解决的问题）

小程序做了一个登录页面，开发版本中进行登录操作，希望登录成功

代码：

formSubmit: function (e) {

// 获取表单数据

var objData = e.detail.value;

// console.log(“formSubmit >>>>>>>>>>>>>>>>”);

// console.log(objData);

if (objData.username && objData.password) {

var url = app.globalData.http + “/clientApi/wxapi/login”;

var openid = wx.getStorageSync(‘openId’);

// console.log(openid);

wx.request({

url: url,

method: “POST”,

data: {

openId: openid,

loginName: objData.username,

password: objData.password

},

header: {

‘content-type’: ‘application/x-www-form-urlencoded’

},

success: function (res) {

console.log(res.data);

var myInfo = res.data.manager

if (res.data.code == 200) {

wx.showToast({

title: “登录成功”,

icon: ‘success’,

duration: 3000

})

wx.setStorageSync(‘mchtId’, res.data.mchtUserInfo[0].mchtId);

wx.setStorageSync(‘mchtName’, res.data.mchtUserInfo[0].mchtName);

wx.setStorageSync(‘mchtTypeName’, res.data.mchtUserInfo[0].mchtTypeName);

// 成功后跳转

wx.reLaunch({

url: ‘…/index/index’,

})

} else {

wx.showModal({

title: ‘提示’,

showCancel: false,

content: res.data.msg,

})

}

},

fail: function (res) {

wx.showToast({

title: ‘登录超时’,

icon: ‘loading’,

duration: 3000

})

}

})

} else {

wx.showToast({

title: ‘登录失败’,

icon: ‘loading’,

duration: 1000

})

}

}

开发版本，不打开调试，小程序一直进fail。打开调试，程序正常运行。请问这是什么原因？？？