我有一个Mongo数据库,在users集合中我只有1个文档.我使用用户名过滤器执行find()和findOne()操作.我认为我认为是find()操作的错误结果.
MongoDB shell version: 3.2.10
connecting to: test
Server has startup warnings:
2016-10-20T20:37:32.681-0700 I CONTROL [initandlisten]
2016-10-20T20:37:32.681-0700 I CONTROL [initandlisten] ** WARNING: /sys/kernel/mm/transparent_hugepage/enabled is 'always'.
2016-10-20T20:37:32.681-0700 I CONTROL [initandlisten] ** We suggest setting it to 'never'
2016-10-20T20:37:32.681-0700 I CONTROL [initandlisten]
2016-10-20T20:37:32.681-0700 I CONTROL [initandlisten] ** WARNING: /sys/kernel/mm/transparent_hugepage/defrag is 'always'.
2016-10-20T20:37:32.681-0700 I CONTROL [initandlisten] ** We suggest setting it to 'never'
2016-10-20T20:37:32.681-0700 I CONTROL [initandlisten]
> use lab2
switched to db lab2
> db.users.find()
{ "_id" : ObjectId("5807ac0765f24dd0660e4332"), "username" : "avtrulzz", …
Run Code Online (Sandbox Code Playgroud) 我希望我的网页在密码字段中显示一个复选框。用户单击复选框并以文本形式查看密码。取消选中,再次输入密码。
这就是我得到的
我想要密码字段中的这个复选框。我在网上找不到有关此主题的任何内容。
这是我的代码:
<!-- Set a password for the account -->
<div class="col-md-12" style="margin: 7px;">
<input type="password" id="reg_password" name="reg_password" style="height: 35px;" class="form-control input-lg" placeholder="Password" ng-model="register_password" />
</div>
<input type="checkbox" id="eye" onclick="if(reg_password.type=='text')reg_password.type='password'; else reg_password.type='text';" />
Run Code Online (Sandbox Code Playgroud) 我正在尝试对 python 中的列表执行二进制搜索。列表是使用命令行参数创建的。用户输入他想在数组中查找的数字,然后返回元素的索引。出于某种原因,程序只输出 1 和 None。代码如下。非常感谢任何帮助。
import sys
def search(list, target):
min = 0
max = len(list)-1
avg = (min+max)/2
while (min < max):
if (list[avg] == target):
return avg
elif (list[avg] < target):
return search(list[avg+1:], target)
else:
return search(list[:avg-1], target)
print "The location of the number in the array is", avg
# The command line argument will create a list of strings
# This list cannot be used for numeric comparisions
# This list has to be converted into …
Run Code Online (Sandbox Code Playgroud)