有什么办法可以将下面的日期HH:MM:SS
格式化为HH:MM:SS:MMM
时间js中的时间格式?
2016-01-10T00:00:00+05:30
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可以将其转换为以下格式吗?
2016-01-10T00:00:00.000+05:30
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简单的JS代码:
2016-01-10T00:00:00+05:30
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2016-01-10T00:00:00.000+05:30
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我正在尝试采用我在下面使用的 3 个命令,并将其变成单个命令,或者将其放入我可以从另一个程序调用的脚本中。
这就是我目前正在做的事情:
$ sudo docker exec -it <my-app container id> bash
$ chmod +r /var/opt/mssql/data/myFile.bak
$ exit
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有没有更简单的方法来 chmod docker 容器内的文件?如果没有,我如何将其放入单个命令或单个脚本中?
这不起作用:
$ sudo docker exec -it <my-app container id> bash; chmod +r /var/opt/mssql/data/myFile.bak; exit
Run Code Online (Sandbox Code Playgroud) 我有一个Express应用程序,我正在尝试将所有中间件放在自己的文件中.一些中间件函数需要该db
对象,而另一些则不需要.
这对于那些不需要的功能非常简单的db
对象,但由于我下面的代码结构,我怎么可以参照db
对象doesNotNeedDbParam
,因为它已经拥有PARAMS req
,res
和next
?
somefile.js:
const router = express.Router()
const doesNotNeedDbParam = require('./middleware')().doesNotNeedDbParam
function foo () {
// Currently I have to call require and pass in the db object here b/c
// it's not set when requiring the function doesNotNeedDbParam
router.use(require('./middleware')(db).needsDbParam // <-- Is there a better way to do this so that I can require the file above and pass the db object in when it's …
Run Code Online (Sandbox Code Playgroud) 我正在尝试在Postgres中执行一个非常简单的删除查询
查询:
delete from "Tasks" where id = "fc1f56b5-ff41-43ed-b27c-39eac9354323";
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结果:
ERROR: column "fc1f56b5-ff41-43ed-b27c-39eac9354323" does not exist
LINE 1: delete from "Tasks" where id = "fc1f56b5-ff41-43ed-...
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我有一个简单的表,其中id是该值的记录。为什么"fc1f56b5-ff41-43ed-b27c-39eac9354323"
是列名呢?
我正在尝试创建一个可重用的函数,您可以传入一个 joi 模式并针对req.body
. 我想验证我的变量schema
以确保它是一个有效的 joi 模式。
有没有办法做到这一点?
function validatePayload(schema) {
return (req, res, next) => {
const valid = joi.validate(req.body, schema)
if (valid.error) {
return handleErr(res, HttpStatus.BAD_REQUEST, valid.error.details[0].message)
}
return next()
}
}
Run Code Online (Sandbox Code Playgroud) 我正在尝试学习 koa 并且无法弄清楚为什么我会收到错误消息:
server error TypeError: ctx.body is not a function
at getHandler (/Users/tomcaflisch/Sites/learn-koa/server.js:32:7)
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当我运行此代码时:
server error TypeError: ctx.body is not a function
at getHandler (/Users/tomcaflisch/Sites/learn-koa/server.js:32:7)
Run Code Online (Sandbox Code Playgroud) 我正在尝试foo/bar
在目录中使用 Composer开发一个 PHP 库(称为)/work/a
,composer.json
内容如下:
{
"name": "foo/bar",
"require": {
"php": ">=7.2"
}
}
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/work/a
是一个 git 项目,我在分支上 mybranch
我正在尝试testing/foobar
在 dir 中使用 Composer 在本地的另一个项目(称为)中使用此库,work/b
其中包含以下composer.json
内容:
{
"name": "testing/foobar",
"type": "project",
"repositories": [
{
"type": "vcs",
"url": "/work/a"
}
],
"require": {
"php": "^7.4",
"foo/bar": "mybranch"
}
}
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运行$ composer install
时/work/b
出现错误:
[UnexpectedValueException]
Could not parse version constraint mybranch: Invalid version string "mybranch"
Run Code Online (Sandbox Code Playgroud) 我正在编写这个脚本,每隔3秒旋转一个背景图像,但它根本不起作用,我很难过为什么不这样做.
$(document).ready(function() {
var inc = 0;
var bgImages = new Array();
bgImages.push("Images/backgroundDog-small.jpg");
bgImages.push("Images/backgroundDog1-small.jpg");
bgImages.push("Images/backgroundDog2-small.jpg");
bgImages.push("Images/backgroundDog3-small.jpg");
setInterval(change, 3000);
function change() {
//if we're not at the end of the array
if (inc < (bgImages.length)) {
var image = bgImages[inc];
$('body').css('backgroundImage', image);
console.log(image);
inc++;
//reset the counter inc and go through the array again
} else {
inc = 0;
}
}
});
Run Code Online (Sandbox Code Playgroud) 我遇到了一个问题,即操作完成后我的页面没有刷新.
我在页面上有这个表格:
<h:form id="fundingResults">
<p:ajaxStatus style="width:16px;height:16px;">
<f:facet name="start">
<p:graphicImage value="/images/loading4.gif" />
</f:facet>
<f:facet name="complete">
<h:outputText value="" />
</f:facet>
</p:ajaxStatus>
<p:dataTable>
// some form elements in the table
</p:dataTable>
<p:commandButton id="btnEdit" value="Submit" type="submit" action="#{fundingBacker.classifyProducts}" style="float: right;margin-top:15px;" />
</h:form>
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我的classifyProducts方法框架是这样的:
public void classifyProducts() {
// a bunch of stuff goes on in here
}
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由于隐式导航,classifyProducts()方法完成后页面是否应该自动刷新?
我尝试这样做时出错:
$ci =& get_instance();
$ci->db->select("CONCAT_WS(' ',users.name_first,users.name_last) AS user_name,CONCAT_WS(' ',advertisers.name_first,advertisers.name_last) AS advertiser_name,image,advertiser_reviews.last_updated");
$ci->db->join('users', 'users.id = advertiser_reviews.user_id');
$ci->db->join('advertisers', 'advertisers.id = advertiser_reviews.advertiser_id');
$ci->db->order_by('advertiser_reviews.id','desc');
$ci->db->limit(1);
$query = $ci->db->get('advertiser_reviews');
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错误如下:
Error Number: 1064
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'FROM (`hc_advertiser_reviews`) JOIN `hc_users` ON `hc_users`.`id` = `hc_advertis' at line 2
SELECT CONCAT_WS(' ', `hc_users`.`name_first`, `hc_users`.`name_last)` AS user_name,
CONCAT_WS(' ', `hc_advertisers`.`name_first`, `hc_advertisers`.`name_last)` AS advertiser_name,
`image`,
`hc_advertiser_reviews`.`last_updated`
FROM …
Run Code Online (Sandbox Code Playgroud) 我在refinerycms中创建了一个引擎,我将数据拉到主页上.Refinerycms自动为此引擎创建页面和菜单项.
如何删除引擎页面以及从菜单栏中删除它?
menu.html.erb
<%
# Collect the root items.
# ::Refinery::Menu is smart enough to remember all of the items in the original collection.
if (roots = local_assigns[:roots] || (collection ||= refinery_menu_pages).roots).present?
dom_id ||= 'menu'
css = [(css || 'menu clearfix')].flatten.join(' ')
hide_children = Refinery::Core.menu_hide_children if hide_children.nil?
-%>
<div id="nav">
<ul>
<%= render :partial => '/refinery/menu_branch', :collection => roots,
:locals => {
:hide_children => hide_children,
:sibling_count => (roots.length - 1),
:menu_levels => local_assigns[:menu_levels],
:apply_css => true #if you don't care …
Run Code Online (Sandbox Code Playgroud) 我正在尝试在基于dataType构建查询的应用程序中构建高级搜索功能.
我的潜在数据类型是 FLOAT, NUMBER, CHAR, VARCHAR2, DATE, TIMESTAMP(6)
我正在尝试"从我的shell中编码",而不是使用ArrayList来存储值,我将尝试使用,enum
因为我只有这6个值,他们永远不会改变.
TIMESTAMP(6)
由于括号,给我带来了困难.
我如何声明枚举值TIMESTAMP(6)
?
目前下面的代码片段给出了错误:构造函数TableBacking.dataTypeOptions(int)未定义.
public enum dataTypeOptions {
FLOAT, NUMBER, CHAR, VARCHAR2, DATE, TIMESTAMP(6)
};
Run Code Online (Sandbox Code Playgroud) 我正在尝试将bash脚本用作git commit-msg钩子.基本上我想确保提交消息包含与正则表达式匹配的字符串.我的正则表达式是正确的,但我不知道如何检查包含的消息是否.git/COMMIT_EDITMSG
与正则表达式匹配.
#!/bin/sh
#
# This hook verifies that the commit message contains a reference to a tfs item
RED=$(tput setaf 1)
NORMAL=$(tput sgr0)
# Regex to validate a string contains "#" followed by 4 or 5 digits anywhere in the commit message
regex="#[0-9]{4,5}($|[^0-9])"
# I NEED TO FIGURE OUT HOW TO READ THE CONTENTS OF THIS FILE AND ENSURE IT MATCHES THE REGEX
echo "MSG = $1" # This prints "MSG = .git/COMMIT_EDITMSG"
# If the commit …
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