我正在尝试使用谷歌地图找到两点之间的位置.这是我正在使用的代码:
function initialize() {
var myOptions = {
center: new google.maps.LatLng(36.8813329,-103.6975488),
zoom: 4,
mapTypeId: google.maps.MapTypeId.ROADMAP
};
var map = new google.maps.Map(document.getElementById("map-canvas"), myOptions);
var impactCoordinates = [
new google.maps.LatLng(37.772323, -122.214897),
new google.maps.LatLng(34.1633766,-81.6487862),
];
var ImpactPath = new google.maps.Polyline({
path: impactCoordinates,
strokeColor: "#FF0000",
strokeOpacity: 1.0,
strokeWeight: 2
});
ImpactPath.setMap(map);
var loc1 = new google.maps.LatLng(37.772323, -122.214897);
var loc2 = new google.maps.LatLng(34.1633766,-81.6487862);
alert(google.maps.geometry.spherical.computeDistanceBetween(loc1, loc2));
}
这是我从控制台得到的错误:
未捕获的TypeError:无法读取未定义的属性"球形"
我正在尝试将字符串值分配给我的.erb文件中的javascript对象,如下所示:
var data = {
'name': '<%= @product.name %>',
...
};
问题是,如果值name是Tom's small ears,
输出data.name将是Tom's small ears.
有没有办法逃脱特殊字符？
我尝试过,'name': '<%= raw @product.name %>'但Uncaught SyntaxError: Unexpected identifier输出到控制台.
做<%= escape_javascript @product.name %>输出Tom\'s small ears
编辑
@Stefan在MrYoshiji's回答中的评论对我有用.
给出这个函数的伪代码
f(0) = 1;
f(1) = 3;
f(n) = 3 * f(n - 1) - f(n - 2); // for n >= 2.
有这种非递归方式吗？
我在尝试安装django项目的要求时遇到了问题.
我无法弄清楚为什么我会收到这个错误.
有人能引导我朝着正确的方向前进吗？
>Running setup.py egg_info for package django-staticfiles
Traceback (most recent call last):
File "<string>", line 14, in <module>
File "/Users/glosseyvilly/Desktop/Workspace/chris_dev/mysite-env/build/django-staticfiles/setup.py", line 128, in <module>
'versiontools >= 1.6',
File "/System/Library/Frameworks/Python.framework/Versions/2.6/lib/python2.6/distutils/core.py", line 113, in setup
_setup_distribution = dist = klass(attrs)
File "build/bdist.linux-i686/egg/setuptools/dist.py", line 260, in __init__
File "build/bdist.linux-i686/egg/setuptools/dist.py", line 284, in fetch_build_eggs
File "build/bdist.linux-i686/egg/pkg_resources.py", line 563, in resolve
plugin_projects = list(plugin_env)
File "build/bdist.linux-i686/egg/pkg_resources.py", line 799, in best_match
File "build/bdist.linux-i686/egg/pkg_resources.py", line 811, in obtain
File "build/bdist.linux-i686/egg/setuptools/dist.py", line 327, in fetch_build_egg
File …我有一个 keydown 处理函数，定义如下：
handleOnKeyDown = (e: React.KeyboardEvent<HTMLButtonElement>) => {
if (key.isEscape(e.key)) {
stopEvent(e);
this.props.handleClose(e);
}
}
但是，在我的代码的另一部分中，将此函数传递给removeEventListenerand addEventListener：
componentDidMount () {
const { current: modal } = this.modalRef;
if (modal) {
modal.addEventListener('keydown', this.handleOnKeyDown);
}
}
componentWillUnmount () {
const { current: modal } = this.modalRef;
if (modal) {
modal.removeEventListener('keydown', this.handleOnKeyDown);
}
}
我收到以下打字稿错误：
handleOnKeyDown = (e: React.KeyboardEvent<HTMLButtonElement>) => {
if (key.isEscape(e.key)) {
stopEvent(e);
this.props.handleClose(e);
}
}
什么会解决这个问题？
我正在研究extjs + php中的文件上传功能.我想从extjs上传文件或图像,需要将它发送到我在PHP设计的服务器端.在extjs中,我有查看文件的代码 -
Ext.define('Balaee.view.kp.dnycontent.Content',
{
extend:'Ext.panel.Panel',
requires:[
'Balaee.view.kp.dnycontent.ContentView'
],
id:'ContentId',
alias:'widget.Content',
title:'This day in a history',
items:[
{
xtype: 'fileuploadfield',
hideLabel: true,
emptyText: 'Select a file to upload...',
id: 'upfile',
//name:'file',
width: 220
}],
buttons:[
{
text: 'Upload',
handler: function () {
var file = Ext.getCmp('upfile').getEl().down('input[type=file]').dom.files[0]; console.log(file);
var reader = new FileReader();
filecontent=reader.readAsBinaryString(file);
console.log(filecontent);
}
}
]
});
因此,当我尝试使用上面的代码上传animage文件.上面一行:
Ext.getCmp('upfile').getEl().down('input[type=file]').dom.files[0];将文件的信息存储为:
File {
webkitRelativePath: "",
lastModifiedDate: Tue Jul 14 2009 11:02:31 GMT+0530 (India Standard Time),
name: "Koala.jpg",
type: "image/jpeg", …当我尝试启动ExtJS 4.2应用程序时出现此错误:
Uncaught TypeError: Cannot read property 'buffered' of undefined
我已经仔细检查了所有的storeID和名称,以确保它们拼写正确,但我仍然有这个问题:(.还有什么我可以尝试解决这个错误吗？
提前致谢.
假设我有一个名为user_table的表:
user_table
id
first_name
last_name
gender
email
created_at
如何编写查询列出多个用户使用的电子邮件？
编辑:
使用LEFT JOIN而不是GROUP BY可以获得相同的结果吗？
我正在尝试使用haskell删除列表中项目的所有实例.我得到一个我不太懂的错误.任何人都可以帮助我,让我知道我做的正确吗？
deleteAllInstances :: (a, [l]) => a -> [l] -> [l]
deleteAllInstances (a, []) = []
deleteAllInstances (i, (x:xs))
| i == x = tail
| otherwise = x ++ tail
where tail = deleteAllInstances i xs
我试图将PHP变量的值传递给javascript,但我不能让它工作.我做错了吗？下面是我正在使用的代码行.
var dist = parseInt("<?php echo json_encode($distance); ?>");