我有以下javascript:
#!/usr/bin/env node
var fs = require(’fs’);
var outfile = "hello.txt";
var out = "Modify this script to write out something different.\n";
fs.writeFileSync(outfile, out);
console.log("Script: " + __filename + "\nWrote: " + out + "To: " + outfil
我用api/services/SomeServices.js编写了一个服务函数
getCreditDebitNotes:function(vid){
console.log('resolving credit and debits');
var deferred=sails.q.defer();
CreditDebitNotes.find({vendorID:vid,status:1},{select:['soid','statementID','amount']})
.exec(funct
我正在为Greasemonkey (FX7)编写一个脚本,试图删除某些链接,我发现由于某种原因,源代码中存在的某个链接(不是由JS隐藏或构造的)没有出现在该函数返回的数组中。
如果这个链接是在运行该页面时通过JS构建的,我不会感到奇怪,但它就在找到的另一个链接后面。
那么,有没有人知道为什么会发生这种情况,以及我如何解决它?
var links = document.getElementsByTagName("a");
for (var l in links){
if (links[l].href == "blah"){ ... }
}
这就是我试图使用它们
我浏览了网页/文档,找不到为什么这不起作用。 我正在尝试从JSON文件插入数据(引号):{ "quotesArray":[{
"personName": "Albert Einstein",
"quote": "Look deep into nature, and then you will understand everything better."
},
{
"personName": "Stephen Hawking",
"quote&
Rails应用,学习CoffeeScript和Backbone.js:
Uncaught TypeError: Cannot use 'in' operator to search for 'id' in easy
_.extend.setbackbone.js:205
Backbone.Modelbackbone.js:142
Gamegames.js:13
(anonymous function)games.js:34
jQuery.Callbacks.firejquery.js:1047
jQuery.Callbacks.self.fireWithjque