ages = {paul: 11, rick: 7 }
for n,a in ages
console.log("names: #{n}, ages: #{a}")
So nothing is being printed in the console. What I'm doing wrong?
That translates into the following JavaScript:
var a, ages, n, _i, _len;
ages = {
paul: 11,
rick: 7
};
for (a = _i = 0, _len = ages.length; _i < _len; a = ++_i) {
n = ages[a];
console.log("names: " + n + ", ages: " + a);
}
Note that it's trying to loop through the ages object using the length property, which doesn't exist.
Change your script so the loop uses for n,a of ages syntax instead (relevant documentation), and your code translates into valid JavaScript and prints to the console.
var a, ages, n;
ages = {
paul: 11,
rick: 7
};
for (n in ages) {
a = ages[n];
console.log("names: " + n + ", ages: " + a);
}
Related
I want to extract all the data from column B, then assign it to variables using for loop. I'm using this code, but it is looping wrong.
Ex:
Total row count = 10
Variable 1 to variable 10 are having the same data; then it will loop 10x. So the final values of array variables are all the same :(
for (int getAllAccts = 1; getAllAccts <= TotalRowCount.size(); getAllAccts++) {
try{
String[] accName = new String[TotalRowCount.size()]
for(int accNameCount=1; accNameCount < TotalRowCount.size(); accNameCount++){
accName[accNameCount] = driver.findElement(By.xpath("//*[#id='tbl-table1']/tbody/tr[" + (getAllAccts)+ "]/td[2]/a")).getText();
log.logWarning('Customer Name' + accNameCount + ' ' + accName[accNameCount])
}
}
catch(org.openqa.selenium.StaleElementReferenceException ex) {
String[] accName
for(int accNameCount=1; accNameCount < TotalRowCount.size(); accNameCount++){
accName[accNameCount] = driver.findElement(By.xpath("//*[#id='tbl-table1']/tbody/tr[" + (getAllAccts)+ "]/td[2]/a")).getText();
log.logWarning('Customer Name' + accNameCount + ' ' + accName[accNameCount])
}
}
}
The getAllAccts is always the same inside of the inner (nested) for loop. So all the accNames will have the same value.
Change
accName[accNameCount] = driver.findElement(By.xpath("//*[#id='tbl-table1']/tbody/tr[" + (getAllAccts)+ "]/td[2]/a")).getText();
to
accName[accNameCount] = driver.findElement(By.xpath("//*[#id='tbl-table1']/tbody/tr[" + (accNameCount)+ "]/td[2]/a")).getText();
Google implemented a captcha to block people from accessing the TTS translate API https://translate.google.com/translate_tts?ie=UTF-8&q=test&tl=zh-TW. I was using it in my mobile application. Now, it is not returning anything. How do I get around the captcha?
Add the qualifier '&client=tw-ob' to the end of your query.
https://translate.google.com/translate_tts?ie=UTF-8&q=test&tl=zh-TW&client=tw-ob
This answer no longer works consistently. Your ip address will be blocked by google temporarily if you abuse this too much.
there are 3 main issues:
you must include "client" in your query string (client=t seems to work).
(in case you are trying to retrieve it using AJAX) the Referer of the HTTP request must be https://translate.google.com/
"tk" field changes for every query, and it must be populated with a matching hash:
tk = hash(q, TKK), where q is the text to be TTSed, and TKK is a var in the global scope when you load translate.google.com: (type 'window.TKK' in the console). see the hash function at the bottom of this reply (calcHash).
to summarize:
function generateGoogleTTSLink(q, tl, tkk) {
var tk = calcHash(q, tkk);
return `https://translate.google.com/translate_tts?ie=UTF-8&total=1&idx=0&client=t&ttsspeed=1&tl=${tl}&tk=${tk}&q=${q}&textlen=${q.length}`;
}
generateGoogleTTSLink('ciao', 'it', '410353.1336369826');
// see definition of "calcHash" in the bottom of this comment.
=> to get your hands on a TKK, you can open Google Translate website, then type "TKK" in developer tools' console (e.g.: "410353.1336369826").
NOTE that TKK value changes every hour, and so, old TKKs might get blocked at some point, and refreshing it may be necessary (although so far it seems like old keys can work for a LONG time).
if you DO wish to periodically refresh TKK, it can be automated pretty easily, but not if you're running your code from the browser.
you can find a full NodeJS implementation here:
https://github.com/guyrotem/google-translate-server.
it exposes a minimal TTS API (query, language), and is deployed to a free Heroku server, so you can test it online if you like.
function shiftLeftOrRightThenSumOrXor(num, opArray) {
return opArray.reduce((acc, opString) => {
var op1 = opString[1]; // '+' | '-' ~ SUM | XOR
var op2 = opString[0]; // '+' | '^' ~ SLL | SRL
var xd = opString[2]; // [0-9a-f]
var shiftAmount = hexCharAsNumber(xd);
var mask = (op1 == '+') ? acc >>> shiftAmount : acc << shiftAmount;
return (op2 == '+') ? (acc + mask & 0xffffffff) : (acc ^ mask);
}, num);
}
function hexCharAsNumber(xd) {
return (xd >= 'a') ? xd.charCodeAt(0) - 87 : Number(xd);
}
function transformQuery(query) {
for (var e = [], f = 0, g = 0; g < query.length; g++) {
var l = query.charCodeAt(g);
if (l < 128) {
e[f++] = l; // 0{l[6-0]}
} else if (l < 2048) {
e[f++] = l >> 6 | 0xC0; // 110{l[10-6]}
e[f++] = l & 0x3F | 0x80; // 10{l[5-0]}
} else if (0xD800 == (l & 0xFC00) && g + 1 < query.length && 0xDC00 == (query.charCodeAt(g + 1) & 0xFC00)) {
// that's pretty rare... (avoid ovf?)
l = (1 << 16) + ((l & 0x03FF) << 10) + (query.charCodeAt(++g) & 0x03FF);
e[f++] = l >> 18 | 0xF0; // 111100{l[9-8*]}
e[f++] = l >> 12 & 0x3F | 0x80; // 10{l[7*-2]}
e[f++] = l & 0x3F | 0x80; // 10{(l+1)[5-0]}
} else {
e[f++] = l >> 12 | 0xE0; // 1110{l[15-12]}
e[f++] = l >> 6 & 0x3F | 0x80; // 10{l[11-6]}
e[f++] = l & 0x3F | 0x80; // 10{l[5-0]}
}
}
return e;
}
function normalizeHash(encondindRound2) {
if (encondindRound2 < 0) {
encondindRound2 = (encondindRound2 & 0x7fffffff) + 0x80000000;
}
return encondindRound2 % 1E6;
}
function calcHash(query, windowTkk) {
// STEP 1: spread the the query char codes on a byte-array, 1-3 bytes per char
var bytesArray = transformQuery(query);
// STEP 2: starting with TKK index, add the array from last step one-by-one, and do 2 rounds of shift+add/xor
var d = windowTkk.split('.');
var tkkIndex = Number(d[0]) || 0;
var tkkKey = Number(d[1]) || 0;
var encondingRound1 = bytesArray.reduce((acc, current) => {
acc += current;
return shiftLeftOrRightThenSumOrXor(acc, ['+-a', '^+6'])
}, tkkIndex);
// STEP 3: apply 3 rounds of shift+add/xor and XOR with they TKK key
var encondingRound2 = shiftLeftOrRightThenSumOrXor(encondingRound1, ['+-3', '^+b', '+-f']) ^ tkkKey;
// STEP 4: Normalize to 2s complement & format
var normalizedResult = normalizeHash(encondingRound2);
return normalizedResult.toString() + "." + (normalizedResult ^ tkkIndex)
}
// usage example:
var tk = calcHash('hola', '409837.2120040981');
console.log('tk=' + tk);
// OUTPUT: 'tk=70528.480109'
You can also try this format :
pass q= urlencode format of your language
(In JavaScript you can use the encodeURI() function & PHP has the rawurlencode() function)
pass tl = language short name (suppose bangla = bn)
Now try this :
https://translate.google.com.vn/translate_tts?ie=UTF-8&q=%E0%A6%A2%E0%A6%BE%E0%A6%95%E0%A6%BE+&tl=bn&client=tw-ob
First, to avoid captcha, you have to set a proper user-agent like: "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:46.0) Gecko/20100101 Firefox/46.0"
Then to not being blocked you must provide a proper token ("tk" get parameter) for each single request.
On the web you can find many different kind of scripts that try to calculate the token after a lot of reverse engineering...but every time the big G change the algorithm you're stuck again, so it's much easier to retrieve your token just observing in deep similar requests to translate page (with your text in the url).
You can read the token time by time grepping "tk=" from the output of this simple code with phantomjs:
"use strict";
var page = require('webpage').create();
var system = require('system');
var args = system.args;
if (args.length != 2) { console.log("usage: "+args[0]+" text"); phantom.exit(1); }
page.onConsoleMessage = function(msg) { console.log(msg); };
page.onResourceRequested = function(request) { console.log('Request ' + JSON.stringify(request, undefined, 4)); };
page.open("https://translate.google.it/?hl=it&tab=wT#fr/it/"+args[1], function(status) {
if (status === "success") { phantom.exit(0); }
else { phantom.exit(1); }
});
so in the end you can get your speech with something like:
wget -U "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:46.0) Gecko/20100101 Firefox/46.0"
"http://translate.google.com/translate_tts?ie=UTF-8&tl=it&tk=52269.458629&q=ciao&client=t" -O ciao.mp3
(token are probably time based so this link may not work tomorrow)
I rewrote Guy Rotem's answer in Java, so if you prefer Java over Javascript, feel free to use:
public class Hasher {
public long shiftLeftOrRightThenSumOrXor(long num, String[] opArray) {
long result = num;
int current = 0;
while (current < opArray.length) {
char op1 = opArray[current].charAt(1); // '+' | '-' ~ SUM | XOR
char op2 = opArray[current].charAt(0); // '+' | '^' ~ SLL | SRL
char xd = opArray[current].charAt(2); // [0-9a-f]
assertError(op1 == '+'
|| op1 == '-', "Invalid OP: " + op1);
assertError(op2 == '+'
|| op2 == '^', "Invalid OP: " + op2);
assertError(('0' <= xd && xd <= '9')
|| ('a' <= xd && xd <='f'), "Not an 0x? value: " + xd);
int shiftAmount = hexCharAsNumber(xd);
int mask = (op1 == '+') ? ((int) result) >>> shiftAmount : ((int) result) << shiftAmount;
long subresult = (op2 == '+') ? (((int) result) + ((int) mask) & 0xffffffff)
: (((int) result) ^ mask);
result = subresult;
current++;
}
return result;
}
public void assertError(boolean cond, String e) {
if (!cond) {
System.err.println();
}
}
public int hexCharAsNumber(char xd) {
return (xd >= 'a') ? xd - 87 : Character.getNumericValue(xd);
}
public int[] transformQuery(String query) {
int[] e = new int[1000];
int resultSize = 1000;
for (int f = 0, g = 0; g < query.length(); g++) {
int l = query.charAt(g);
if (l < 128) {
e[f++] = l; // 0{l[6-0]}
} else if (l < 2048) {
e[f++] = l >> 6 | 0xC0; // 110{l[10-6]}
e[f++] = l & 0x3F | 0x80; // 10{l[5-0]}
} else if (0xD800 == (l & 0xFC00) &&
g + 1 < query.length() && 0xDC00 == (query.charAt(g + 1) & 0xFC00)) {
// that's pretty rare... (avoid ovf?)
l = (1 << 16) + ((l & 0x03FF) << 10) + (query.charAt(++g) & 0x03FF);
e[f++] = l >> 18 | 0xF0; // 111100{l[9-8*]}
e[f++] = l >> 12 & 0x3F | 0x80; // 10{l[7*-2]}
e[f++] = l & 0x3F | 0x80; // 10{(l+1)[5-0]}
} else {
e[f++] = l >> 12 | 0xE0; // 1110{l[15-12]}
e[f++] = l >> 6 & 0x3F | 0x80; // 10{l[11-6]}
e[f++] = l & 0x3F | 0x80; // 10{l[5-0]}
}
resultSize = f;
}
return Arrays.copyOf(e, resultSize);
}
public long normalizeHash(long encondindRound2) {
if (encondindRound2 < 0) {
encondindRound2 = (encondindRound2 & 0x7fffffff) + 0x80000000L;
}
return (encondindRound2) % 1_000_000;
}
/*
/ EXAMPLE:
/
/ INPUT: query: 'hola', windowTkk: '409837.2120040981'
/ OUTPUT: '70528.480109'
/
*/
public String calcHash(String query, String windowTkk) {
// STEP 1: spread the the query char codes on a byte-array, 1-3 bytes per char
int[] bytesArray = transformQuery(query);
// STEP 2: starting with TKK index,
// add the array from last step one-by-one, and do 2 rounds of shift+add/xor
String[] d = windowTkk.split("\\.");
int tkkIndex = 0;
try {
tkkIndex = Integer.valueOf(d[0]);
}
catch (Exception e) {
e.printStackTrace();
}
long tkkKey = 0;
try {
tkkKey = Long.valueOf(d[1]);
}
catch (Exception e) {
e.printStackTrace();
}
int current = 0;
long result = tkkIndex;
while (current < bytesArray.length) {
result += bytesArray[current];
long subresult = shiftLeftOrRightThenSumOrXor(result,
new String[] {"+-a", "^+6"});
result = subresult;
current++;
}
long encondingRound1 = result;
//System.out.println("encodingRound1: " + encondingRound1);
// STEP 3: apply 3 rounds of shift+add/xor and XOR with they TKK key
long encondingRound2 = ((int) shiftLeftOrRightThenSumOrXor(encondingRound1,
new String[] {"+-3", "^+b", "+-f"})) ^ ((int) tkkKey);
//System.out.println("encodingRound2: " + encondingRound2);
// STEP 4: Normalize to 2s complement & format
long normalizedResult = normalizeHash(encondingRound2);
//System.out.println("normalizedResult: " + normalizedResult);
return String.valueOf(normalizedResult) + "."
+ (((int) normalizedResult) ^ (tkkIndex));
}
}
I have been using the followng Macro and it works fine:
Sub PremTable()
Dim i, m, j As Integer
Dim PDFDiv, PDFClass, PDFSex, PDFPlan, LimAge As Variant
Dim FlagD, FlagC, Band, FlagP, FlagB, IssAge, Dur As Integer
PDFClass = Array("N", "S")
PDFSex = Array("M", "F")
PDFDiv = Array("G", "E")
PDFPlan = Array(10, 20, 30)
LimAge = Array(70, 60, 50)
j = 0
For FlagD = 1 To 2
Range("div").Value = PDFDiv(FlagD)
For FlagP = 1 To 3
Range("plan").Value = PDFPlan(FlagP)
For Band = 1 To 3
Range("band").Value = Band
For FlagS = 1 To 2
Range("sex").Value = PDFSex(FlagS)
For FlagC = 1 To 2
Range("class").Value = PDFClass(FlagC)
m = 18
For i = 1 To Range("LimAge").Value - 17
Range("IssAge").Offset(i + j, 0) = m
Range("age").Value = Range("IssAge").Offset(i + j, 0)
Worksheets("input").Range("J4:J76").Copy
Worksheets("Premium Tables").Range("M1").Offset(i + j, 0).PasteSpecial xlPasteValues, Transpose:=True
Range("DIV2").Offset(i + j, 0) = Range("Div")
Range("PLAN2").Offset(i + j, 0) = Range("plan")
Range("BAND2").Offset(i + j, 0) = Range("band")
Range("SEX2").Offset(i + j, 0) = Range("sex")
Range("CLASS2").Offset(i + j, 0) = Range("class")
m = m + 1
Next i
j = j + i - 1
Next FlagC
Next FlagS
Next Band
Next FlagP
Next FlagD
End Sub
Now I have another very similar spreatsheet that I want to use this macro to creat tables, but it always give me the "run time error 9" for all of the arrays having text format variables (for example: Range("class").Value = PDFClass(FlagC) causing an runtime error 9)
Please advise! Thanks very much!
Using Highstock, I've a serie timestamp / value
with different rangeselectors (hour, day, week, month,...) or zoomX
I want to display the average value for the displayed time period.
Now, I can compute the average of the overall series data:
for (i = 0; i < chart.series[0].yData.length; i++) {
total += chart.series[0].yData[i];
}
seriesAvg = (total / chart.series[0].yData.length).toFixed(4); // fix decimal to 4 places
$('#report1').html('<b>Average:</b>: '+ seriesAvg);
How to compute the average only on the displayed datapoints ? And to refresh automatically after zoom ?
Thank you
Use series.processedYData or series.points.
I implemented the same but I notice that processedYdata is loading also one data more than the one showed at the screen.
for that reason i added one additional command to remove the first processedYData:
processedYData.splice(0, 1);
below is the final result of the function
function showStat(objHighStockchart) {
for (j = 0; j < (json_data.length); j++) {
var seriesAvg = 0,
processedYData = objHighStockchart.series[j].processedYData;
processedYData.splice(0, 1);
var seriesMin = Math.min.apply(null, processedYData);
var seriesMax = Math.max.apply(null, processedYData);
var i = 0
var total = 0;
console.log(processedYData);
for (i = 1; i < processedYData.length; i++) {
total += processedYData[i];
}
seriesAvg = (total / processedYData.length).toFixed(2); // fix decimal to 4 places
$('#container_stat' + j).html(
'<br>Statistics for ' + objHighStockchart.series[j].name + '<br>' +
'Total: ' + total + ' logs<br>' +
'Min: ' + seriesMin + ' logs<br>' +
'Avg: ' + seriesAvg + ' logs<br>' +
'Max: ' + seriesMax + ' logs<br>'
+ '---' + processedYData
);
}
};
I was puzzled with one of the question in Microsoft interview which is as given below:
A function should accept a range( 3 - 21 ) and it should print all the consecutive numbers combinations to form each number as given below:
3 = 1+2
5 = 2+3
6 = 1+2+3
7 = 3+4
9 = 4+5
10 = 1+2+3+4
11 = 5+6
12 = 3+4+5
13 = 6+7
14 = 2+3+4+5
15 = 1+2+3+4+5
17 = 8+9
18 = 5+6+7
19 = 9+10
20 = 2+3+4+5+6
21 = 10+11
21 = 1+2+3+4+5+6
could you please help me in forming this sequence in C#?
Thanks,
Mahesh
So here is a straightforward/naive answer (in C++, and not tested; but you should be able to translate). It uses the fact that
1 + 2 + ... + n = n(n+1)/2,
which you have probably seen before. There are lots of easy optimisations that can be made here which I have omitted for clarity.
void WriteAsSums (int n)
{
for (int i = 0; i < n; i++)
{
for (int j = i; j < n; j++)
{
if (n = (j * (j+1) - i * (i+1))/2) // then n = (i+1) + (i+2) + ... + (j-1) + j
{
std::cout << n << " = ";
for (int k = i + 1; k <= j; k++)
{
std::cout << k;
if (k != j) // this is not the interesting bit
std::cout << std::endl;
else
std::cout << " + ";
}
}
}
}
}
This is some pseudo code to find all the combinations if any exists:
function consecutive_numbers(n, m)
list = [] // empty list
list.push_back(m)
while m != n
if m > n
first = list.remove_first
m -= first
else
last = list.last_element
if last <= 1
return []
end
list.push_back(last - 1)
m += last - 1
end
end
return list
end
function all_consecutive_numbers(n)
m = n / 2 + 1
a = consecutive_numbers(n, m)
while a != []
print_combination(n, a)
m = a.first - 1
a = consecutive_numbers(n, m)
end
end
function print_combination(n, a)
print(n + " = ")
print(a.remove_first)
foreach element in a
print(" + " + element)
end
print("\n")
end
A call to all_consecutive_numbers(21) would print:
21 = 11 + 10
21 = 8 + 7 + 6
21 = 6 + 5 + 4 + 3 + 2 + 1
I tested it in ruby (code here) and it seems to work. I'm sure the basic idea could easily be implemented in C# as well.
I like this problem. Here is a slick and slightly mysterious O(n) solution:
void DisplaySum (int n, int a, int b)
{
std::cout << n << " = ";
for (int i = a; i < b; i++) std::cout << i << " + ";
std::cout << b;
}
void WriteAsSums (int n)
{
N = 2*n;
for (int i = 1; i < N; i++)
{
if (~(N%i))
{
int j = N/i;
if (j+i%2)
{
int a = (j+i-1)/2;
int b = (j-i+1)/2;
if (a>0 & a<b) // exclude trivial & negative solutions
DisplaySum(n,a,b);
}
}
}
}
Here's something in Groovy, you should be able to understand what's going on. It's not the most efficient code and doesn't create the answers in the order you cite in your question (you seem to be missing some though) but it might give you a start.
def f(a,b) {
for (i in a..b) {
for (j in 1..i/2) {
def (sum, str, k) = [ 0, "", j ]
while (sum < i) {
sum += k
str += "+$k"
k++
}
if (sum == i) println "$i=${str[1..-1]}"
}
}
}
Output for f(3,21) is:
3=1+2
5=2+3
6=1+2+3
7=3+4
9=2+3+4
9=4+5
10=1+2+3+4
11=5+6
12=3+4+5
13=6+7
14=2+3+4+5
15=1+2+3+4+5
15=4+5+6
15=7+8
17=8+9
18=3+4+5+6
18=5+6+7
19=9+10
20=2+3+4+5+6
21=1+2+3+4+5+6
21=6+7+8
21=10+11
Hope this helps. It kind of conforms to the tenet of doing the simplest thing that could possibly work.
if we slice a into 2 digit, then a = b + (b+1) = 2*b + (0+1)
if we slice a into 3 digit, then a = b + (b+1) + (b+2) = 3*b + (0+1+2)
...
if we slice a into n digit, then a = b + (b+1) +...+ (b+n) = nb + (0+1+n-1)
the last result is a = nb + n*(n-1)/2, a,b,n are all ints.
so O(N) Algorithm is:
void seq_sum(int a)
{
// start from 2 digits
int n=2;
while(1)
{
int value = a-n*(n-1)/2;
if(value < 0)
break;
// meet the quotation we deduct
if( value%n == 0 )
{
int b=value/n;
// omit the print stage
print("......");
}
n++;
}
}