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Urukhayy2016-02-23 17:26:34
Asterisk
Urukhayy, 2016-02-23 17:26:34

Are there any resources that provide virtual telephone exchanges for free?

Are there any resources that provide virtual telephone exchanges for free?

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6 answer(s)
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AntonMZ, 2016-02-24
@AntonMZ

If you mean a virtual PBX, then as an option zadarma.com, but in fact all companies that provide IP telephony already have this service, and many of them on a free basis.

V
vm916, 2016-02-23
@vm916

Put an asterisk on a cheap vds c ds in the region of residence and potential use and configure iptables

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tukoev, 2016-02-24
@tukoev

What does "resources providing free virtual telephone exchanges" mean? I have never seen such a thing. There are shareware PBXs, like Virtua.pro , for example.

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andreloster, 2014-11-16
@habrazhitel

The question was originally asked crazy -_-
If I understand you correctly, then here:

var array = [{a:"word",b:"world"},{a:"2015",b:"people"}];
var result = -1;
for(var i=0; i<array.length; i++) {
    for(var key in array[i]) {
        if(array[i][key] == "people")    result = i;
    }
    if(result > -1)    break;
}

document.write(result);

B
barkalov, 2014-11-16
@barkalov

There is no specific tree search method in Javascript.
You need to recursively traverse the tree (width or height, it doesn't matter) and check each value.

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xmoonlight, 2014-11-16
@xmoonlight

here it is easier to use a non-standard solution:
1. split by the symbol "," (comma)
2. take the remainder of the division of the found element by 2 (the number of elements in the subarray)
As a result, there will be one array of 4 elements
(3rd: 2015) and the 4th (people-entry) - the required one:
2015: 3%2===1 - so it is the 1st.
people: 4%2===0 - so he is 2nd.

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