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I would like to know if it is possible to take a string and turn it an array, and also check its 5th letter. Like "I am walking on the street", check if its 5th letter is "z",for example.

theMagicOne
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    You can do this without converting to an array. There are multiple ways to do this without making an array. One of them is `.charAt()`. https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/charAt – gen_Eric Jan 14 '15 at 19:03
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    `var mystring = "I am walking"; alert(mystring[5]);` –  Jan 14 '15 at 19:04
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    http://stackoverflow.com/help/how-to-ask – Ruan Mendes Jan 14 '15 at 19:05
  • possible duplicate of [How do I split a string, breaking at a particular character?](http://stackoverflow.com/questions/96428/how-do-i-split-a-string-breaking-at-a-particular-character) – Ryan Jan 14 '15 at 20:22

3 Answers3

2

You can use String charAt. But it's zero based so, the first position is zero.

"I am walking on the street".charAt(4); // Get 5th letter

About the array transformation. You can use split function

"I am walking on the street".split(''); // will return ['I',' ','a','m', ...]
nanndoj
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  • Yeah Thanks! I've made a mistake! – nanndoj Jan 14 '15 at 19:07
  • this yields a " " character...cause it takes into account spaces. OP wants letters only...the 5th letter – indubitablee Jan 14 '15 at 19:07
  • @indubitablee The question is not clear enough to indicate that spaces should be skipped, a space is a character too – Ruan Mendes Jan 14 '15 at 19:08
  • @JuanMendes OP said "and also check its 5th letter" – indubitablee Jan 14 '15 at 19:09
  • @JuanMendes actually the string I would use has no spaces. Does it changes the way .split() will be used? – theMagicOne Jan 14 '15 at 19:15
  • No. It doesn't! You can just try – nanndoj Jan 14 '15 at 19:16
  • @indubitablee The OP's last comment proves that spaces are irrelevant :) – Ruan Mendes Jan 14 '15 at 19:17
  • @JuanMendes just because OP changed the scope of the requirements doesnt prove that spaces are irrelevant. The code written and suggested should be solid code and maintainable. With your high reputation, you would agree. This answer only works without spaces. It doesn't prove to be true with the example it uses "I am walking on the street" as .charAt(4) results in a space character and the 5th letter is "a" – indubitablee Jan 14 '15 at 19:26
  • @indubitablee My point is that the OP said letter, but really meant character. The question was kind of vague and you were making a requirement about something that the OP wasn't requiring. – Ruan Mendes Jan 14 '15 at 19:29
  • @nanndoj That's doubtful as the OP has no spaces in their string. – Spencer Wieczorek Jan 14 '15 at 19:35
1

You don't need to use an array, we can just check the index:

if(str.charAt(5) == 'z') 
    // It is
else
    // It's not

Note that this includes a 0th letter. If you start counting with 1 then use (4) instead. This would also count spaces, if you want to only include counting letters where for example "H e l l o" where o is the 8th letter then you need to remove the spaces from the string first:

var stringWithoutSpace = str.replace(/ /g,'');

if(stringWithoutSpace.charAt(5) == 'z') 
    // It is
else
    // It's not
Spencer Wieczorek
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-1

First, its not necessary to create an array, but if you want to you can do 

var string = "This is my string";
string.split("");

And then check the letter with

string[4];//5th letter

You can also use the charAt() statement, like

string.charAt(4);//5th letter
gabzerbinato
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