Change a single character in a string

[Tom de Neef]

I need to change one character at a known position in a string.
In Pascal I would change the P's character of a string S into 'x' with
S[P]:='x';
In JavaScript I come no further than S = S.substr(0,P-2)+'x'+S.substr (P)

Is there really no more trivial way? I'm looking for the write equivalent of
S.charAt(P).
TIA
Tom

[Evertjan.]

Tom de Neef wrote on 17 feb 2008 in comp.lang.javas cript:

I need to change one character at a known position in a string.
In Pascal I would change the P's character of a string S into 'x'

with

S[P]:='x';
In JavaScript I come no further than S =
S.substr(0,P-2)+'x'+S.substr (P)

Is there really no more trivial way? I'm looking for the write
equivalent of S.charAt(P).

A string cannot be changed in JS, only replaced.

<script type='text/javascript'>

function replaceOneChar( s,c,n){
var re = new RegExp('^(.{'+ --n +'}).(.*)$','') ;
return s.replace(re,'$ 1'+c+'$2');
};

alert( replaceOneChar( 'abcde','X',3) ); // abXde

</script>
--
Evertjan.
The Netherlands.
(Please change the x'es to dots in my emailaddress)

[Tom de Neef]

"Evertjan." <ex************ **@interxnl.net schreef in bericht
news:Xn******** ************@19 4.109.133.242.. .

Tom de Neef wrote on 17 feb 2008 in comp.lang.javas cript:

>I need to change one character at a known position in a string.
In Pascal I would change the P's character of a string S into 'x'

with

>S[P]:='x';
In JavaScript I come no further than S =
S.substr(0,P-2)+'x'+S.substr (P)

Is there really no more trivial way? I'm looking for the write
equivalent of S.charAt(P).

A string cannot be changed in JS, only replaced.

Thank you EJ
Tom

[Bart Van der Donck]

Tom de Neef wrote:

I need to change one character at a known position in a string.
In Pascal I would change the P's character of a string S into 'x' with
S[P]:='x';
In JavaScript I come no further than S = S.substr(0,P-2)+'x'+S.substr (P)
Is there really no more trivial way? I'm looking for the write equivalent of
S.charAt(P).

S = S.replace(S.cha rAt(P),'x');

--
Bart

[T. Rex]

In article <47************ ***********@new s.xs4all.nl>, td*****@qolor.n l
says...

I need to change one character at a known position in a string.
In Pascal I would change the P's character of a string S into 'x' with
S[P]:='x';
In JavaScript I come no further than S = S.substr(0,P-2)+'x'+S.substr (P)

You couldn't possibly have tried that, and found it even remotely close
to satisfactory. The second parameter to string.substr() is a length
parameter, not a position index.

As coded, that will:
a) fail, if P < 2
b) delete character P-2 and replace character P-1 with 'x', if P >= 2

If you want to use a position index instead of a length, look at
string.substrin g() or string.slice().

Is there really no more trivial way? I'm looking for the write equivalent of
S.charAt(P).

If by that you mean some means of modifying it directly as in Pascal --
no.

[Bart Van der Donck]

"Evertjan." wrote:

Bart Van der Donck wrote on 17 feb 2008 in comp.lang.javas cript:

>Tom de Neef wrote:

>>I need to change one character at a known position in a string.
In Pascal I would change the P's character of a string S into 'x'
with S[P]:='x';
In JavaScript I come no further than S =
S.substr(0, P-2)+'x'+S.substr (P) Is there really no more trivial way?
I'm looking for the write equivalent of S.charAt(P).

>S = S.replace(S.cha rAt(P),'x');

No that would not work right, Bart,
as it would replace the first appearance of that letter.

You're right. Trying to adapt my code, I come to exactly the same
result as you.

--
Bart

[Evertjan.]

Bart Van der Donck wrote on 18 feb 2008 in comp.lang.javas cript:

"Evertjan." wrote:

>Bart Van der Donck wrote on 17 feb 2008 in comp.lang.javas cript:

>>Tom de Neef wrote:

>>>I need to change one character at a known position in a string.
In Pascal I would change the P's character of a string S into 'x'
with S[P]:='x';
In JavaScript I come no further than S =
S.substr(0 ,P-2)+'x'+S.substr (P) Is there really no more trivial way?
I'm looking for the write equivalent of S.charAt(P).

>>S = S.replace(S.cha rAt(P),'x');

No that would not work right, Bart,
as it would replace the first appearance of that letter.

You're right. Trying to adapt my code, I come to exactly the same
result as you.

I already gave this regex in another branch of this tread:

function replaceOneChar( s,c,n){
var re = new RegExp('^(.{'+ --n +'}).(.*)$','') ;
return s.replace(re,'$ 1'+c+'$2');
};

A non regex solution would be:

function replaceOneChar( s,c,n){
(s = s.split(''))[--n] = c;
return s.join('');
};
--
Evertjan.
The Netherlands.
(Please change the x'es to dots in my emailaddress)

[Dr J R Stockton]

In comp.lang.javas cript message <Xn************ ********@194.10 9.133.242>
, Mon, 18 Feb 2008 08:01:18, Evertjan. <ex************ **@interxnl.net >
posted:

>
I already gave this regex in another branch of this tread:

function replaceOneChar( s,c,n){
var re = new RegExp('^(.{'+ --n +'}).(.*)$','') ;
return s.replace(re,'$ 1'+c+'$2');
};

A non regex solution would be:

function replaceOneChar( s,c,n){
(s = s.split(''))[--n] = c;
return s.join('');
};

There is an overhead to the construction of a RegExp and to the
commencement of each use, but after that the scanning and replacement
will be reasonably fast.

Method split requires the creation of a number of Objects for short-term
use, but after that the replacement will be quick.

With XP sp2 IE6, I find that the two methods are of similar speed for
8-character strings; for a 2-character string, RegExp takes about half
as long again as split; for a 30-character string, split takes about
twice as long as RegExp; for a 90-character string, split takes over
five times as long as RegExp.

--
(c) John Stockton, nr London UK. ?@merlyn.demon. co.uk IE6 IE7 FF2 Op9 Sf3
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