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Javascript Regular Expression with Replace

I want to use the .replace() method with the regular expression /^ %VAR
% =,($|&)/. The following DOESN'T replace the "^default.aspx= ,($|&)"
regular expression with "":
---------------------------------
myStringVar = myStringVar.rep lace("^" + iName + "=,($|&)", "");
---------------------------------

The following DOES replace it though:
---------------------------------
var match = myStringVar.mat ch("^" + iName + "=,($|&)");
if(match != null && match.length 0) myStringVar =
myStringVar.rep lace(match[0], "");
---------------------------------

Am I using the replace method wrong? The following DOES work when
using the replace method for me:
---------------------------------
myStringVar = myStringVar.rep lace(/^default.aspx=, ($|&)/, "");
---------------------------------

But I need to use the "iName" variable instead of "default.as px".
Does replace not let me use strings as regular expressions
like .match() and .search()? Anyone know how I would get .replace()
to work the way I want it to? Reasion I'm curious is because the 2nd
method above takes about 2 seconds to complete on my machine when I
have a feeling the single ".replace() " would probably take 1/2 the
time.

Thanks

NB
Dec 14 '07 #1
1 3407
NvrBst wrote:
I want to use the .replace() method with the regular expression /^ %VAR
% =,($|&)/. The following DOESN'T replace the "^default.aspx= ,($|&)"
regular expression with "":
---------------------------------
myStringVar = myStringVar.rep lace("^" + iName + "=,($|&)", "");
---------------------------------

The following DOES replace it though:
---------------------------------
var match = myStringVar.mat ch("^" + iName + "=,($|&)");
if(match != null && match.length 0) myStringVar =
myStringVar.rep lace(match[0], "");
---------------------------------

Am I using the replace method wrong?
Yes, you do.
The following DOES work when using the replace method for me:
---------------------------------
myStringVar = myStringVar.rep lace(/^default.aspx=, ($|&)/, "");
---------------------------------
The reason for that is that String.prototyp e.replace() takes both a RegExp
object reference and a string value as first argument. However, when passed
a string value, it works differently. While String.prototyp e.match() always
expects a RegExp object reference and converts the argument into one if it
is not one already.

From the ECMAScript Specification, Edition 3 Final:

| 15.5.4.10 String.prototyp e.match (regexp)
|
| If `regexp' is not an object whose [[Class]] property is "RegExp",
| it is replaced with the result of the expression `new RegExp(regexp)' .
| [...]
|
| 15.5.4.11 String.prototyp e.replace (searchValue, replaceValue)
|
| [...]
| If `searchValue' is a regular expression (an object whose [[Class]]
| property is "RegExp"), do the following: If searchValue.glo bal is
| false, then search `string' for the first match of the regular
| expression `searchValue'. If `searchValue.gl obal' is true, then
| search string for all matches of the regular expression
| `searchValue'. [...]
|
| If searchValue is not a regular expression, let `searchString' be
| `ToString(searc hValue)' and search `string' for the first
| occurrence of `searchString'. [...]

So the string argument in your String.prototyp e.replace() call denotes a
*literal* string, including the literal `($|&)', which does not match.
But I need to use the "iName" variable instead of "default.as px".
Does replace not let me use strings as regular expressions
like .match() and .search()?
A string value is not a RegExp object reference, so you can not use it as
one. However, if the implementation does not already do it for you (as in
this case), you can convert it into a Regular Expression manually, provided
you escape potential special characters if they are meant to be literally in
the match.
Anyone know how I would get .replace() to work the way I want it to?
myStringVar = myStringVar.rep lace(
new RegExp("^" + iName + "=,($|&)"),
"");

I don't know what you actually want to match, but you should consider
escaping the `$' with `\\$' if it is not meant as end-of-input.
Reasion I'm curious is because the 2nd method above takes about 2 seconds
to complete on my machine when I have a feeling the single ".replace() "
would probably take 1/2 the time.
ISTM you should consider other factors as reasons for this.
PointedEars
--
realism: HTML 4.01 Strict
evangelism: XHTML 1.0 Strict
madness: XHTML 1.1 as application/xhtml+xml
-- Bjoern Hoehrmann
Dec 14 '07 #2

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