|
Subject: [xsl] how to get detailed information of XSL by javascript From: shiwudao <shiwudao@xxxxxxxxxxx> Date: Wed, 30 Jan 2008 03:04:13 +0000 |
Hi, Anybody knows how to get detailed error message when I use Javascript/XSLTProcessor() under Firefox.
For example, I wrote a XSL1.0 XSL file like:
... < xsl:if test=" $ level & g t 2 "> ...
There should be " & gt; ", but I wrote " & g t "
When I call
var xsltProcessor = new XSLTProcessor();
xsltProcessor.importStylesheet(myxsl);
There will throw an exception with the error message: XSLT-Component returned failure code: 0x80600001 [nsIXSLTProcessor.importStylesheet]
I do not think this information can give more about the error reason.
But in the error console of FireFox, it will display an error message like:
error: source:http s:// xxx.xxx.xxx.com:xxx/xxx/xsl/test.xsl
line:45 column:27
source: --------------------------^
The question is how can get this error message by javascript? thank you.
_________________________________________________________________
LlA9AK#,LmRBAK#,PD6/AK#,!0F_<~!1AK
http://get.live.cn
| Current Thread |
|---|
|
| <- Previous | Index | Next -> |
|---|---|---|
| RE: [xsl] transforming xml with nam, Michael Kay | Thread | Re: [xsl] how to get detailed infor, Martin Honnen |
| RE: [xsl] transforming xml with nam, Michael Kay | Date | Re: [xsl] how to get detailed infor, Martin Honnen |
| Month |