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Thank you. It works for me.
M.Sworna Vidhya
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use this javascript code, but this gives following error:
"'mappedimg' is an unexpected token. Expecting white space. Line 1, position 27."
code is here:
<body onLoad='LoadImgMap('mappedimg','cng');ChangeMappedImgSize('fitinframe','mappedimg');' onResize='LoadImgMap('mappedimg','cng');ChangeMappedImgSize('fitinframe','mappedimg');'>";
thanx in advance
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Instead of your line try following line.
<body onLoad='LoadImgMap("mappedimg","cng");ChangeMappedImgSize("fitinframe","mappedimg");' onResize='LoadImgMap("mappedimg","cng");ChangeMappedImgSize("fitinframe","mappedimg");'>
See that I have converted all the single quote inside the outer single quote into double quote
so that it will interpret your arguments that you have passed inside the function correctly.
HTH
Jinal Desai - LIVE
Experience is mother of sage....
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thnx for ur attention.
but it gives same 16 errors:
the error is : ;expected
and if i remove the code it gives is erron on "onResize". said : not closed
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Now try this one. I think you have just copy pasted my code. By mistakenly I have
printed 'o' of önResize with some different language.
So you just need to replace önResize with onResize.
HTH
Jinal Desai - LIVE
Experience is mother of sage....
modified on Saturday, July 17, 2010 2:29 AM
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When you submit a form via javascript, (it is already done), the value of Submit button is not set. I want this to be set for security reason and as good programming practice. I tried something like this, that should work but it does not.
document.hiddenform.school.value = 'My School';
document.hiddenform.submitbutton.value = 'My Submit';
document.hiddenform.submit() ;
That is I am setting value of of submit button in javascript. While "My School" text is correctly assigned to the school text field, "My Submit" text is no assigned to submit button. Any one know what is the best way to go about it? I am using link to submit form and i want the submit button to be set as well. Well actually I might go without the submit button but just curious if anyone can help in this case, just for testing.
Thanks
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I think I solved the problem. I introduced a hidden field that acts as a button which accomplishes the purpose.
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CodeProject is currently working with an Assistant Professor from Oregon State University to understand better the trends in Javascript development, and our Javascript articles on CodeProject itself.
The survey will ask you to identify 3 pieces of JavaScript code on the CodeProject.com website that you have tried to use, and ask you to describe whether it was easy or hard to use each piece of code.
That's it. No personally-identifiable information will be requested or shared (by Oregon State University rules), and it won't take long.
If you've got a few minutes of spare time and are a Javascript developer, then the survey is at http://experiments.eecs.oregonstate.edu/general/surveyor/[^]
CodeProject and Oregon State University thank you for your help!
Thanks,
Sean Ewington
The Code Project
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I'm trying to create a tableless website. When using external stylesheet, the background-image property alone is not working with my external stylesheet. Other properties are working fine with the external stylesheet. I'm getting fuzzed to continue.
My external stylesheet (style.css) and is placed in folder 'css' code is:
The image is in folder 'images'.
When surfng the google with 'background-image not working in external stylesheet' i found that if we place css in different folder and made a call for an image in differnt folder we've to give mention as
background-image:url("../images/disdat-gold-red.gif");
- even this failed.
Again when searched, i found that instead of saying 'background-image' give code as follows for which they answered it worked fine. And the code is:
background: #000000 url(images/disdat-gold-red.gif) repeat-x; - even this get failed for me.
I'm attaching my html file along with my css file.
Coding part in "style.css"
#header
{
text-align:center;
font-family:arial, verdana, 'times new roman';
font-size:15pt;
font-weight:bold;
color:#0ac0af;
padding:2% 0 15% 0;
background:url(images/disdat-gold-red.gif) no-repeat;
}
HTML code is:
<html>
<head>
<meta name="GENERATOR" content="Microsoft FrontPage 6.0">
<meta name="ProgId" content="FrontPage.Editor.Document">
<meta http-equiv="Content-Type" content="text/html; charset=windows-1252">
<title>Try Web Page</title>
<link type=text/css rel=stylesheet href=css/style.css>
</head>
<body>
<div id="container">
<div id="top">
<div id="header">Inside Header Top Container</div>
<div id="nav">Inside Nav Top Container</div>
</div>
<div id="main">
<div id="left">Inside Left Main Container</div>
<div id="right">Inside Right Main Container</div>
</div>
<div id="footer">Inside footer Container</div>
</div>
</body>
</html>
M.Sworna Vidhya
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I tried your code it is working here fine at my end.
Keep the images in image folder.
Keep style sheet file in css folder.
Both of these folders (image and css) should be
at the same level of your folder hierarchy.
Now when you are referring image file try to refer it as follows.
background-image:url("../images/disdat-gold-red.gif");
HTH
Jinal Desai - LIVE
Experience is mother of sage....
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Thank you, desai. Now it works for me.
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My Pleasure.
Jinal Desai - LIVE
Experience is mother of sage....
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Make all of your references relative to the root (start with /). Or use a base tag.
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Hello Friends
I am using object tag in .html file and getting its Id in javascript code bu using document.getElemenById("view");
When I run in IE,it is returning the exact element.
but when I run the same in Firefox then It is just returning Object html Element.Now the exact elemt whose Id i am passing.
Do I need to change some internet option for browser or do I ned to change in java script code to get exact element.
Thanks In Advance.
Regards
Yogesh
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Try following options..
window.document["objectid"]
window.document.objectid
document.embeds.objectid
document.embeds["objectid"]
document.getElementsByName("objectid")[0]
Refer this[^] link for more information and understanding.
HTH
Jinal Desai - LIVE
Experience is mother of sage....
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Thanks A LOt Mr. Desai.
Very Helpful for me.Thanks Once again.![Rose | [Rose]](https://www.codeproject.com/script/Forums/Images/rose.gif)
Regards
Yogesh
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My Pleasure.
Jinal Desai - LIVE
Experience is mother of sage....
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Watch out for spelling! Try checking HTTP_REFERER instead of HTTP_REFERRER.
Software rusts. Simon Stephenson, ca 1994.
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Yes I am using HTTP_REFERER. It still doesn't work.
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HTTP REFERER is not a good way to do this as it can be faked. If your can you may want to look at using a session stored in a database that can be accessed from both server A and B. This will also require that the servers have a common domain www.mysite.com and users.mydomain.com.
Some people may have another way to handle this too.
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Connection con = DriverManager.getConnection("jdbc:odbc:employee");
please tell me that the methode of defining this "jdbc:odbc:employee" string.
As mentioned in getConnnection detail in netbeans the first part jdbc is fix and the second and the third is subProtocol and subName ... so tell me please what are these three things?
I'm accessing MySQL database employee.sql using java servlet ... but i dont know this above mentioned statment mechanism so i m still failed to complete.
I've also read some bloges/etc but cant ound detail that clear my queation.
So plz me.
Also plz tell me that in most of the examples and also in above given statment they defines database in two forms,
"jdbc:mysql:employee"
or
"jdbc:odbc://localhost/employee"
so what are these location on computer or on my local mechine so shere i keep database(employee.sql) to refer them like these two ways.
THANKS TO ALL HELPERS ...
modified on Wednesday, July 14, 2010 12:23 AM
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Thanks .. But thing still creates problem to me.
In this program they uses database test which is already exist ... where i dont know ..... but in case i create my own database then where i keep it and how to give database path in statment con = DriverManager.getConnection("jdbc:mysql:///test", "root", "secret");
package com.stardeveloper.example;
import java.sql.Connection;
import java.sql.DriverManager;
import java.sql.SQLException;
public class JdbcExample2 {
public static void main(String args[]) {
Connection con = null;
try {
Class.forName("com.mysql.jdbc.Driver").newInstance();
con = DriverManager.getConnection("jdbc:mysql:///test",
"root", "secret");
if(!con.isClosed())
System.out.println("Successfully connected to " +
"MySQL server using TCP/IP...");
} catch(Exception e) {
System.err.println("Exception: " + e.getMessage());
} finally {
try {
if(con != null)
con.close();
} catch(SQLException e) {}
}
}
}
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