WebDesigns | Various Web Designs layout using plain HTML and CSS | Theme library
kandi X-RAY | WebDesigns Summary
kandi X-RAY | WebDesigns Summary
Each folder of this project contains various Web Designs, patterns and also complete Websites. Most are made using pure HTML and CSS, without using any CSS library like bootstrap etc.
Support
Quality
Security
License
Reuse
Top functions reviewed by kandi - BETA
Currently covering the most popular Java, JavaScript and Python libraries. See a Sample of WebDesigns
WebDesigns Key Features
WebDesigns Examples and Code Snippets
Community Discussions
Trending Discussions on WebDesigns
QUESTION
In my project im getting such kind of error when im trying to redirect to another page when the person submit the contact form then it should redirect to some other page but im getting such kind of error. the error what im getting is this
warning: Cannot modify header information - headers already sent by (output started at C:\xampp\htdocs\grade\header.php:89) in C:\xampp\htdocs\grade\contact.php on line 64
header.php
...ANSWER
Answered 2018-Oct-17 at 12:49There are lots of reason for the error: Cannot modify header information - headers already sent below are some of them:
printandechostatements will terminate the opportunity to send HTTP headersbefore anyheader()call.- Whitespace before
- Whitespace after ?>
- Preceding error messages can also be a cause of this error.
If You want to check all reasons in detail then here is referenced stackoveflow answer
But for you the reason of this error is that, you have included Raw sections prior
to code, you can not include html before php code.
In your code you are including header.php and
which is html, before php code therefor you are getting this error, so for solving it change your
contact.php file as below
QUESTION
As the title suggest, I've tried different set-ups and had several other issues.
...ANSWER
Answered 2017-Sep-07 at 10:02It's bind_param when working with mysqli.
It's bindParam when working with PDO.
With MySQLi bind_param, you have to set the variable type, 1,2,3 are not variable types.
Check here for variable types: http://php.net/manual/en/mysqli-stmt.bind-param.php#refsect1-mysqli-stmt.bind-param-parameters
QUESTION
I am new to php. I have a sign up page that takes in few user details and makes an account. On running the code I get this error:
...ANSWER
Answered 2017-May-12 at 02:24Add a closing bracket to line 141, so that the code is:
QUESTION
on the button click of calculate ; i want to calculate all the cost. i'm trying btn the alert is not working on the click.
i'm trying without the variable it is working. but when i calculate all the values and pass it in the alert function . it just doesn't display anything. i have also used intParse() method to typecast..
Help Needed. Much appreciated
...ANSWER
Answered 2017-Apr-09 at 10:39The problem is all your variables are defined inside the functions causing them to cease existing when function ends.
You need to define these variables outside the functions.
Community Discussions, Code Snippets contain sources that include Stack Exchange Network
Vulnerabilities
No vulnerabilities reported
Install WebDesigns
Support
Reuse Trending Solutions
Find, review, and download reusable Libraries, Code Snippets, Cloud APIs from over 650 million Knowledge Items
Find more librariesStay Updated
Subscribe to our newsletter for trending solutions and developer bootcamps
Share this Page
