How to get first item 100% and rest in 50% i.e in two columns if the total number of list is an odd number
I am working on a blog listing page. There is a condition required:
If the total number of items is odd then the first item should be in
100% width and the rest in 50%. So that means in two columns.If the total number is even then it would be in two columns (50%
width each).
javascript jquery
New contributor
Vikash Singh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
I am working on a blog listing page. There is a condition required:
If the total number of items is odd then the first item should be in
100% width and the rest in 50%. So that means in two columns.If the total number is even then it would be in two columns (50%
width each).
javascript jquery
New contributor
Vikash Singh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Please provide your attempt codes.
– Shailesh Rathod
2 days ago
Use two different class names and toggle between them? Or an odd/even CSS rule.
– Shilly
2 days ago
Please show your work and where exactly you have an issue to implement the condition you describe.
– trincot
2 days ago
add a comment |
I am working on a blog listing page. There is a condition required:
If the total number of items is odd then the first item should be in
100% width and the rest in 50%. So that means in two columns.If the total number is even then it would be in two columns (50%
width each).
javascript jquery
New contributor
Vikash Singh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I am working on a blog listing page. There is a condition required:
If the total number of items is odd then the first item should be in
100% width and the rest in 50%. So that means in two columns.If the total number is even then it would be in two columns (50%
width each).
javascript jquery
javascript jquery
New contributor
Vikash Singh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Vikash Singh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 days ago
trincot
118k1480111
118k1480111
New contributor
Vikash Singh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
Vikash Singh
11
11
New contributor
Vikash Singh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Vikash Singh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Vikash Singh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Please provide your attempt codes.
– Shailesh Rathod
2 days ago
Use two different class names and toggle between them? Or an odd/even CSS rule.
– Shilly
2 days ago
Please show your work and where exactly you have an issue to implement the condition you describe.
– trincot
2 days ago
add a comment |
Please provide your attempt codes.
– Shailesh Rathod
2 days ago
Use two different class names and toggle between them? Or an odd/even CSS rule.
– Shilly
2 days ago
Please show your work and where exactly you have an issue to implement the condition you describe.
– trincot
2 days ago
Please provide your attempt codes.
– Shailesh Rathod
2 days ago
Please provide your attempt codes.
– Shailesh Rathod
2 days ago
Use two different class names and toggle between them? Or an odd/even CSS rule.
– Shilly
2 days ago
Use two different class names and toggle between them? Or an odd/even CSS rule.
– Shilly
2 days ago
Please show your work and where exactly you have an issue to implement the condition you describe.
– trincot
2 days ago
Please show your work and where exactly you have an issue to implement the condition you describe.
– trincot
2 days ago
add a comment |
active
oldest
votes
Your Answer
StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "1"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Vikash Singh is a new contributor. Be nice, and check out our Code of Conduct.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53945079%2fhow-to-get-first-item-100-and-rest-in-50-i-e-in-two-columns-if-the-total-numbe%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Vikash Singh is a new contributor. Be nice, and check out our Code of Conduct.
Vikash Singh is a new contributor. Be nice, and check out our Code of Conduct.
Vikash Singh is a new contributor. Be nice, and check out our Code of Conduct.
Vikash Singh is a new contributor. Be nice, and check out our Code of Conduct.
Thanks for contributing an answer to Stack Overflow!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53945079%2fhow-to-get-first-item-100-and-rest-in-50-i-e-in-two-columns-if-the-total-numbe%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Please provide your attempt codes.
– Shailesh Rathod
2 days ago
Use two different class names and toggle between them? Or an odd/even CSS rule.
– Shilly
2 days ago
Please show your work and where exactly you have an issue to implement the condition you describe.
– trincot
2 days ago