javascript - How do you create a table using jQuery? -


what i'm trying create table (7 columns , 5 rows) using jquery. days of week (eg. monday, tuesday, etc.)

this html version of i'm trying create in jquery:

<table id="table">   <tr>     <th>sunday</th><th>monday</th><th>tuesday</th><th>wednesday</th><th>thursday</th><th>friday</th><th>saturday</th>   </tr>   <tr>     <td>1</td><td>1</td><td>1</td><td>1</td><td>1</td><td>1</td><td>1</td>   </tr>   <tr>     <td>2</td><td>2</td><td>2</td><td>2</td><td>2</td><td>2</td><td>2</td>   </tr>   <tr>     <td>3</td><td>3</td><td>3</td><td>3</td><td>3</td><td>3</td><td>3</td>   </tr>   <tr>     <td>4</td><td>4</td><td>4</td><td>4</td><td>4</td><td>4</td><td>4</td>   </tr>

i need style background-color 1 color. rest of rows alternating colors. please me solve problem.

this have:

<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.11.2/jquery.min.js"></script> <script>     $(document).ready(function() {       for(var i=0; < 5; i++) {            for(var j=0; j < 7; j++) {           $("#container").append( "sunday" + "monday" + "tuesday" + "wednesday" + "thursday" + "friday" + "saturday" + );           if (i == 0) {              $("th").css("background-color", "#ccc");           } else if {             $(this).css("background-color", "red");           }         }       }       $("#container").append('</table>');     }); </script>

here's 1 way...

var days = ["sunday", "monday", "tuesday", "wednesday", "thursday", "friday", "saturday"],      rowcount = 4;    $(document).ready(function() {      var table = $('<table/>'),          head = $('<thead/>'),          body = $('<tbody/>'),          headrow = $('<tr/>');        $.each(days, function(index, day) {          headrow.append('<th>' + day + '</th>');      });      head.append(headrow);        (var = 0; < rowcount; i++) {          var brow = $('<tr/>');          (var j = 0; j < days.length; j++) {              brow.append('<td>' + (i + 1) + '</td>');          }          body.append(brow);      }      table.append(head).append(body).appendto('body');  });
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>


Comments

Post a Comment

Popular posts from this blog

c++ - Difference between pre and post decrement in recursive function argument -

i have following sample code used pre-decrement void function(int c) { if(c == 0) { return; } else { cout << "dp" << c << endl; function(--c); cout << c << endl; return; } } this function give output input 4 : dp3 dp2 dp1 dp0 0 1 2 3 but when use post decrement void function(int c) { if(c == 0) { return; } else { cout << "dp" << c << endl; function(c--); cout << c << endl; return; } } output goes in infinite loop dp4 can explain me in detail why happens ? this happens because function(c--) called same value of c , when finishes c decremented. recursively called, hence called same value no chance of return , hit stack overflow error. assume this, int x = 1, y = 0; y = x++ result y == 1 , x == 2. if y = ++x , bot x , y 2.
Read more

php - Nothing but 'run(); ' when browsing to my local project, how do I fix this? -

i added new project. made mistake in projectname.conf file in sites-available apache2, led me reinstalling apache2. after reinstalling , re-enabling projects on every page of every project .. <body> run(); </body> .. i've run composer update, restarted service, restarted pc, checked config files, tried disabling new project. there nothing in error log, , access.log: 127.0.0.1 - - [15/may/2015:01:40:47 +0200] "get / http/1.1" 304 181 "-" "mozilla/5.0 (x11; ubuntu; linux x86_64; rv:38.0) gecko/20100101 firefox/38.0" maybe i'm forgetting apache2 reinstall? don't think did different last time installed it. looks haven't setup php correctly. take @ this great guide , maybe you're forgetting link php config apache2.
Read more

php - How can I echo out this array? -

i don't know how go grabbing data in variable , put foreach loop. class tree{ private $info = array( array('name'=>'jim','state'=>'nu','company'=>'nu','phone'=>array('cell'=>'5615111111','office'=>'5611111111'),'email'=>array('primary'=>'exs@example.com','ex@there.com')), array('name'=>'joe smith','city'=>'phoenix','phone'=>'4805551111','email'=>'jsmith@some_email.com'), array('name'=>'john doe','city'=>'chandler','company'=>'doe co.','email'=>array('jdoe@gmail.com','personal'=>'email@email.com'),'phone'=>'6025550002') ); } if you're inside of class, can access private variable using $this->variablename . example, i...
Read more