jQuery match part of class with hasClass
You can use the startswith CSS3 selector to get those divs:
$('div[class^="project"]')
To check one particular element, you'd use .is()
, not hasClass
:
$el.is('[class^="project"]')
For using the exact /project\d/
regex, you can check out jQuery selector regular expressions or use
/(^|\s)project\d(\s|$)/.test($el.attr("class"))
$('div[class*="project"]')
will not fail with something like this:
<div class="some-other-class project1"></div>
A better approach for your html would be: I believe these div's share some common properties.
<div class="project type1"></div>
<div class="project type2"></div>
<div class="project type3"></div>
<div class="project type4"></div>
Then you can find them using:
$('.project')
$('div[class^="project"]')
will fail with something like this:
<div class="some-other-class project1"></div>
Here is an alternative which extends jQuery:
// Select elements by testing each value of each element's attribute `attr` for `pattern`.
jQuery.fn.hasAttrLike = function(attr, pattern) {
pattern = new RegExp(pattern)
return this.filter(function(idx) {
var elAttr = $(this).attr(attr);
if(!elAttr) return false;
var values = elAttr.split(/\s/);
var hasAttrLike = false;
$.each(values, function(idx, value) {
if(pattern.test(value)) {
hasAttrLike = true;
return false;
}
return true;
});
return hasAttrLike;
});
};
jQuery('div').hasAttrLike('class', 'project[0-9]')
original from sandinmyjoints: https://github.com/sandinmyjoints/jquery-has-attr-like/blob/master/jquery.hasAttrLike.js (but it had errrors so I fixed it)