CODEWITHSUNDEEP
jquery
Daniel
Daniel

Reputation: 4342

find all id with a specific string

I am trying to get all the href tags of all links on a page, and put them in an array. I have the following code, but when I check the console I get the error Uncaught TypeError: Object has no method 'attr' I'm not to sure where to go from here. Any ideas?

Code

function videoLinks() {
    var videoLinks = $("a[id^=a_l_]").each(function() {
        var linkArray = jQuery.makeArray(videoLinks);
        console.log(linkArray.attr("href"));
    });
}

Upvotes: 2

Views: 221

Answers (2)

Ravi Ram
Ravi Ram

Reputation: 24488

This worked for me .. (from 'Matt Ball' code above)

$("a[id^=a_l_]").each(function() {
        console.log(this.value);
    });

Upvotes: 0

Matt Ball
Matt Ball

Reputation: 359786

$.makeArray returns a native JavaScript array, not a jQuery object. Native JavaScript arrays do not have jQuery methods like .attr(). Is this starting to make sense?

Passing videoLinks to $.makeArray simply does not make sense, since you're either passing the function videoLinks, or the function-local videoLinks which is already a jQuery object. So, I think this is more along the lines of what you're trying to do:

function videoLinks() {
    $("a[id^=a_l_]").each(function() {
        console.log(this.href);
    });
}

That will log the href attribute of every <a> element with an id that starts with 'a_l_'. Perhaps instead you'd like to build up an array of those href attributes instead of logging them. Then you'd use .map() and .get():

function videoLinks() {
    var hrefs = $("a[id^=a_l_]").map(function() {
        return this.href;
    }).get(); // ← note the .get() call
}

my ultimate goal is to return one of the links at random

Then you're almost there. Just get a random element from the hrefs array:

function videoLinks() {
    var hrefs = $("a[id^=a_l_]").map(function() {
        return this.href;
    }).get(); // ← note the .get() call

    var randHref = hrefs[Math.floor(Math.random() * hrefs.length)];
    console.log(randHref);
}

Upvotes: 4

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