The jQuery .hasClass() function does not work properly with SVG elements

Question

The jQuery .hasClass() function does not work properly with SVG elements

I am working with a group of SVG elements that are assigned the classes node and link. My goal is to determine whether an element contains either the node or link class when hovering over any of the SVG components. However, I am encountering an issue where the .hasClass() method does not seem to be functioning correctly:

$(".node").hover(function(evt){
    console.log($(this).attr("class")); //outputs "node"
    console.log($(this).hasClass('node')); //returns false
}, function(){console.log("Done");});

Even though the element being hovered on possesses the class node, as evidenced by

console.log($(this).attr("class"));

, the .hasClass() evaluation fails inexplicably. What could be causing this discrepancy? Could it be related to the use of SVG?

javascript jquery html css svg

Answer 1

Answer №1

In SVG, the class attribute of HTML elements has a different interpretation.

$("<b></b>").addClass($(this).attr("class")).hasClass("node")

Alternatively,

/(^|\s)node(\s|$)/.test($(this).attr("class"))

can be used for SVG elements.

UPDATE .hasClass appears to be functioning correctly (at least in IE9 and FF) http://jsfiddle.net/X6BPX/1/

The issue could potentially stem from a syntax error, utilization of an outdated browser, or working with an older version of jQuery.

Answer 2

In SVG, the class attribute of HTML elements has a different interpretation.

$("<b></b>").addClass($(this).attr("class")).hasClass("node")

Alternatively,

/(^|\s)node(\s|$)/.test($(this).attr("class"))

can be used for SVG elements.

UPDATE .hasClass appears to be functioning correctly (at least in IE9 and FF) http://jsfiddle.net/X6BPX/1/

The issue could potentially stem from a syntax error, utilization of an outdated browser, or working with an older version of jQuery.

Answer 3

Answer №2

In the discussions, Bergi highlighted that jQuery has a silent failure when it comes to SVG elements due to the fact that className returns an SVGAnimatedString object instead of a regular DOMString.

To see a comparison, check out this JSFiddle link.

I thought about submitting a pull request for this issue, but after searching through the project, it seems that the jQuery team's position on SVG problems is "won't fix": https://github.com/jquery/jquery/pull/1511

If you are using D3, you can utilize d3.select(this).classed('node'). It's worth noting that D3 handles both HTML and SVG elements correctly.

Answer 4

In the discussions, Bergi highlighted that jQuery has a silent failure when it comes to SVG elements due to the fact that className returns an SVGAnimatedString object instead of a regular DOMString.

To see a comparison, check out this JSFiddle link.

I thought about submitting a pull request for this issue, but after searching through the project, it seems that the jQuery team's position on SVG problems is "won't fix": https://github.com/jquery/jquery/pull/1511

If you are using D3, you can utilize d3.select(this).classed('node'). It's worth noting that D3 handles both HTML and SVG elements correctly.

Answer 5

Answer №3

While not the quickest option available, this method presents a viable solution. Rather than utilizing jQuery's hasClass, one could extract the class attribute as a string and utilize indexOf to search within it. It is worth noting that there may be scenarios where this approach may fall short, thus it is advisable to use it primarily for straightforward projects.

Below is a functional demonstration:

var s = $(this).attr('class');
if( s.indexOf('node')!==-1 ){
    // perform desired action
}

It is important to keep in mind that indexOf yields a result of -1 when no match is found, rather than 0. The latter value is only returned when the substring commences at index 0.

Answer 6

While not the quickest option available, this method presents a viable solution. Rather than utilizing jQuery's hasClass, one could extract the class attribute as a string and utilize indexOf to search within it. It is worth noting that there may be scenarios where this approach may fall short, thus it is advisable to use it primarily for straightforward projects.

Below is a functional demonstration:

var s = $(this).attr('class');
if( s.indexOf('node')!==-1 ){
    // perform desired action
}

It is important to keep in mind that indexOf yields a result of -1 when no match is found, rather than 0. The latter value is only returned when the substring commences at index 0.

Answer 7

Answer №4

In order to extend the functionalities of addClass, removeClass, and hasClass jQuery methods for versions prior to 3.x.x, this hack can be used.

$.fn.extend({
    addSVGClass: function (cls) {
        return this.each(function () {
            var classList = $(this).attr('class');
            if (classList) {
                var classListArr = classList.split(" ");
                if (classListArr.indexOf(cls) === -1) {
                    classListArr.push(cls);
                    classList = classListArr.join(" ").trim();
                    $(this).attr('class', classList);
                }
            } else {
                $(this).attr('class', cls);
            }
        });
    },
    removeSVGClass: function (cls) {
        return this.each(function () {
            var classList = $(this).attr('class');
            if (classList) {
                var classListArr = classList.split(" ");
                if (classListArr.indexOf(cls) !== -1) {
                    delete classListArr[classListArr.indexOf(cls)];
                    classList = classListArr.join(" ").trim();
                    $(this).attr('class', classList);
                }
            }

        });
    },
    hasSVGClass: function (cls) {
        var el = this[0];
        var classList = $(el).attr('class');
        if (classList) {
            var classListArr = classList.split(" ");
            if (classListArr.indexOf(cls) !== -1) {
                return true;

            } else {
                return false;
            }
        }
        return false;
    }
});

Example usage:

$('.svg-element').addSVGClass('selected');

Answer 8

In order to extend the functionalities of addClass, removeClass, and hasClass jQuery methods for versions prior to 3.x.x, this hack can be used.

$.fn.extend({
    addSVGClass: function (cls) {
        return this.each(function () {
            var classList = $(this).attr('class');
            if (classList) {
                var classListArr = classList.split(" ");
                if (classListArr.indexOf(cls) === -1) {
                    classListArr.push(cls);
                    classList = classListArr.join(" ").trim();
                    $(this).attr('class', classList);
                }
            } else {
                $(this).attr('class', cls);
            }
        });
    },
    removeSVGClass: function (cls) {
        return this.each(function () {
            var classList = $(this).attr('class');
            if (classList) {
                var classListArr = classList.split(" ");
                if (classListArr.indexOf(cls) !== -1) {
                    delete classListArr[classListArr.indexOf(cls)];
                    classList = classListArr.join(" ").trim();
                    $(this).attr('class', classList);
                }
            }

        });
    },
    hasSVGClass: function (cls) {
        var el = this[0];
        var classList = $(el).attr('class');
        if (classList) {
            var classListArr = classList.split(" ");
            if (classListArr.indexOf(cls) !== -1) {
                return true;

            } else {
                return false;
            }
        }
        return false;
    }
});

Example usage:

$('.svg-element').addSVGClass('selected');

Answer 9

Answer №5

Make sure to remember to close the function properly

$(".node").hover(function(evt){
    console.log($(this).attr("class")); //will display "node"
    console.log($(this).hasClass('node')); //will return false
}, function(){console.log("Done");});

http://jsfiddle.net/X6BPX/

Answer 10

Make sure to remember to close the function properly

$(".node").hover(function(evt){
    console.log($(this).attr("class")); //will display "node"
    console.log($(this).hasClass('node')); //will return false
}, function(){console.log("Done");});

http://jsfiddle.net/X6BPX/

The jQuery .hasClass() function does not work properly with SVG elements

Answer №1

Answer №2

Answer №3

Answer №4

Answer №5

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