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I wanted on different events, different display modi:

.on("enter leave", function (e) {
    $("#artwork figcaption").style.display(e.type == "none" ? "block" : "none");
});
Jai
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Taki Kiometzis
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1 Answers1

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Instead of putting styles this way i suggest you to use .css(): 

.on("mouseenter", function (e) {
    $("#artwork figcaption").css("display", e.type == "mouseenter" ? "block" : "none");
});

or using a callback in .css(): 

.on("mouseenter", function (e) {
    $("#artwork figcaption").css("display", function(){
        return e.type == "mouseenter" ? "block" : "none";
    });
});
Jai
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  • to use .css was correct, you were right with the event condition, I haven't seen it. I figured out as a result now: .on("enter leave", function (e) { $("#artwork figcaption").css(e.type == "enter" ? {display:"block"} : {display:"none"}); }); – Taki Kiometzis Sep 24 '15 at 14:32