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css - How to showhide multiple elements at the same time with javascript - Stack Overflow

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So, I have this between my head tags

<script type="text/javascript">
hidden_links = document.getElementsByName("javascript_needed");
    for (i = 0; i < hidden_links.length; i++) {
        hidden_links[i].style.display = "visible";
    }
</script>

And my divs are all similar to

    <div name="javascript_needed" style="display: none;">stuff</div>

the overall goal here, is to have these divs hide when javascript is disabled, and re-enable them when javascript is enabled.... but for whatever reason, my code doesn't work. I ever tried it in the webkit console, and nothing errored =\

So, I have this between my head tags

<script type="text/javascript">
hidden_links = document.getElementsByName("javascript_needed");
    for (i = 0; i < hidden_links.length; i++) {
        hidden_links[i].style.display = "visible";
    }
</script>

And my divs are all similar to

    <div name="javascript_needed" style="display: none;">stuff</div>

the overall goal here, is to have these divs hide when javascript is disabled, and re-enable them when javascript is enabled.... but for whatever reason, my code doesn't work. I ever tried it in the webkit console, and nothing errored =\

Share Improve this question edited Jun 11, 2010 at 14:52 VMAtm 28.4k17 gold badges83 silver badges132 bronze badges asked Jun 11, 2010 at 14:48 NullVoxPopuliNullVoxPopuli 65.2k76 gold badges214 silver badges361 bronze badges 0
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5 Answers 5

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The JavaScript is executed before the divs are in the DOM. The standard way to do something after the DOM is ready is to use jQuery's $(document).ready(function () { });, but there are other ways as well.

The oldschool way is to use <body onload="myfunction()">.

Here's a newer way (edit: put display:none into CSS):

HTML:

<p class='javascript_needed'>hello</p>​

CSS:

.javascript_needed {display:none;}

JavaScript:

​$(document).ready(function () {
    $('.javascript_needed').show();
});
​

Your JS should be setting the div's display to "block" ("visible" isn't a valid value for display).

Also, from the looks of things your elements aren't in the DOM at the time the code is fired (your code doesn't see them yet). Do any of the following:

Place your code anywhere in the document body below the divs

or, use an unobtrusive strategy to fire your function on window load, a la:

function addLoadEvent(func) {
  var oldonload = window.onload;
  if (typeof window.onload != 'function') {
    window.onload = func;
  } else {
    window.onload = function() {
      if (oldonload) {
        oldonload();
      }
      func();
    }
  }
}

addLoadEvent(nameOfSomeFunctionToRunOnPageLoad);

or, Use a JS framework's "ready" functionality, a la jQuery's:

$(function () {
    nameOfSomeFunctionToRunOnPageLoad();
});

"visible" is not a valid value for "display". You're after "inline" or "block".

"visible" and "hidden" are valid values for the "visibility" CSS property.

Difference between display and visible:

  • An element that is visible still takes up space on the page. The adjacent content is not rearranged when the element is toggled between visible and hidden.
  • An element that is display=none will not take up any space on the page. Other display values will cause the element to take up space. For example, display=block not only displays the element, but adds line breaks before and after it.

The disadvantage of showing elements on ready is that they will only flicker in after the page has finished loading. This usually looks odd.

Here's what I usually do. In a script in the <head> of the document (which runs before the body begins to render), do this:

document.documentElement.className = "JS";

Then, any CSS selectors that descend from .JS will only match if JavaScript is enabled. Let's say you give your links a class of javascriptNeeded (a class is more appropriate than a name here). Add this to your CSS:

.javascriptNeeded{
 display: none;
}
.JS .javascriptNeeded{

 display: inline;
}

…and the elements will be there from the start, but only if JavaScript is enabled.

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