javascript - Why won't querySelectorAll select all of my elements? - Stack Overflow

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In this example, the querySelectorAll selects td elements 2 through 4:

document.querySelectorAll("td + td");

<table>
    <tr>
        <td>1</td>
        <td>2</td>
        <td>3</td>
        <td>4</td>
    </tr>
</table>

So, it's going to select all td elements greater than or equal to 2.

But in the following example, 'div' is not selecting both of my div elements when I use:

document.querySelectorAll('div').style.color = 'red';

<div>foo</div>
<div>bar</div>

Demo

Why doesn't querySelectorAll work in this basic scenario?

In this example, the querySelectorAll selects td elements 2 through 4:

document.querySelectorAll("td + td");

<table>
    <tr>
        <td>1</td>
        <td>2</td>
        <td>3</td>
        <td>4</td>
    </tr>
</table>

So, it's going to select all td elements greater than or equal to 2.

But in the following example, 'div' is not selecting both of my div elements when I use:

document.querySelectorAll('div').style.color = 'red';

<div>foo</div>
<div>bar</div>

Demo

Why doesn't querySelectorAll work in this basic scenario?

• javascript
• dom

Share Improve this question Follow edited Nov 19, 2014 at 19:17 TylerH 21.1k7979 gold badges7979 silver badges114114 bronze badges asked Jul 7, 2014 at 10:42 Navin RauniyarNavin Rauniyar 10.6k1414 gold badges4747 silver badges7070 bronze badges 3

• querySelectorAll – rogaldh Commented Jul 7, 2014 at 10:43
• 2 See: developer.mozilla/en/docs/Web/API/Document.querySelectorAll. It returns a NodeList, which acts like an array. – Evan Trimboli Commented Jul 7, 2014 at 10:43
• I already checked that but couldn't understand and now seeing the answer below I got concept. – Navin Rauniyar Commented Jul 7, 2014 at 10:45

Add a ment  | 

1 Answer 1

Sorted by: Reset to default 5

querySelectorAll returns an array-like object (NodeList), so you can't set the property of all matched elements like that (like jQuery). You have to iterate over them:

var divs = document.querySelectorAll('div');

for(var i=0; i<divs.length; i++){
    divs[i].style.color = "red";
}

Note: this differs from querySelector, which by contrast always returns a single element.

You can write a helper function like:

function qsa(selector, styleProperty, value){
    var divs = document.querySelectorAll('div');

    for(var i=0; i<divs.length; i++){
        divs[i].style[styleProperty] = value;
    }
    return divs;
}

Usage:

var divs = qsa('div', 'color', 'red');

In this example, the querySelectorAll selects td elements 2 through 4:

document.querySelectorAll("td + td");

<table>
    <tr>
        <td>1</td>
        <td>2</td>
        <td>3</td>
        <td>4</td>
    </tr>
</table>

So, it's going to select all td elements greater than or equal to 2.

But in the following example, 'div' is not selecting both of my div elements when I use:

document.querySelectorAll('div').style.color = 'red';

<div>foo</div>
<div>bar</div>

Demo

Why doesn't querySelectorAll work in this basic scenario?

In this example, the querySelectorAll selects td elements 2 through 4:

document.querySelectorAll("td + td");

<table>
    <tr>
        <td>1</td>
        <td>2</td>
        <td>3</td>
        <td>4</td>
    </tr>
</table>

So, it's going to select all td elements greater than or equal to 2.

But in the following example, 'div' is not selecting both of my div elements when I use:

document.querySelectorAll('div').style.color = 'red';

<div>foo</div>
<div>bar</div>

Demo

Why doesn't querySelectorAll work in this basic scenario?

• javascript
• dom

Share Improve this question Follow edited Nov 19, 2014 at 19:17 TylerH 21.1k7979 gold badges7979 silver badges114114 bronze badges asked Jul 7, 2014 at 10:42 Navin RauniyarNavin Rauniyar 10.6k1414 gold badges4747 silver badges7070 bronze badges 3

• querySelectorAll – rogaldh Commented Jul 7, 2014 at 10:43
• 2 See: developer.mozilla/en/docs/Web/API/Document.querySelectorAll. It returns a NodeList, which acts like an array. – Evan Trimboli Commented Jul 7, 2014 at 10:43
• I already checked that but couldn't understand and now seeing the answer below I got concept. – Navin Rauniyar Commented Jul 7, 2014 at 10:45

Add a ment  | 

1 Answer 1

Sorted by: Reset to default 5

querySelectorAll returns an array-like object (NodeList), so you can't set the property of all matched elements like that (like jQuery). You have to iterate over them:

var divs = document.querySelectorAll('div');

for(var i=0; i<divs.length; i++){
    divs[i].style.color = "red";
}

Note: this differs from querySelector, which by contrast always returns a single element.

You can write a helper function like:

function qsa(selector, styleProperty, value){
    var divs = document.querySelectorAll('div');

    for(var i=0; i<divs.length; i++){
        divs[i].style[styleProperty] = value;
    }
    return divs;
}

Usage:

var divs = qsa('div', 'color', 'red');

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