Javascript：在表中查找所有“输入”
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Javascript: find all "input" in a table
提问by sid_com
Is there a shorter way to write this in JavaScript?
有没有更短的方法在 JavaScript 中编写它？
var data = [];
var table = document.getElementById( 'address' );
var rows = table.getElementsByTagName( 'tr' );
for ( var x = 0; x < rows.length; x++ ) {
var td = rows[x].getElementsByTagName( 'td' );
for ( var y = 0; y < td.length; y++ ) {
var input = td[y].getElementsByTagName( 'input' );
for ( var z = 0; z < input.length; z++ ) {
data.push( input[z].id );
}
}
}
采纳答案by Jon
element.getElementsByTagNamefinds all descendants, not only children, so:
element.getElementsByTagName找到所有后代，而不仅仅是孩子，所以：
<script type="text/javascript>
var data = [];
var table = document.getElementById( 'address' );
var input = table.getElementsByTagName( 'input' );
for ( var z = 0; z < input.length; z++ ) {
data.push( input[z].id );
}
</script>
回答by nnnnnn
Yep.
是的。
var data = [],
inputs = document.getElementById('address').getElementsByTagName('input');
for (var z=0; z < inputs.length; z++)
data.push(inputs[z].id);
Note, even in your longer version with three loops you can also say:
请注意，即使在具有三个循环的较长版本中，您也可以说：
var rows = table.rows;
// instead of
var rows = table.getElementsByTagName('tr');
// and, noting that it returns <th> cells as well as <td> cells,
// which in many cases doesn't matter:
var tds = rows[x].cells;
// instead of
var tds = rows[x].getElementsByTagName('td');
回答by ?ime Vidas
For modern browsers :)
对于现代浏览器 :)
var table, inputs, arr;
table = document.getElementById( 'test' );
inputs = table.querySelectorAll( 'input' );
arr = [].slice.call( inputs ).map(function ( node ) { return node.id; });
Live demo:http://jsfiddle.net/HHaGg/
现场演示：http : //jsfiddle.net/HHaGg/
So instead of a forloop, I make use of the mapmethod - each array element (each INPUT node) is replaced with its ID value.
因此for，我使用了该map方法而不是循环- 每个数组元素（每个 INPUT 节点）都替换为其 ID 值。
Also note that `inputs.map(...doesn't work since inputsis a NodeListelement - it's an array-like object, but not an standard array. To still use the mapmethod on it, we just have to transform it into an array which is what [].slice.call( inputs )does for us.
另请注意，`inputs.map(...它不起作用，因为它inputs是一个NodeList元素 - 它是一个类似数组的对象，但不是标准数组。要仍然map在其上使用该方法，我们只需要将其转换为一个数组，这[].slice.call( inputs )对我们有用。
