按同名顺序，然后按字母顺序 [英] Order by same name and then alphabetical

问题描述

我有一个Мебельноепроизводство"值，并且我有一个表，其中包含所有与Мебельноепроизводство"巧合有关的项目.

I have a this value 'Мебельное производство' And I have a table which contains all the related items that has 'Мебельное производство' as a coincidence.

到目前为止，我已按字母顺序对它进行排序，但我需要在表上返回所有与值Мебельноепроизводство"相符的巧合，而其余的均应按字母顺序返回.

So far I sort it in Alphabetical way, but I need to return on the table all the coincidence with the value 'Мебельное производство' and the rest on alphabetical way.

到目前为止，我被困在这里:

So far I'm stucked here:

jQuery(".orderings tr").sort(sort).appendTo('.orderings');
function sort(a, b){
    return (jQuery(b).text()) < (jQuery(a).text()) ? 1 : -1;    
}

有什么想法吗?

推荐答案

您需要先进行比较，然后进行alpha排序

You need to do the first comparison, then the alpha sorting

尽管如此，该匹配项仍未排序:

This one does not sort the matches though:

jQuery(".orderings tr").sort(sort).appendTo('.orderings');
var match = "Мебельное производство";
function sort(a, b) {
    var at = jQuery(a).text();
    var bt = jQuery(b).text();
    if (at.indexOf(match) >= 0)
    {
        return -1;  // force to the top
    }
    return (bt < at) ? 1 : -1;
}

这也对匹配项进行了初始排序:

This one also does the initial sorting of the matches:

function sort(a, b) {
    var at = jQuery(a).text();
    var bt = jQuery(b).text();
    console.log(at.indexOf(match));
    if (at.indexOf(match) >= 0 && bt.indexOf(match) < 0) {
        return -1;
    } else if (bt.indexOf(match) >= 0 && at.indexOf(match) < 0) {
        return 1;
    }
    return (bt < at) ? 1 : -1;
}

var match = "Мебельное производство";
jQuery(".orderings tr").sort(sort).appendTo('.orderings');

JSFiddle:: http://jsfiddle.net/s7nb5o6g/5/

注意:由于排序顺序的副作用，因此不需要进行else if测试，但保留它不会有任何危害.例如 http://jsfiddle.net/s7nb5o6g/6/仍然有效

Note: the else if test should not be required, due to a side-effect of the sort order, but it does no harm to leave it. e.g. http://jsfiddle.net/s7nb5o6g/6/ still works

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