javascript - 删除 'onKeyUp' 搜索的多个请求

Question

这是中止使用“onkeyup”搜索时创建的多个请求的正确方法吗？

(注意:对于多个请求，我指的是为字符串的每个单独字符创建的顺序请求。例如:“Hello”创建 'H'、'He'、...'Hello' - 5 个不同的请求)

var insideReq = false;                      /* Initialise to false */
$('#search').keyup( function() {
    var SearchString = $('#search').val();
    if((SearchString.length) >= 3) {
        if(insideReq == true )              /* Check if somebody is inside */
            ajaxReq.abort();                /* If yes, then throw her out */
        insideReq = true;                   /* Inform I am inside */
        var ajaxReq = $.get('search.exec.php', {q: SearchString}, function(ajaxContent) {
            $('#container-list').html("<img src=\"img\\busy.gif\" class=\"busy-indicator\"/>");
            $('#container-list').fadeOut(5);
            $('#container-list').html(ajaxContent);
            $('#container-list').fadeIn(1500);
            insideReq = false;          /* Work done; I am going */
        });
    }
});

Answer 1

看起来不错。很少有代码优化技术可以使您的代码快速执行。

var insideReq = false;                      /* Initialise to false */
$('#search').keyup( function() {
    var SearchString = $('#search').val();
    if((SearchString.length) >= 3) {
        if(insideReq == true )              /* Check if somebody is inside */
            ajaxReq.abort();                /* If yes, then throw her out */
        insideReq = true;                   /* Inform I am inside */
        var ajaxReq = $.get('search.exec.php', {q: SearchString}, function(ajaxContent) {
            $('#container-list')
            .html("<img src=\"img\\busy.gif\" class=\"busy-indicator\"/>");
            .fadeOut(5);
            .html(ajaxContent);
            .fadeIn(1500);
            insideReq = false;          /* Work done; I am going */
        });
    }
});