php - 使用向上和向下键突出显示从 php ajax 搜索中获取的搜索文本

Question

我正在尝试切换从 ajax 结果集中获取的搜索结果。实际上，问题在于 onkeyup 和 onkeydown 它不会在搜索结果中上下移动，这非常令人沮丧。但是，我尝试的代码如下:-

代码:-

<input type="search" class="searchnews" name="news" autocomplete="off" />
<div id="suggestion-list"></div>
<style>
 .selected {background: #EAEAEA;}
</style>
<script>
  $('.searchnews').keyup(function(e){
    $('input[name="hidden_news"]').val('');                         
    $('input[name="hidden_news"]').val($.trim($(this).val()));
    if(e.keyCode!=40 && e.keyCode!=38){
            var str = $(this).val();
        $.ajax({
            type:'POST',
            url:'ajax/getnews.php',
            data:'str='+str,
            success: function(data){
                if(data.length <= 2){
                    $("#suggestion-list").css('display','none');
                }
                else{
                    $("#suggestion-list").html('');
                    $("#suggestion-list").css('display','block');
                    $("#suggestion-list").append(data);
                }
            }
        });
    }

});
  $(document).keydown(function(e){
    if (e.keyCode == 38) { 
        var selected = $(".selected");
        $("#suggestion-list ul li").removeClass("selected");
        if (selected.prev().length == 0) {
            selected.siblings().last().addClass("selected");
        } else {
            selected.prev().addClass("selected");
        }
    }
    if (e.keyCode == 40) {
        var selected = $(".selected");
        $("#suggestion-list ul li").removeClass("selected");
        selected.next().addClass("selected");
        if (selected.next().length == 0) {
            selected.siblings().first().addClass("selected");
        } else {
            selected.next().addClass("selected");
        }
    }   
  });
</script>

**getnews.php**

<?php
    if(isset($_POST['str']) && $_POST['str']!=''){
        $str=$_POST['str'];
        $str1 = explode(' ',$str);
        foreach($str1 as $st){
        $sql[] =" title like '%".$st."%'";
        }
        $result=mysql_query("SELECT * FROM news WHERE ".implode('and',$sql)." or  title like '%".$str."%'");
        //$result=mysql_query("SELECT * FROM newsupdate WHERE title LIKE \"%$str%\" and status='1' ORDER BY published_date DESC");
        $i=0;
        if(mysql_num_rows($result) > 0){
                    echo '<ul>';
        while($row=mysql_fetch_array($result)){
        $i++;
        if($i==1){
        ?>
                <li class="selected"><a style="color:#030303;" target="_blank" >
                <span class="text"><?php echo preg_replace("/$str/i",'<span class="highlight">'.$str.'</span>',$row['title']); ?></span>
                </a></li>
        <?php   
        }
        else{
        ?>
            <li><a style="color:#030303;" target="_blank">
            <span class="text"><?php echo preg_replace("/$str/i",'<span class="highlight">'.$str.'</span>',$row['title']); ?>
            </span>
            </a></li>
        <?php
        }   
        }
        echo '</ul>';
        }
    }
?>

请帮我解决这个问题。谢谢。

Answer 1

Jason，能否请您删除我从您的代码中注释掉的行，并告诉我您这样做会发生什么？

$(document).keydown(function(e){
    if (e.keyCode == 38) { 
        var selected = $(".selected");
        $("#suggestion-list ul li").removeClass("selected");
        if (selected.prev().length == 0) {
            selected.siblings().last().addClass("selected");
        } else {
            selected.prev().addClass("selected");
        }
    }
    if (e.keyCode == 40) {
        var selected = $(".selected");
        $("#suggestion-list ul li").removeClass("selected");
        //selected.next().addClass("selected");
        if (selected.next().length == 0) {
            selected.siblings().first().addClass("selected");
        } else {
            selected.next().addClass("selected");
        }
    }   
});