使用php和ajax的Livesearch不工作

Question

我有一个，它基本上是从一个MySQL数据库中提取制造项目，并显示在一个下拉列表中，当用户输入他们的搜索词时，每一行一个项目，就像在Google中搜索一样。

我需要的是一种允许用户单击特定链接项的方法，并为此在同一页面上显示数据，就在单击的项目(链接)下方。

任何帮助都将不胜感激。

1.HTML表单

<form class="navbar-form navbar-left" action="javascript:">
  <div class="input-group">
    <input type="text" class="form-control" id="searchbox1" name="q" token="<?=$csrf->token()?>" action='search1' placeholder="Search for Templates" autocomplete="off">
      <div class="input-group-btn">
        <button class="btn btn-default "  id="searchbtn1" type="submit">
        <i class="fa fa-search"></i></button>
      </div>
  </div>
  <div id="livesearch1"></div>
</form>

2. AJAX调用分离的.js文件

    $('#searchbox1').on('keyup focus', function(e) {    
        var b = $(this).attr();
        delete b.class, delete b.placeholder, delete b.id, delete b.name, delete b.type, delete b.autocomplete;

        b.q = $(this).val();

        if (b.q != '' && b.q.length > 0) {
            $.ajax({

                type: "POST",
                url: api,                            
                data: b,
                cache: false,
                success: function(a) {

                    $("#livesearch1").html(a);
                    $("#livesearch1").fadeIn();
                }
            });

        } else {
            $("#livesearch1").fadeOut();
        }
    });
    $('#searchbox1').on('blur', function(e) {

        $('#livesearch1').fadeOut();
    });

3.api调用

case 'search':

                 if($app->isAdmin() || $app->isEditor() || $app->isUser())
                {           
                          $app->escape('q');

                          ob_start();

                          ajaxsearch($q);

                          echo $result = ob_get_clean();

                         // json('success','true','results',$result);
                }

            break;

4. .php文件

    function ajaxsearch1($q){ 

         $db = MysqliDb::getInstance();
        $csrf = new Csrf_Protect(); 

        $q = removeWhiteSpace($q);
        $q = htmlspecialchars_decode($q,ENT_QUOTES);
        $q = preg_replace('/[^a-zA-Z0-9.-. .).(]/', '', $q);

        if(strlen($q) >0 )
        {

          $term = $q; 
          $searchterm = explode(' ',$term);

          $searchColumns = array("name","slug");
          $searchCondition = '';

            for($i = 0; $i < count($searchColumns); $i++)
            {
                $searchFieldName = $searchColumns[$i];
                $searchCondition .= "($searchFieldName LIKE '%" . implode("%' AND $searchFieldName LIKE '%", $searchterm) . "%')";
                if($i+1 < count($searchColumns)) $searchCondition .= " OR ";
            }

            $res = $db->rawQuery("SELECT * FROM tbl_templates WHERE ($searchCondition) AND (version='1') order by id desc Limit 10 ");

            foreach($res as $sr)
                {  ?><a href="response/<?=$sr['slug']?>" title="<?=ucfirst($sr['name'])?>"><li><?=ucfirst($sr['name'])?></li></a><?php
                }

            }

    }
    ?>

Answer 1

这是一个参数问题。

数据不在$q中，而是在$name代码中，应该是这样的。

case 'search':

     if($app->isAdmin() || $app->isEditor() || $app->isUser())
     {           
         $app->escape('name');
         ob_start();
         ajaxsearch($name);
         echo $result = ob_get_clean();
     }
     break;