php - 通过 jquery 传递搜索参数

Question

我有一个表单，如果用户输入搜索查询，它的参数应该通过 jquery 传递，并且在获得结果后应该将结果加载到 div 容器中。由于我不太精通 jquery，我该怎么做？

html:

    //currently the data is being displayed on pageload:
    $(document).ready(function() {
    $("#bquote").load("quotes_in.php")
   });  

   $(".bsearch")
    .keydown(function() {
    //pass the parameter to the php page

       //display in div #bquote

    });


 <!-- Begin Search -->
        <div id="search" style="display:none;">
                <input type="text" class="bsearch" name="search" value="Search" />
        </div>
<!-- End Search -->

php [使用 OOP]:

if (isset($_POST["search"])) 
{
    $quotes->searchQuotes();
}

PHP 类:

  $search = $_POST['search'];

  $sql = "SELECT * FROM thquotes WHERE cQuotes LIKE '%"
  .  mysql_real_escape_string($search) ."%' ORDER BY idQuotes DESC";


  try {

  $query = $this->_db->prepare($sql);
  $query->execute();

  if(!$query->rowCount()==0)
  {
   while($row = $query->fetch())
    {
    echo $this->formatSearch($row);
}

Answer 1

我会这样做:

$(".bsearch").keydown(function() {
  //create post data
  var postData = { 
    "bsearch" : $(this).val(), 
    "anotherParam" : "anotherValue" 
  };

  //make the call
  $.ajax({
    type: "POST",
    url: "yourAjaxUrl.php",
    data: postData, //send it along with your call
    success: function(response){
      alert(response); //see what comes out
      //fill your div with the response
      $("#bquote").html(response);
    }
  });
});

编辑:

要放置加载程序，您需要在此处检查:

http://api.jquery.com/category/ajax/global-ajax-event-handlers/

并显示加载器图像，例如:

$("#loading").ajaxStart(function(){
   $(this).show();
});

并在 ajax 调用完成时隐藏它:

$("#loading").ajaxComplete(function(){
   $(this).hide();
 });