Question

我有一个AJAX函数，它调用php函数来搜索mysql数据库。该脚本在keyup上触发，问题出在第一次按键时html内容已更新，但在初始keyup event

之后不会更新

如何使用每个键盘后输入的新数据使页面不断更新html内容。

我的AJAX功能，

var searchPath = "<?php echo $searchPath ?>"; 

$("#itemID").keyup(function (){
    var itemID = $(this).val();
    var url = searchPath;
    $.ajax({
        type  : "GET",
        async : false,
        url   : url,
        data  : "itemID=" + encodeURIComponent(itemID),
        cache : false,
        success: function(html) {
              $('#loader_image').hide();
           $( "#productResults" ).replaceWith( html );

              if (html === "") {
                  $("#loader_message").html('<p>There were no results that match your search criteria</p>').show();
              } else {
                  $("#loader_message").html('Searching... Please wait <img src="http://www.example.com/monstroid/wp-content/uploads/2016/02/LoaderIcon.gif" alt="Loading">').show();
              }
              window.busy = false;
        }
      });
});

这就是背后的PHP，

<?php
require_once ('Dbconfig.php');
  $sql=" SELECT * FROM wuno_inventory WHERE wuno_product like '%".$itemID."%' OR wuno_alternates like '%".$itemID."%' ORDER BY wuno_product ";
try {
  $stmt = $DB_con->prepare($sql);
  $stmt->execute();
  $results = $stmt->fetchAll();
} catch (Exception $ex) {
  echo $ex->getMessage();
}
if (count($results) > 0) {
  foreach ($results as $res) {
    echo '<tr class="invent">';  
    echo '<td>' . $res['wuno_product'] . '</td>';  
    echo '<td>' . $res['wuno_alternates'] . '</td>';  
    echo '<td>' . $res['wuno_description'] . '</td>';  
    echo '<td>' . $res['wuno_onhand'] . '</td>';  
    echo '<td>' . $res['wuno_condition'] . '</td>';  
    echo '</tr>';   
  }
}
?>

AJAX搜索功能不更新答案

0 个答案: