Question

此代码有问题..个人资料搜索有效正确地只是“查看个人资料”按钮不起作用，我希望我可以显示每个成员的按钮并进入特定页面，你可以帮帮我吗？而且，如何不单击任何内容就无法显示整个查询？非常感谢，对不起，我的英语水平

index.php

 <head>
 <meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
 <title>test search</title>
 <script 
 src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.3/jquery.min.js"> 
 </script>
  <script 


src="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.6/js/bootstrap.min.js"> 
</script>
<link href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.6/css/bootstrap.min.css" 
rel="stylesheet" />
</head>
<body>
  <div class="container">
    <br />
          <h2 align="center">   Ajax Live Data Search</h2> 
  <br />
    <div class="form-group">
        <div class="input-group">
            <span class="input-group-addon">Search</span>
            <input type="text" name="search_text" id="search_text" 
placeholder="Search by Customer Details" class="form-control" />
        </div>
    </div>
    <br />
    <div id="result"></div>
</div>

</html>
  <script>
  $(document).ready(function() {

    load_data();

    function load_data(query) {
        $.ajax({
            url: "fetch.php",
            method: "POST",
            data: {
                query: query
            },
            success: function(data) {
                $('#result').html(data);
            }
        });
    }
    $('#search_text').keyup(function() {
        var search = $(this).val();
        if (search != '') {
            load_data(search);
        } else {
            load_data();
        }
    });
});

fetch.php

 <?php

  $connect = mysqli_connect("localhost", "xxx", "xxx", 
"testing");
$output = '';
if (isset($_POST["query"])) {
$search = mysqli_real_escape_string($connect, $_POST["query"]);
$query = "
SELECT * FROM tbl_customer 
WHERE CustomerName LIKE '%" . $search . "%'
OR Address LIKE '%" . $search . "%' 
OR City LIKE '%" . $search . "%' 
OR PostalCode LIKE '%" . $search . "%' 
OR Country LIKE '%" . $search . "%'
";
 } else {
 $query = "
 SELECT * FROM tbl_customer ORDER BY CustomerID
 ";
}
$result = mysqli_query($connect, $query);
if (mysqli_num_rows($result) > 0) {
$output .= '
<div class="table-responsive">
 <table class="table table bordered">
 <tr>
 <th>Customer Name</th>
 <th>Address</th>
 <th>City</th>
 <th>Postal Code</th>
 <th>Country</th>
 <th> <a class="viewProfile" href="index.php?id=' . ['CustomerID'] . '"> 
 <btn btn-primary>View Profile</btn btn-primary></a><th>
  </tr>
 ';
  while ($row = mysqli_fetch_array($result)) {
  $output .= '
  <tr>
  <td>' . $row["CustomerName"] . '</td>
  <td>' . $row["Address"] . '</td>
  <td>' . $row["City"] . '</td>
  <td>' . $row["PostalCode"] . '</td>
  <td>' . $row["Country"] . '</td>
  <td>' . $row["CustomerID"] . '</td>
  </tr>
   ';
   }
  echo $output;
   }  else {
    echo 'Data Not Found';
   }

在ajax搜索中链接视图配置文件

0 个答案: