“找不到匹配的记录”未显示在“引导表”中

Question

我正在为我的网站使用“ bootstrap-table”（ https://github.com/wenzhixin/bootstrap-table ）插件，目前正在使用它的服务器端分页属性。

该表已正确填充，并且搜索功能也起作用。 但是，当搜索不存在的内容时，它将显示所有记录， 而不是显示“未找到匹配的记录”。

这是我正在使用的html代码...

<table data-toggle="table"
       data-url="1.php"
       data-pagination="true"
       data-side-pagination="server"
       data-page-list="[5, 10, 20, 50, 100, 200]"
       data-search="true"
       data-height="300">
    <thead>
    <tr>
        <th data-field="state" data-checkbox="true"></th>
        <th data-field="memberID" data-align="right" data-sortable="true">Member ID</th>
        <th data-field="name" data-align="center" data-sortable="true"> Name</th>
        <th data-field="dob" data-sortable="true">Date of Birth</th>
    </tr>
    </thead>


</table>

这是我用来创建JSON响应的php脚本...

<?php

require_once('db-connect.php');

if(isset($_GET["limit"])) {
    $limit = $_GET["limit"];
} else {
    $limit = 10;
}

if(isset($_GET["offset"])) {
    $offset = $_GET["offset"];
} else {
    $offset = 0;
}

if(isset($_GET["sort"])) {
    $sort = $_GET["sort"];
} else {
    $sort = "";
}

if(isset($_GET["order"])) {
    $order = $_GET["order"];
} else {
    $order = "asc";
}

 if(isset($_GET["search"])) {
    $search = $_GET["search"];
 } else {
    $search = "";
 }


if($search == "") {
    $result = mysqli_query($connection,"select memberID, name, dob from memberdetails" );
} else {

    $result = mysqli_query($connection,"select memberID, name, dob from memberdetails WHERE memberID LIKE '%$search%'"  );
}


$row = array();

if ( mysqli_num_rows($result) > 0 ) {
            while($row = mysqli_fetch_assoc($result)) {
                $result_2d_arr[] = array (  'memberID' => $row['memberID'],
                                            'name' => $row['name'],
                                            'dob' => $row['dob']);
            }




//get the result size
$count = sizeof($result_2d_arr);

//order the array
if($order != "asc") {
    $result_2d_arr = array_reverse($result_2d_arr);
}

//get the subview of the array
$result_2d_arr = array_slice($result_2d_arr, $offset, $limit);



echo "{";
echo '"total": ' . $count . ',';
echo '"rows": ';
echo json_encode($result_2d_arr);
echo "}";

}

?>

JSON响应如下...

{"total": 23,"rows": [{"memberID":"1","name":"asd","dob":"2015-06-03"},{"memberID":"2","name":"asd","dob":"2015-06-03"},{"memberID":"3","name":"asd","dob":"2015-06-03"},{"memberID":"4","name":"asd","dob":"2015-06-03"},{"memberID":"5","name":"asd","dob":"2015-06-03"},{"memberID":"6","name":"asd","dob":"2015-06-03"},{"memberID":"7","name":"asd","dob":"2015-06-03"},{"memberID":"8","name":"asd","dob":"2015-06-03"},{"memberID":"9","name":"asd","dob":"2015-06-03"},{"memberID":"10","name":"asd","dob":"2015-06-03"}]}

Answer 1

终于...让它工作了。 我将此代码放在其他部分。

else {
$count=0;
echo "{";
echo '"total": ' . $count . ',';
echo '"rows": ';
echo json_encode(array());
echo "}";
}