Question

我正在尝试为php运行jtable演示，但我总是遇到一个错误： “与服务器通信时发生错误”。

我知道我可以列出json编码，因为我试图打印它，但由于该错误而无法获取表内的数据。

提前致谢

的index.html：

    <!DOCTYPE html>
    <html>
    <head>
    <meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Datatable with mysql</title>
 <link href="themes/redmond/jquery-ui-1.8.16.custom.css" rel="stylesheet" type="text/css" />
<link href="Scripts/jtable/themes/lightcolor/blue/jtable.css" rel="stylesheet" type="text/css" />
<script src="scripts/jquery-1.6.4.min.js" type="text/javascript"></script>
    <script src="scripts/jquery-ui-1.8.16.custom.min.js" type="text/javascript"></script>
<script src="Scripts/jtable/jquery.jtable.js" type="text/javascript"></script>


<div class="container">
  <div class="">
    <h1>jTable Demo Server Side</h1>
    <div class="col-sm-8" id="employee_grid">

    </div>
  </div>
</div>
<script type="text/javascript">
$( document ).ready(function() {
$('#employee_grid').jtable({
        title: 'List of Employees',
               defaultSorting: 'Name ASC',
        actions: {
            listAction: 'response.php?action=list',
            createAction: 'response.php?action=create',
            updateAction: 'response.php?action=update',
            deleteAction: 'response.php?action=delete'
        },
        fields: {
            personID: {
                title: 'EMPId',
                width: '10%',
                edit: false
            },
            name: {
                title: 'Employee Name',
                width: '40%'
            },
            age: {
                title: 'Employee Salary',
                width: '20%'
            },
            recordDate: {
                title: 'Age',
                width: '30%'
            }
        }
    });
    $('#employee_grid').jtable('load');
});
</script>

response.php

   <?php
   try
  {


//Open database connection
$con = mysqli_connect('localhost', 'root', '','aaa') or die("Connection failed: " . mysqli_connect_error());
mysqli_select_db( $con,"aaa");

//Getting records (listAction)
if($_REQUEST["action"] == "list")
{
    //Get records from database
    $result = mysqli_query($con,"SELECT * FROM `people`;");

    //Add all records to an array
    $rows = array();
        while($row = mysqli_fetch_array($result))
       {
        $rows[] = $row;
    }

    //Return result to jTable
    $jTableResult = array();
    $jTableResult['Result'] = "OK";
    $jTableResult['Records'] = $rows;
    echo json_encode($jTableResult);
}
else if($_REQUEST["action"] == "create")
{
    //Insert record into database
    $result = mysqli_query($con, "INSERT INTO people(Name, Age, RecordDate)    VALUES('" . $_POST["Name"] . "', " . $_POST["Age"] . ",now());");

    //Get last inserted record (to return to jTable)
    $result = mysqli_query($con , "SELECT * FROM people WHERE PersonId = LAST_INSERT_ID();");
    $row = mysqli_fetch_array($result);

    //Return result to jTable
    $jTableResult = array();
    $jTableResult['Result'] = "OK";
    $jTableResult['Record'] = $row;
    print json_encode($jTableResult);
}


//Creating a new record (createAction)

//Close database connection
mysqli_close($con);

}
catch(Exception $ex)
{
//Return error message
$jTableResult = array();
$jTableResult['Result'] = "ERROR";
$jTableResult['Message'] = $ex->getMessage();
echo json_encode($jTableResult);
}

?>

那是我不能让这个演示工作。

JTables php给出错误

0 个答案: