我在点击单选按钮发送AJAX请求的页面上工作。有三个无线电可以列出“未批准”,“已批准”和“全部”的用户。根据无线电选择,我想在表格中显示各自的数据。目前我没有从AJAX函数中的请求页面获得响应。我将非常感谢并感谢任何帮助。 我的代码,
<script>
$('#selection').change
(
function()
{
var selected_value = $("input[name='users']:checked").val();
//till here the code works fine.
$.ajax
(
{
url: "approval_ajax.php",
dataType : "json",
type: "POST",
cache: false,
data: { selected_value : selected_value },
success: function(response)
{
console.log(response);
var len = response.length;
for(var i=0; i<len; i++){
var id = response[i].id;
var email = response[i].email;
var employee_id = response[i].employee_id;
var first_name = response[i].first_name;
var middle_name = response[i].middle_name;
var last_name = response[i].last_name;
var mobile = response[i].mobile;
var created_on = response[i].created_on;
var disabled = response[i].disabled;
var tr_str = "<tr>" +
"<td>" + (i+1) + "</td>" +
"<td>" + email + "</td>" +
"<td>" + employee_id + "</td>" +
"<td>" + first_name + " " + middle_name + " " + last_name + "</td>" +
"<td>" + mobile + "</td>" +
"<td>" + created_on + "</td>" +
"<td><input type='checkbox' name='check[]'" + disabled + "value= '" + id + "' class='checkbox' id='select_all' ></td>" +
"<input type='hidden' value='" + id + "' name='user_id' id='user_id' >" +
"</tr>";
$("#example").append(tr_str);
}
alert("AJAX was a success");
}
}
);
}
);
</script>
现在我的 approval_ajax.php
<?php
session_start();
require("../includes/config.php");
require("../classes/Database.class.php");
$db = new Database(DB_SERVER, DB_USER, DB_PASS, DB_DATABASE);
$return_arr = array();
$status='';
if($_SERVER['REQUEST_METHOD'] == 'POST')
{
$value = filter_input(INPUT_POST, "selected_value");
if (isset($value))
{
$users=$value;
}else{
$users='';
}
switch ($users)
{
case "all":
$sqlQuery = "SELECT * FROM tbl_user WHERE type =3";
break;
case "approved":
$sqlQuery = "SELECT * FROM tbl_user WHERE type =3 AND status =1";
break;
case "unapproved":
$sqlQuery = "SELECT * FROM tbl_user WHERE type =3 AND status =0";
break;
}
$sq = $db->query($sqlQuery);
if ($db->affected_rows > 0) {
while ($row = mysql_fetch_array($sq)) {
$disabled = '';
if ($status == '1') {
$disabled = "disabled = 'disabled' checked='checked' ";
}
$id = $row['id'];
$email = $row['email'];
$employee_id = $row['employee_id'];
$first_name = $row['first_name'];
$middle_name = $row['middle_name'];
$last_name = $row['last_name'];
$mobile = $row['mobile'];
$created_on1 = $row['created_on'];
$created_on = date("d-m-Y", strtotime($created_on1));
$return_arr[] = array("id" => $id,
"email" => $email,
"employee_id" => $employee_id,
"first_name" => $first_name,
"middle_name" => $middle_name,
"last_name" => $last_name,
"mobile" => $mobile,
"created_on" => $created_on
"disabled" => $disabled
);
}
}
header('Content-Type: application/json', true, 200);
echo json_encode($return_arr);
}
答案 0 :(得分:2)
$return_arr
数组中存在语法错误。您应该在created_on => value
上使用逗号。将数组修改为此
$return_arr[] = array("id" => $id,
"email" => $email,
"employee_id" => $employee_id,
"first_name" => $first_name,
"middle_name" => $middle_name,
"last_name" => $last_name,
"mobile" => $mobile,
"created_on" => $created_on,
"disabled" => $disabled
另请不要使用mysql_*
功能。使用mysqli_*
或pdo