Question

我是Ajax的初学者。我想从主题表中获取数据行，只包含一列主题为varchar（100），在MySQL DB中定义。以下是我的PHP代码。

Data.php

<?php

$con=mysqli_connect("","root","root","DBTemp") or die("</br> Error: " .mysqli_connect_error());

$sql="select * from Subject";
$result = mysqli_query($con,$sql);
while($row = mysqli_fetch_assoc($result))
 {
    echo $row["SUBJECT"];
    //I Want This Value to Be received in my Jquery Page
    //So that i can take certain action based on each Subject.
    //For example creating a select box child elements,options.
 }
?>

的jquery.js

$(document).ready(function()
 {
    var response='';
    $("body").ready(function()
     {
       $.ajax(
          {
            url: '/Data.php',
            type: 'GET'
            success: function(text)
                {
                     response=text;
                }
          });
     });
    $("body").append("<select> /*Get values here as options*/ </select>");
 });

但是所欲望的行动正在逐行获得价值： - 第一行值 - ＆gt;在jquery中采取某些行动; 第二行值来了 - ＆gt;采取行动..; 。。等等。

Answer 1

1）您需要像数组一样使用数据结构，并将其作为json响应传递给您的ajax调用。 2）您需要遍历json数组，这是您可以单独处理每一行并创建嵌套选择选项的地方。

更新

$con=mysqli_connect("","root","root","DBTemp") or die("</br> Error: " 
.mysqli_connect_error());
$sql="select * from Subject";
$result = mysqli_query($con,$sql);
$jsonResult = [];
while($row = mysqli_fetch_assoc($result))
{
$jsonResult[] = $row["SUBJECT"];
}
echo json_encode($jsonResult);

jquery应该看起来像这样

$(document).ready(function()

{
var response='';
$("body").ready(function()
 {
   $.ajax(
      {
        url: '/Data.php',
        type: 'GET'
        dataType : 'JSON',
        success: function(data)
            {
                 //Alert should return an array of your subjects
                 //If it does then you need to iterate through this array and create options manually.
                 alert(data);
            }
      });
 });
$("body").append("<select> /*Get values here as options*/ </select>");

}）;

Answer 2

我会让php函数返回一个json响应。您可以通过两种方式执行此操作：通过while语句服务器端手动构造JSON，或使用json_encode PHP函数并回显服务器端。这样，当您的ajax响应中的客户端返回数据时，您可以将JSON数据解析为JSON对象JSON.parse(json);，然后以结构化方式逐行控制数据。

希望这有帮助！

Answer 3

Data.php

<?php

$con=@mysqli_connect("","root","root","DBTemp");

# Instead of that use header 500 to let javascript side know there is a real error.

if (mysqli_connect_errno())
{
    echo "Could not connect to database : ". mysqli_connect_error();
    header($_SERVER['SERVER_PROTOCOL'] . ' 500 Internal Server Error', true, 500);
    exit();
}


$sql="select * from Subject";

$result = mysqli_query($con,$sql);

if (mysqli_error($con))
{
    echo "Query failed : ".mysqli_error($con);
    header($_SERVER['SERVER_PROTOCOL'] . ' 500 Internal Server Error', true, 500);
    exit();
}

$options = array();

# populate options arrey with using table's id field as key 
# and subject field as value.

while($row = mysqli_fetch_assoc($result))
{
    $options[$row['id']] = $row['subject'];

}

# return json encoded array to parse from javascript.
echo json_encode($options);

Data.php将输出：

{"1":"Subject ID 1","2":"Subject ID 3"}

的jquery.js

$(document).ready(function()
{

    $("body").ready(function()
    {
        $.ajax(
                {
                    url: '/Data.php',
                    type: 'GET',
                    dataType: 'json',  // Let jQuery know returned data is json.
                    success: function(result)
                    {
                        $.each(result, function(id, subject) {
                            # Loop through results and add an option to select box.
                            $("#ajaxpopulate").append( new Option(subject,id) )
                        });

                    }
                });
    });

});

Page p Page.html，身体内部。此选择框将从ajax请求中填充。

 <select id="ajaxpopulate"></select>

如何从mysql表中逐行获取数据，使用jquery（AJAX）在php中编码的查询？

3 个答案: