Question

如何将以下php更改为ajax？

<?php

    $selectMSGs = 'SELECT * FROM group_messages WHERE group_id="'.$_GET["id"].'" ORDER BY msg_id DESC';
    $selectMSGs2 = $connect->query($selectMSGs);
    while ($rowMSGs = $selectMSGs2->fetch_assoc()) {

        echo "<div class='lead MessageOfPerson'>";
        echo "<div class='NameOfPerson'>";
        echo "<p>";
        echo $rowMSGs['msg_sender'];
        echo "</div>";
        echo "<div class='MessageText'>";
        echo nl2br($rowMSGs['group_msg']);
        echo "</p>";
        echo "</div>";
        echo "</div>";
        echo "<hr>";

    }

?>

我已经有一些已经在这个页面上运行的其他ajax课程（不是我自己的），是否可以将这个php与我当前的ajax和php文件结合起来？文件如下所示：

AJAX

<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.1.1/jquery.min.js"></script>

  <script type="text/javascript">

    function updateChat(){

      $.ajax({

        //TODO set localhost to the strato livechat file
        url: "http://localhost:8080/LiveChat/LiveChat/code/php/groups/liveChat.php",
        type: "get",
        data: "msg_id="+$('#msg_id').val(),
        dataType: "json",
        success: function(response, status, http){

          $.each(response, function(index, item){

            $('#msg_id').val(item.msg_id);

            //TODO update messages

          });
        },
        error: function(http, status, error){

          alert( 'Error updating chat, ' + error);

        }
      });

    }

    //auto update chat
    setInterval( updateChat, 2000 );

</script>

PHP

<?php

    // Lets fetch the value of msg_id
    $data = $_REQUEST;
    $msg_id = $data['msg_id'];

    // Connect to MySQL Server TODO change to strato
    $con = mysqli_connect( "localhost" , "root" , "" , "db2723249" );

    // If msg_sender and group_msg is available then
    // add it in table chats
    if(
        isset( $data['msg_sender'] ) &&
        isset( $data['group_msg'] )
    ) {

        $insert = "
            INSERT INTO group_messages (group_id, group_msg, msg_sender)
            VALUES( '".$data['msg_sender']."' , '".$data['group_msg']."' )
        ";
        $insert_result = mysqli_query( $con , $insert );
    }

    $select = "SELECT *
                FROM group_messages
                WHERE msg_id > '".$msg_id."'
            ";
    $result = mysqli_query( $con , $select );

    $arr = array();
    $row_count = mysqli_num_rows( $result );

    if( $row_count > 0 ) {
        while( $row = mysqli_fetch_array( $result ) ) {
            array_push( $arr , $row );
        }
    }

    // Close the MySQL Connection
    mysqli_close( $con );

    // Return the response as JSON
    echo json_encode( $arr );

?>

Answer 1

好的，经过大约20个小时的工作后，我终于找到了怎么做（没有帮助:)！我能够合并新旧的ajax和php。

新的ajax成了

php成了：

require 'prime'

def max_number_primes(n)
  return 0 if n < 2
  t = 1
  Prime.each_with_object([]) do |p, a|
    tmp = p*t
    return a if tmp > n
    a << p
    t = tmp
  end
end

max_number_primes(50)              #=> [2, 3, 5]
max_number_primes(50_000)          #=> [2, 3, 5, 7, 11, 13] 
max_number_primes(10_000_000_000)  #=> [2, 3, 5, 7, 11, 13, 17, 19, 23, 29]

我怎么能把这个PHP改成ajax？

1 个答案: