ajax数据响应在php mysql中始终为0 [英] ajax data response always 0 in php mysql

问题描述

我想通过ajax php将一些数据从mysql加载到div.自动递增的id字段始终等于零.

I want to load some data from mysql to a div via ajax php. The id field which is auto increment is always equal to zero.

我尝试使用get，post等，但是没有任何作用

I have tried using get, post etc but none is working

<script>
  $(document).ready(function(){

   $(document).on('click', '#getUser', function(e){

    e.preventDefault();

    var uid = $(this).data("customer_id");   // it will get id of clicked row

    $('#txtHint').html(''); // leave it blank before ajax call
    $('#mySidenav').show();      // load ajax loader

    $.ajax({
      url: 'getCustomerDetails.php',
      type: 'GET',
      data: 'customer_id='+uid,
      dataType: 'html'
    })
    .done(function(data){
      console.log(data);  
      $('#txtHint').html('');    
      $('#txtHint').html(data); // load response 
      $('#modal-loader').hide();      // hide ajax loader 
    })
    .fail(function(){
      $('#txtHint').html('<i class="glyphicon glyphicon-info-sign"></i> Something went wrong, Please try again...');
      $('#modal-loader').hide();
    });

  });

 });

</script>


        <?php

include_once('../../config/dbconfig.php');


    if (isset($_REQUEST['customer_id'])) {

        $id = intval($_REQUEST['customer_id']);
        $query = "SELECT * FROM customers WHERE customer_id=:id";
        $stmt = $pdo->prepare( $query );
        $stmt->execute(array(':id'=>$id));
        $row=$stmt->setFetchMode(PDO::FETCH_ASSOC);


        ?>
        <div class="row">
            <div class="col-md-1">
            <div class="col-md-4" >
                <?php echo $row['first_name'];?>
            </div>
            <div class="col-md-6">

            </div>
        </div>
    </div>


    <?php echo $id;?><br/>


<?php
}

?>

由于$ id始终为0，因此结果没有回显firs_name. 我想获取单个$ id(auto_increment) 完成后应显示用户记录

The result is not echo the firs_name because the $id is always 0. I want to get the individual $id(auto_increment) when is done it should display user record

推荐答案

从调试数据库的实际结果开始.

Start with debugging your actual result from the database.

if (isset($_REQUEST['customer_id'])) {

        $id = intval($_REQUEST['customer_id']);
        $query = "SELECT * FROM customers WHERE customer_id=:id";
        $stmt = $pdo->prepare( $query );
        $stmt->execute(array(':id'=>$id));
        $row=$stmt->setFetchMode(PDO::FETCH_ASSOC);

您不检查错误.

两个建议:

1)您正在使用<?php echo $row['first_name'];?>.如果您检查了结果集，您会发现问题出在哪里.只需使用print_r()等在(错误命名的)$ row变量中输出结果.我相信您会发现问题所在.

1) You are using <?php echo $row['first_name'];?>. If you inspected the resultset you could see what is wrong with that. Just output the result in the (wrongly named) $row variable with print_r() and the like. I am sure you will see what went wrong.

2)我强烈建议再次使用$ _REQUEST.这是懒惰且容易出错.您知道'customer_id'的来源吗?会议?曲奇饼?邮政?还是得到?如果要通过GET =>传递信息，请使用GET

2) I strong advise AGAINST using $_REQUEST. It is lazy and errorprone. Do you know where the 'customer_id' came from? Session? Cookie? POST? Or Get? If you are passing information via GET => use GET

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