Question

我想使用jQuery将一些数据从表单发送到PHP文件。我进行了搜索，发现我必须以JSON形式发送数据并以数组形式接收多个变量。但是我完全感到困惑，这是行不通的。

<input id="username" type="text" class="inputBox">
<input id="password" type="password" class="inputBox">
<button id="submitLogin" class="submitLogin">login</button>
<div id="test"></div>
<div id="test1"></div>

$(document).ready(function() {
  $("#submitLogin").click(function() {
    var superuser = $("#username").val();
    var superpass = $("#password").val();
    $.ajax({
      type: "POST",
      url: 'http://localhost/mainclinic/controllers/login/login.php',
      dataType: 'application/json',
      data: {
        loginid: superuser,
        loginpass: superpass
      },
      cache: false,
      success: function(result) {
        $('#test').html(result[0]);
        $('#test1').html(result[1]);
      }
    })
  })
})

<?php
  include "../config.php";

  if (!$db) 
  {
    die("Connection failed: " . mysqli_connect_error());
  }

  if ($_SERVER['REQUEST_METHOD'] === 'POST') 
  {
    $username = mysqli_real_escape_string($db, $_POST['loginid']);
    $password = mysqli_real_escape_string($db, $_POST['loginpass']);
    $sql = "SELECT id FROM superusers WHERE docid = '$username' and doccpass = '$password'";
    $result = mysqli_query($db, $sql);
    $row = mysqli_fetch_array($result, MYSQLI_ASSOC);

    if(mysqli_num_rows($result) > 0)
    {
      $array = array(success, $username);
      echo json_encode($array);
    }
    else
    {
      $array = array(failed, nousername);
      echo json_encode($array);
    }
  }
?>

Answer 1

将此部分更改为此：

$.ajax
    ({
        type:"POST",
        url:'http://localhost/mainclinic/controllers/login/login.php',
        dataType: "json",
        data:{loginid:superuser,loginpass:superpass},
        cache: false,
        success:function (result) {
            $('#test').html(result.status);
            $('#test1').html(result.result);
        }
    })

这样的PHP代码：

if(mysqli_num_rows($result) > 0)
{
    $array = array('status' => 'success', 'result' => $username);
    echo json_encode($array);
}else
{
    $array = array('status' => 'failed', 'result' => 'nousername');
    echo json_encode($array);
}

如何使用JSON和数组将多变量发送和接收到PHP

1 个答案: