jQuery Ajax從php返回一些值

Question

我正在使用ajax從php文件中選擇錯誤，我遇到了一些麻煩。 像在php文件中一樣，我將錯誤放入$ email_error和$ password_error中，因此我想將錯誤報告返回給ajax，並將$ email_error分配給id =“ email_errors”，將$ password_error分配給id =“ password_errors”。 也許有人可以解釋我如何指定要返回的變量以及應使用的ID。我將在下面留下一些注釋的代碼。 謝謝！

的PHP

<?php


if (isset($_POST['email']) && isset($_POST['password1']) && isset($_POST['password2'])) {

$email = trim ($_POST['email']);
$password1 = trim ($_POST['password1']);
$password2 = trim ($_POST['password2']);

}

$email_error = 'No errors<br>';
$password_error = 'No errors<br>';

if (empty($email))
$email_error = ('Pleas fill in email field<br>');

if ($email == 'example')
$email_error =('There already is user with this email<br>');

if (empty($password1))
$password_error =  ('Please fill in password fields<br>'); 

if (empty($password2))
$password_error = ('Please fill in password fields<br>');

$email_error; //this variable must be returned to ajax and assigned to id "email_errors"
$password_error; //this variable must be returned to ajax and assigned to id "password_errors"

?>

javascript

$(document).ready(function ()   {

$('#push_button').click(function() {

$.post('php.php',
{
email : $('#email').val(), // i take these variables to php
password1 : $('#password1').val(),
password1 : $('#password2').val()
} ,
function ( data ) { //what do i return here?

$('#email_errors').val(data); //how to assign $emaill_error to #email_errors
$('#password_errors').val(data); //how to assign $password_error to #password_errors

}
)
})


})

Answer 1

要返回值，只需使用json_encode()來回顯變量，例如

$return_array = new array();
$return_array['email_error'] = $email_error;
$return_array['password_errors'] = $password_errors;
echo json_encode($return_array);

在javascript函數（數據）{}中：

function ( data ) { //what do i return here?

    $('#email_errors').val(data['email_error']); //how to assign $emaill_error to #email_errors
    $('#password_errors').val(data['password_errors']); //how to assign $password_error to #password_errors

}

Answer 2

如果要向ajax返回多個變量，則必須返回一些json

PHP的：

// .. your php code
$ret = array("email_error" => $email_error, "password_error" => $password_error);
echo json_encode($ret);

請注意，json_encode需要PHP> = 5.2

JS：

$.ajax({
  url: "php.php",
  type: "POST",
  dataType: "json", // The type of data that you're expecting back from the server
  data: {
    email: $("#email").val(),
    password1: $("#password1").val(),
    password2: $("#password2").val() // careful, you had "password1" as variable name 2 times
  },
  success: function(obj) {
    // obj is your php array, transformed into a js object
    // you may want to use .html() instead of .val() since your errors are strings with html tags - depends if #email_errors / #password_errors are inputs or divs
    $("#email_errors").html(obj.email_error);
    $("#password_errors").html(obj.password_error);
  }
});

Answer 3

在PHP中，以下內容不會返回任何內容：

$email_error; 
$password_error;

您不回顯值或任何東西。 如果要傳遞兩個不同的值，我將返回一個JSON對象，如下所示（在PHP中）：

echo json_encode(array(
    'email_error' => $email_error,
    'password_error' => $password_error
));

然后在JavaScript中，您的數據現在應該是一個JavaScript對象，因為jQuery應該解析JSON對象並將其理解為對象。 這樣您就可以在JavaScript中執行以下操作：

$('#email_errors').val(data.email_error); 
$('#password_errors').val(data.password_error);

如果您不想使用數組，則可以創建一個新對象，然后將該對象傳遞給json_encode 。

$obj = new stdClass;
$obj->email_error = $email_error;
$obj->password_error = $password_error;

echo json_encode($obj);