如何使用json将响应从php返回到javascript？
[英]how to get the response back to javascript from php using json?
问题描述
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此问题已得到解决
THE BELOW CODE IS THE CORRECTED AND WORKING CODE 下面的代码是正确的工作代码
I have created a html form which sends certain variables to a php file and returns the success through json back to the javascript. 我创建了一个html表单，该表单将某些变量发送到php文件，并通过json将成功返回给javascript。 But the problem is am not able to get any response back to the javascript file. 但是问题是无法获得对javascript文件的任何响应。 I donno what is the reason. 我不知道是什么原因。 So can some help me with this. 所以有些可以帮助我。 Thank you 谢谢
my form is 我的表格是
<!doctype html>
<html>
<head>
<script src="http://ajax.googleapis.com/ajax/libs/jquery/2.0.3/jquery.min.js"> </script>
<script src="scriptj.js"></script>
</head>
<body>
<form action="http://localhost/donotdel/process.php" method="POST">
<div id="name-group" class="form-group">
<label for="name">Name</label>
<input type="text" class="form-control" name="name" placeholder="name">
</div>
<div id="email-group" class="form-group">
<label for="email">Email</label>
<input type="text" class="form-control" name="email" placeholder="email">
</div>
<button type="submit" class="btn btn-success">Submit <span class="fa fa-arrow-right"></span></button>
</form>
</body>
</html>
my javascript file is 我的JavaScript文件是
$(document).ready(function () {
$('form').submit(function (event) {
$('.form-group').removeClass('has-error');
$('.help-block').remove();
var formData = {
'name': $('input[name=name]').val(),
'email': $('input[name=email]').val(),
};
$.ajax({
type: 'POST',
url: 'http://localhost/donotdel/process.php',
data: formData,
dataType: 'json',
encode: true
}).done(function (data) {
console.log(data);
if (!data.success) {
if (data.errors.name) {
$('#name-group').addClass('has-error');
$('#name-group').append('<div class="help-block">' + data.errors.name + '</div>');
}
if (data.errors.email) {
$('#email-group').addClass('has-error');
$('#email-group').append('<div class="help-block">' + data.errors.email + '</div>');
}
}
else {
$('form').append('<div class="alert alert-success">' + data.message + '</div>');
}
}).fail(function (data) {
console.log(data);
});
event.preventDefault();
});
});
and my php file is 和我的PHP文件是
<?php
$errors = array();
$data = array();
if (empty($_POST['name']))
$errors['name'] = 'Name is required.';
if (empty($_POST['email']))
$errors['email'] = 'Email is required.';
if ( ! empty($errors)) {
$data['success'] = false;
$data['errors'] = $errors;
} else {
$data['success'] = true;
$data['message'] = 'Success!';
}
header ('Content-Type: application/json');
header("Access-Control-Allow-Origin: *");
echo json_encode($data);
?>
the error that i am getting is instead of passing the json to javascript file, the json is printed in the screen. 我得到的错误不是将json传递给javascript文件，而是将json打印在屏幕上。 But whereas i need the javascript to show an alert. 但是，尽管我需要JavaScript来显示警报。
2 个解决方案
解决方案1
1 2015-06-03 10:46:53
Try setting the proper header in php: 尝试在php中设置适当的标头：
header ('Content-Type: application/json');
Failing that - can you post a plunker or jsfiddle? 失败了-您可以发布一个punker或jsfiddle吗？
解决方案2
1 2015-06-03 11:20:39
As discussed with OP. 正如与OP讨论的那样。
-> First error is jQuery library not loaded from goole server so suggest to download latest jQuery library
-> second error is
'Access-Control-Allow-Origin'.'Access-Control-Allow-Origin'。 suggest to addheader ('Content-Type: application/json');header("Access-Control-Allow-Origin: *");header ('Content-Type: application/json');header("Access-Control-Allow-Origin: *");in php file or make sure to call ajax request from same origin domain
header ('Content-Type: application/json');
header("Access-Control-Allow-Origin: *");
echo json_encode($data);
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