如何在成功 AJAX/jQuery POST 时返回 PHP 变量
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/13212882/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
How to return PHP variables on success AJAX/jQuery POST
提问by fabbb
How do I use AJAX to return a variable in PHP? I am currently using echo in my controller to display a price on dropdown .changein a divcalled price.
如何在 PHP 中使用 AJAX 返回变量？我目前使用我的控制器回波上显示的价格dropdown .change在一个div叫价。
However I have a hidden field which I need to return the row id to on change. How do I assign the return var in jQuery so that I can echo it in my hidden field?
但是，我有一个隐藏字段，我需要在更改时将行 ID 返回到该字段。我如何在 jQuery 中分配返回变量，以便我可以在我的隐藏字段中回显它？
jQuery
jQuery
$(document).ready(function() {
$('#pricingEngine').change(function() {
var query = $("#pricingEngine").serialize();
$('#price').fadeOut(500).addClass('ajax-loading');
$.ajax({
type: "POST",
url: "store/PricingEngine",
data: query,
success: function(data)
{
$('#price').removeClass('ajax-loading').html('$' + data).fadeIn(500);
}
});
return false;
});
});
Controller
控制器
function PricingEngine()
{
//print_r($_POST);
$this->load->model('M_Pricing');
$post_options = array(
'X_SIZE' => $this->input->post('X_SIZE'),
'X_PAPER' => $this->input->post('X_PAPER'),
'X_COLOR' => $this->input->post('X_COLOR'),
'X_QTY' => $this->input->post('X_QTY'),
'O_RC' => $this->input->post('O_RC')
);
$data = $this->M_Pricing->ajax_price_engine($post_options);
foreach($data as $pData) {
echo number_format($pData->F_PRICE / 1000,2);
return $ProductById = $pData->businesscards_id;
}
}
View
看法
Here is my hidden field I want to pass the VAR to every-time the form is changed. " />
这是我的隐藏字段，我想在每次更改表单时将 VAR 传递给。" />
Thanks for the help!
谢谢您的帮助！
回答by Lix
Well.. One option would be to return a JSON object. To create a JSON object in PHP, you start with an array of values and you execute json_encode($arr). This will return a JSON string.
嗯.. 一种选择是返回一个 JSON 对象。要在 PHP 中创建 JSON 对象，您需要从一组值开始，然后执行json_encode($arr). 这将返回一个 JSON 字符串。
$arr = array(
'stack'=>'overflow',
'key'=>'value'
);
echo json_encode($arr);
{"stack":"overflow","key":"value"}
Now in your jQuery, you'll have to tell your $.ajaxcall that you are expecting some JSON return values, so you specify another parameter - dataType : 'json'. Now your returned values in the successfunction will be a normal JavaScript object.
现在在您的 jQuery 中，您必须告诉您的$.ajax调用您需要一些 JSON 返回值，因此您指定另一个参数 - dataType : 'json'。现在success函数中的返回值将是一个普通的 JavaScript 对象。
$.ajax({
type: "POST",
url: "...",
data: query,
dataType: 'json',
success: function(data){
console.log(data.stack); // overflow
console.log(data.key); // value
}
});
回答by Svetoslav
echo json_encode($RESPONDE);
exit();
The exit is not to display other things except answer. RESPONDE is good to be array or object. You can access it at
退出是不显示除回答之外的其他内容。RESPONDE 最好是数组或对象。你可以访问它
success: function(data)
{ data }
data is the responde array or whatever you echo.. For example...
数据是响应数组或您回显的任何内容..例如...
echo json_encode(array('some_key'=>'yesss')); exit();
at jquery
在 jquery
success: function(data){ alert(data.some_key); }
回答by mohan.gade
if u are returning only single value from php respone to ajax then u can set it hidden feild using val method
如果您只从 php 响应返回单个值到 ajax，那么您可以使用 val 方法将其设置为隐藏字段
$("#hidden_fld").val(return_val);
