如何成功AJAX / jQuery的POST返回PHP变量 [英] How to return PHP variables on success AJAX/jQuery POST
问题描述
我如何使用AJAX在PHP中返回一个变量?我目前正在使用的回声在我的控制器上显示的 DIV
叫价下拉.change
价格。
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 .change
in a div
called price.
不过,我有一个隐藏字段,我需要行ID返回到上的变化。我该如何分配收益VAR jQuery中,这样我可以在我的隐藏字段呼应呢?
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?
$(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;
});
});
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 = $pData->businesscards_id;
}
}
查看
下面是我的隐藏字段我想在VAR传递到每一个时间的形式发生改变。 />
View
Here is my hidden field I want to pass the VAR to every-time the form is changed. " />
感谢您的帮助!
推荐答案
嗯..一种办法是返回一个JSON对象。要在PHP中JSON对象,你开始与值的数组,并执行 json_en code($ ARR)
。这将返回一个JSON字符串。
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.
$arr = array(
'stack'=>'overflow',
'key'=>'value'
);
echo json_encode($arr);
{"stack":"overflow","key":"value"}
现在在你的jQuery,你必须要告诉你的 $ AJAX
调用你期待一些JSON的返回值,所以你指定的另一个参数 - 数据类型:JSON
。现在在成功的返回值
功能将是一个普通的JavaScript对象。
Now in your jQuery, you'll have to tell your $.ajax
call that you are expecting some JSON return values, so you specify another parameter - dataType : 'json'
. Now your returned values in the success
function will be a normal JavaScript object.
$.ajax({
type: "POST",
url: "...",
data: query,
dataType: 'json',
success: function(data){
console.log(data.stack); // overflow
console.log(data.key); // value
}
});
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