[英]Passing/returning value to/from a function using ajax and callback function
I'm trying to read p_info
array returned from the function getproductInfo
containing a ajax call but I'm getting undefined value. 我正在尝试读取从包含ajax调用的
getproductInfo
函数返回的p_info
数组,但是得到的是不确定的值。 I'm using a callback function to achieve this but still doesn't work. 我正在使用回调函数来实现此目的,但仍然无法正常工作。 Where am I wrong?
我哪里错了?
$(document).ready(function() {
function successCallback(data)
{
var name = data.name;
var image = data.image;
var link = data.link;
var product_info = [name, image, link];
console.log(product_info); // Correct: shows my product_info array
return product_info;
}
function getProductInfo(prodId, successCallback) {
$.ajax({
type: "POST",
url: "getProductInfo.php",
data: "id=" + prodId,
dataType: "json",
success: function(data) {
var p_info = successCallback(data);
console.log(p_info); // Correct: shows my product_info array
return p_info;
},
error: function()
{
alert("Error getProductInfo()...");
}
});
return p_info; // Wrong: shows "undefined" value
}
var p_info = getProductInfo(12, successCallback);
console.log(p_info); // Wrong: shows an empty value
});
The code should speak for itself. 该代码应该说明一切。 But basically, you cant return an upper-level function inside a function.
但基本上,您不能在函数内部返回上级函数。 You must set a variable to be used to return after the ajax is submitted.
提交ajax之后,必须设置一个用于返回的变量。
//This makes the p_info global scope. So entire DOM (all functions) can use it.
var p_info = '';
//same as you did before
function successCallback(data) {
var name = data.name;
var image = data.image;
var link = data.link;
var product_info = [name, image, link];
return product_info;
}
//This takes prodID and returns the data.
function getProductInfo(prodId) {
//sets up the link with the data allready in it.
var link = 'getProductInfo.php?id=' + prodId;
//creates a temp variable above the scope of the ajax
var temp = '';
//uses shorthand ajax call
$.post(link, function (data) {
//sets the temp variable to the data
temp = successCallback(data);
});
//returns the data outside the scope of the .post
return temp;
}
//calls on initiates.
var p_info = getProductInfo(12);
console.log(p_info);
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.