存取json物件值时发生问题 [英] Issue accessing json object values

问题描述

我遇到了一个问题console.logging从我的jquery ajax调用发送回脚本的json对象的值.

I am having an issue console.logging the values from the json object sent back from my jquery ajax call to a script.

脚本:

for(i=0; i<IDs.length; i++)
    {
        var vendor = IDs[i]; 
        $j.ajax({
                type: "POST",
                url: "/ajax_calls/updatePrices.php",
                data: { 'vendorID': vendor, 'product_id': product_id}
                }).done(function(data) {
                    console.log('The data is ' + data);

                        var basePrice = simpleArray[vendor][colorSelected];

                        //if(data.tier2_range_start[i])
                            console.log('Range start is data.tier2_range_start: ' + data.tier2_range_start);

                        if (qty < data.tier2_range_start){
                            simpleArray[vendor][colorSelected]= basePrice * qty;
                        }
                        else if (qty > data.tier2_range_start){
                            simpleArray[vendor][colorSelected]= (basePrice * qty) * data.tier2_discount;
                        }
                        else if (qty > data.tier3_range_start){
                            simpleArray[vendor][colorSelected]= (basePrice * qty) * data.tier3_discount;
                        }
                        else if (qty > data.tier4_range_start){
                            simpleArray[vendor][colorSelected]= (basePrice * qty) * data.tier4_discount;
                        }
                        else if (qty > data.tier5_range_start){
                            simpleArray[vendor][colorSelected]= (basePrice * qty) * data.tier5_discount;
                        }
                        else{
                            console.log('Something went wrong');
                        }


                    $j('.details'+vendor+ ' .priceBlock').empty();
                    $j('.details'+vendor+ ' .priceBlock').append('<span>'+simpleArray[vendor][colorSelected]+'</span>');
                });
        }//end for

脚本名为:

    <?php
require_once('/var/www/Staging/connect.php');

//post variable
$ID= $_POST['vendorID'];
$product_id= $_POST['product_id'];
$echoArray= array();

    //if ( !isset($echoArray[$x]) )
            //$echoArray[$x] = array();
 $sql = 'SELECT * FROM tier_pricing WHERE vendor_id=' . $ID. ' AND product_id=' . $product_id;
    foreach ($con->query($sql) as $row) {
        $echoArray['vendor_id']= $row['vendor_id'];
            $echoArray['tier2_range_start']= $row['tier2_range_start'];
        $echoArray['tier2_range_stop']= $row['tier2_range_stop'];
        $echoArray['tier3_range_start']= $row['tier3_range_start'];
        $echoArray['tier3_range_stop']= $row['tier3_range_stop'];
        $echoArray['tier4_range_start']= $row['tier4_range_start'];
        $echoArray['tier4_range_stop']= $row['tier4_range_stop'];
        $echoArray['tier5_range_start']= $row['tier5_range_start'];
        $echoArray['tier2_discount']= $row['tier2_discount'];
        $echoArray['tier3_discount']= $row['tier3_discount'];
        $echoArray['tier4_discount']= $row['tier4_discount'];
        $echoArray['tier5_discount']= $row['tier5_discount'];
    }

echo json_encode($echoArray); 
?>

数据记录为(例如，在没有空数据对象的循环中):

Data is logging as (for example, on a loop without empty data object):

[15:06:13.397] The data is {"vendor_id":"3","tier2_range_start":"5","tier2_range_stop":"20","tier3_range_start":"20","tier3_range_stop":"100","tier4_range_start":"100","tier4_range_stop":"500","tier5_range_start":"500","tier2_discount":"2","tier3_discount":"3.1","tier4_discount":"4.3","tier5_discount":"5"}

为什么未定义data.tier2_range_start?

推荐答案

您需要告诉jQuery返回的数据是json，除非您在响应中放入适当的mime标头，否则jQuery应该如何知道?

you need to tell jQuery that the returning data is json, how should jQuery know unless you put an appropriate mime header in your response?

PHP开头的某个位置:header('Content-type: text/json');

somewhere in the beginning of your PHP: header('Content-type: text/json');

在您的ajax中:

...
data: { 'vendorID': vendor, 'product_id': product_id},
dataType: "json",
...

然后jQuery为您完成这项工作.而且您的踪迹应该看起来像

then jQuery is doing the job for you. And your trace should look like

[15:06:13.397] The data is Object object

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