Question

我正在通过JSON将JSON数据从jQuery脚本发送到php文件，在该文件中，我试图将其拆分为变量，以便将其全部上传到数据库中。到目前为止，JSON数据已作为JSON对象发送和接收，但我无法提取任何数据。

jquery脚本

<html>
<head>
    <script src="../jquery-3.1.0.min.js"></script>  
<script>
    function update(){
        var x =  {"1":{"item":"leaks", "qty":"2", "store":"fruitVeg"},"2":{"item":"chicken stock", "qty":"1", "store":"pan"}};
        var dataSend = JSON.stringify(x);
        $.ajax({
            url:"updateTest.php",
            type: "POST",
            data: dataSend,
            dataType: "json",
            success: function(data){
                var x = data.item;
                console.log(data);
                //$('#added').text(x +' Added');
            }
        });
    }
    $(document).ready(function(){
        $('#add').click(update);
    });
</script>   
</head>
<body>
    <div> <input type='button' value='Add to Shopping' id='add'/></div>
</body>
</html>

PHP脚本

<?php
$obj = json_decode(file_get_contents('php://input'), true);
$count = count($obj);
$test = $obj[1]['item'];

//echo json_encode ($obj);
echo $count;
echo $test;
?>

目前，我返回的只是行数。如果我注释掉echo并取消注释echo json，则会返回以下内容

{1: {item: "leaks", qty: "2", store: "fruitVeg"}2: {item: "chicken stock", qty: "1", store: "pan"}

Answer 1

我设法使用以下脚本使其工作

<html>
<head>
    <script src="../jquery-3.1.0.min.js"></script>  
<script>
    function update(){
        var x =  [{"item":"leaks", "qty":"2", "store":"fruitVeg"},{"item":"chicken stock", "qty":"1", "store":"pantry"}];
        var dataSend = JSON.stringify(x);
        $.ajax({
            url:"updateTest.php",
            type: "POST",
            data: "data="+dataSend,
            success: function(data){
                var x = data.item;
                console.log(data);
                //$('#added').text(x +' Added');
            }
        });
    }
    $(document).ready(function(){
        $('#add').click(update);
    });
    </script>   
    </head>
    <body>
    <div> <input type='button' value='Add to Shopping' id='add'/></div>
</body>
</html>

php看起来像这样

<?php
$list = stripslashes($_REQUEST["data"]);
$obj = json_decode($list);

echo $item1 = $obj[0]->item;
echo $item2 = $obj[1]->item;
?>

将JSON对象传递到php变量中

1 个答案: