Question

这是以异步方式发送到我的php页面的JSON。它本质上是一个产品列表，将被插入到我的mySQL数据库中。

我的问题是在PHP中解码JSON。我可以使用'eval'函数在js中做到这一点，但在PHP中，我的努力导致了一系列复杂的爆炸和内爆函数。

{
    "Product": [
        {
            "Product_Title": "Cloth",
            "Product_Description": "Here is cloth",
            "Price": "100",
            "Category_ID": "1"
        },
        {
            "Product_Title": "Cloth",
            "Product_Description": "Here is cloth",
            "Price": "100",
            "Category_ID": "1"
        },
        {
            "Product_Title": "Cloth",
            "Product_Description": "Here is cloth",
            "Price": "100",
            "Category_ID": "1"
        }
    ]
}

我知道php有一个内置的json_decode函数，但是在PHP文档中它们只展示了如何处理数组。

非常感谢任何建议或帮助

泰勒

Answer 1

如果您致电json_decode($data,true);，您的数据将是：

Array(
    "Product"=>Array(
        Array(
            "Product_Title"=>"Cloth",
            "Product_Description"=>"Here is cloth",
            "Price"=>"100",
            "Category_ID"=>"1"
        ),
        Array(
.............
        )
    )
);

这有什么问题？

Answer 2

如果要保留stdClass个对象，则需要使用object-property语法。

<?php

$json = '{
    "Product": [
        {
            "Product_Title": "Cloth",
            "Product_Description": "Here is cloth",
            "Price": "100",
            "Category_ID": "1"
        },
        {
            "Product_Title": "Cloth",
            "Product_Description": "Here is cloth",
            "Price": "100",
            "Category_ID": "1"
        },
        {
            "Product_Title": "Cloth",
            "Product_Description": "Here is cloth",
            "Price": "100",
            "Category_ID": "1"
        }
    ]
}
';

$json_decoded = json_decode($json);

// Note, it's usually a bad idea to use use count() like this;
// cache the count before the for() in a variable and use that.
// This is for demo purposes only. :)
for ($i = 0; $i < count($json_decoded->{'Product'}); $i++) {
    echo "Products:
" . $json_decoded->{'Product'}[$i]->{'Product_Title'} . "
" . $json_decoded->{'Product'}[$i]->{'Product_Description'} . "
" . $json_decoded->{'Product'}[$i]->{'Price'} . "
" . $json_decoded->{'Product'}[$i]->{'Category_ID'} . "

";
}

?>

输出：

Products:
Cloth
Here is cloth
100
1

Products:
Cloth
Here is cloth
100
1

Products:
Cloth
Here is cloth
100
1

http://codepad.org/JxYAO5De

如何在PHP中处理JSON？

2 个答案: