PHP函数，获取2个参数，JSON字符串和日期字符串并返回JSON字符串

Question

我正在研究一个PHP问题来写一个函数，它接受两个参数：1）一个包含文章的json字符串2）格式为“2015-09-04”的字符串日期该函数应返回一个json字符串的文章在给定日期之前的两周内发表。但我得到“在第94行的C：\ wamp64 \ www \ phpExercisesW3 \ index.php中为foreach（）提供的无效参数”

<?php  
        $json = '[
            {
                "title": "This is the first title",
                "body": "Lorem ipsum dolor sit amet, consectetur adipiscing elit.",
                "date": "2015-07-30"
            },
            {
                "title": "Another article",
                "body": "Quisqu porttitor dignissim massa rhoncus imperdiet.",
                "date": "2015-08-18"
            },
            {
                "title": "The Final Title",
                "body": "In aliquam auctor ex quis consequat. Praesent non lobortis metus.",
                "date": "2015-08-30"
            },
        ]';    

        $date = "2015-09-04";

        function returnTwoWeeksPrior($jsonString, $dateString) {
            $decode = json_decode($jsonString);
            forEach($decode as $decoded) {
                if ($decode->date < strtotime('-2 weeks')) {
                    $result = json_encode($decoded);
                    echo $result;
                } else {
                    echo "No articles found.";
                }
            }
        }

        echo returnTwoWeeksPrior($json, $date);

        ?>

Answer 1

您的JSON无效。为什么？

看到这些行：

    {
        "title": "The Final Title",
        "body": "In aliquam auctor ex quis consequat. Praesent non lobortis metus.",
        "date": "2015-08-30"
    },
]';    

你有一个额外的'，'需要被删除才能成为一个有效的数组。

应该：

{
    "title": "The Final Title",
    "body": "In aliquam auctor ex quis consequat. Praesent non lobortis metus.",
    "date": "2015-08-30"
}

]';

你遇到类似问题的第一步是var_dump（）或var_export（）你期望的变量。使用$ decode变量执行此操作会向您显示它为null。

像JSONLint这样的工具可以在将来帮助你解决这些类型的问题，但最终，在处理javascript时，chrome控制台总是很棒。

其他错误

您的代码中有一个错误，而不是引用您在foreach循环中创建的变量，而是尝试引用原始对象数组。

看到：

if ($decode->date < strtotime('-2 weeks')) {

应该

if ($decoded->date < strtotime('-2 weeks')) {

PHP错误非常有用：

试图获取非对象的属性“日期”

您应该立即关注您的代码引用“非对象”这一事实并质疑其原因。

Answer 2

错误：

• 你的json上有逗号
• 如果你试图在$ decode上访问日期是数组，你应该使用$ decoding-> date
• 另外，你正在比较一个字符串（$ decoding-> date）和一个int（strtotime（' - 2周'））

这是你的代码工作。

    <?php  
    $json = '[
        {
            "title": "This is the first title",
            "body": "Lorem ipsum dolor sit amet, consectetur adipiscing elit.",
            "date": "2015-07-30"
        },
        {
            "title": "Another article",
            "body": "Quisqu porttitor dignissim massa rhoncus imperdiet.",
            "date": "2015-08-18"
        },
        {
            "title": "The Final Title",
            "body": "In aliquam auctor ex quis consequat. Praesent non lobortis metus.",
            "date": "2015-08-30"
        }
    ]';    

    $date = "2015-09-04";

   function returnTwoWeeksPrior($jsonString, $dateString) {
        $decode = json_decode($jsonString);
        $results = [];
        forEach($decode as $decoded) {
            if (strtotime($decoded->date) < (strtotime($dateString.'-2 weeks')) ) {
                $result = json_encode($decoded);
                echo $result;
                $results[] = $result;
            } 
        }
        if(count($results) == 0){
            echo "No articles found.";
        }
    }

    echo returnTwoWeeksPrior($json, $date);

Answer 3

你的json应该如下

$json = '{
            "items" :[
            {
                "title": "This is the first title",
                "body": "Lorem ipsum dolor sit amet, consectetur adipiscing elit.",
                "date": "2015-07-30"
            },
            {
                "title": "Another article",
                "body": "Quisqu porttitor dignissim massa rhoncus imperdiet.",
                "date": "2015-08-18"
            },
            {
                "title": "The Final Title",
                "body": "In aliquam auctor ex quis consequat. Praesent non lobortis metus.",
                "date": "2015-08-30"
            }
            ]
        }';   

你不应该使用

, 

在列表中的最后一项之后

Answer 4

你去：

<?php  
        $json = '[
            {
                "title": "This is the first title",
                "body": "Lorem ipsum dolor sit amet, consectetur adipiscing elit.",
                "date": "2015-07-30"
            },
            {
                "title": "Another article",
                "body": "Quisqu porttitor dignissim massa rhoncus imperdiet.",
                "date": "2015-08-18"
            },
            {
                "title": "The Final Title",
                "body": "In aliquam auctor ex quis consequat. Praesent non lobortis metus.",
                "date": "2015-08-30"
            }
        ]';    

        $date = "2015-09-04";

        function returnTwoWeeksPrior($jsonString, $dateString) {
            $decode = json_decode($jsonString);
            forEach($decode as $decoded) {
                if (strtotime($decoded->date) < strtotime('-2 weeks', strtotime($date))) {
                    $result = json_encode($decoded);
                    echo $result;
                } else {
                    echo "No articles found.";
                }
            }
        }

        echo returnTwoWeeksPrior($json, $date);

        ?>

首先你的JSON是无效的：有一个尾随的逗号init，你试图比较一个字符串和一个数字，你搞砸了$decoded和$decode。