Question

我在$ jsontest上使用json_decode（出于演示目的，因为它是确切的格式）。

问题是我正在尝试为附件7分配文件名，名称和类型，以将变量分隔为字符串。我尝试过

$attachment7filename = $jsontest2->{'attachment7->filename'};

例如，输出为PHP注意：

试图获取非对象的属性。

任何帮助将不胜感激。

 <?php

    $jsontest = '{"attachment7":{"filename":"small.flv","name":"small.flv","type":"video/x-flv"},"attachment2":{"filename":"sample.png","name":"sample.png","type":"image/png"},"attachment6":    {"filename":"small.3gp","name":"small.3gp","type":"video/3gpp"},"attachment5":    {"filename":"small.webm","name":"small.webm","type":"video/webm"},"attachment3":    {"filename":"small.mp4","name":"small.mp4","type":"video/mp4"},"attachment1":        {"filename":"109-1new.jpeg","name":"109-    1new.jpeg","type":"image/jpeg"},"attachment4":    {"filename":"small.ogv","name":"small.ogv","type":"video/ogg"}}';

    $jsontest2 = json_decode($jsontest, true);

    $attachment7filename = $jsontest2->{'attachment7->filename'}; // should be small.flv for example
    $attachment7name = $jsontest2->{'attachment7->name'};
    $attachment7type = $jsontest2->{'attachment7->type'};

    var_dump($attachment7filename);

    ?>

Answer 1

您正在使用带有{}引用的大括号（object），例如$jsontest2->{'attachment7->filename'};用{'attachment7->filename'}替换['attachment7']['filename'];

这是代码

<?php

    $jsontest = '{"attachment7":{"filename":"small.flv","name":"small.flv","type":"video/x-flv"},"attachment2":{"filename":"sample.png","name":"sample.png","type":"image/png"},"attachment6":    {"filename":"small.3gp","name":"small.3gp","type":"video/3gpp"},"attachment5":    {"filename":"small.webm","name":"small.webm","type":"video/webm"},"attachment3":    {"filename":"small.mp4","name":"small.mp4","type":"video/mp4"},"attachment1":        {"filename":"109-1new.jpeg","name":"109-    1new.jpeg","type":"image/jpeg"},"attachment4":    {"filename":"small.ogv","name":"small.ogv","type":"video/ogg"}}';

    $jsontest2 = json_decode($jsontest, true);
//print_r($jsontest2);
    $attachment7filename = $jsontest2['attachment7']['filename']; // should be small.flv for example
    $attachment7name = $jsontest2['attachment7']['name'];
    $attachment7type = $jsontest2['attachment7']['type'];

    var_dump($attachment7filename);

    ?>

您可以检查所需的output here

json_decode将特定部分分配给字符串

1 个答案: