从JSON字符串/数组中提取第一个图像

I'm trying to get a fake movie website going for a University project. But because I'm a designer, I'm terrible at php (and my terminology is probably way off). Let me know if something doesn't make sense or is badly explained.

我正试图让一个假的电影网站去大学项目。但是因为我是一名设计师,我在php上很糟糕(而且我的术语可能很偏僻)。如果某些事情没有意义或者说得不好,请告诉我。

I'm using http://themoviedb.org's API to bring in movie data, and am struggling on one of the pages.

我正在使用http://themoviedb.org的API来引入电影数据,并在其中一个页面上挣扎。

I'm pulling a JSON string in from their API, and trying to get the first 'backdrop' image to display (there are high ranging numbers of backdrops per different movie, from none up to about twenty). There are different sizes of 'backdrop' as well, ranging from the 'original' size, down to a 'thumb'. I'm trying to pull in just the first 'backdrop' which is called the size 'poster'.

我正在从他们的API中提取一个JSON字符串,并试图获得第一个“背景”图像(每个不同的电影有大量的背景幕,从无到二十几个)。 “背景”也有不同的尺寸,从“原始”尺寸到“拇指”。我试图引入第一个'背景',称为尺寸'海报'。

So here's what I've got. The problem is - to me - it looks perfectly structured, but just returns the fallback 'no-backdrop.gif'.

所以这就是我所拥有的。问题是 - 对我来说 - 它看起来很完美,但只是返回后备'no-backdrop.gif'。

<?php
    //TMDb id for a movie
    $tmdb_id = $_GET["id"];

    //Search Movie with default return format
    $movies_result = $tmdb->getMovie($tmdb_id);

    // Decode JSON into PHP array
    $movies = json_decode($movies_result);
?>
<?php
    // Set default poster image to use if movie doesn't have one
    $backdrop_url = 'no-backdrop.gif';

    if ($movies[0]->backdrops->image->size == 'poster') {
      $backdrop_url = $movies[0]->backdrops->image->url;
    }

    echo '<img src="' . $backdrop_url . '" />';
?>

And here's is the json string. Which is massive unfortunately, so I've had to cut out what seems to be irrelevant to me.

这是json字符串。不幸的是,这是巨大的,所以我不得不削减似乎与我无关的东西。

The full string can be found at the bottom of the page on the link below.

完整的字符串可以在下面链接的页面底部找到。

or to see it failing in action, you could go here - http://rowancavanagh.com/getfilm/tmdb_movie.php?id=44214
(complete with the json string printed at the bottom)

或者看到它失败了,你可以去这里 - http://rowancavanagh.com/getfilm/tmdb_movie.php?id=44214(完整的底部印有json字符串)

1 个解决方案

#1

The backdrops property is an array, you have to access it as an array:

backdrops属性是一个数组,您必须以数组形式访问它:

$backdrop_url = $movies[0]->backdrops[0]->image->url;

Try turning on error_reporting and/or display_errors or log_errors, this is an error that is easily caught with these on: "Trying to get property of non-object".

尝试启用error_reporting和/或display_errors或log_errors,这是一个很容易被这些错误捕获的错误:“试图获取非对象的属性”。

Also, use isset in your if to prevent an E_NOTICE from popping if there is no image:

另外,如果没有图像,请在if中使用isset以防止E_NOTICE弹出:

if (isset($movies[0]->backdrops[0]->image->size) &&
    $movies[0]->backdrops[0]->image->size == 'poster') 
{
    // stuff
}

If you have more than one element in backdrops, and want to fetch only the one with the poster type, use a foreach loop:

如果背景中有多个元素,并且只想获取具有海报类型的元素,请使用foreach循环:

foreach ($movies[0]->backdrops as $backdrop) {
    if (isset($backdrop->image->size) && $backdrop->image->size == 'poster') {
        $backdrop_url = $backdrop->image->url;
        break;
    }
}

1 个解决方案

#1

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