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如何将JSON放入PHP变量?(How do I put JSON into a PHP Variable?)

我试图把JSON-Information放到一个PHP变量中,我一直都是错误的。

这是我使JSON可读的代码:

header('Content-Type: application/json');
$jsondecode1 = json_decode($json1);
print_r ($jsondecode1);

这是$jsondecode1一个片段:

Array
(
    [0] => stdClass Object
        (
            [flightId] => ******
            [******] => stdClass Object
                (
                    [******] => ******
                    [******] => ******
                    [******] => ******
                    [******] => ******
                    [******] => ******                
                )

注意:当我echojsondecode1它会输出:

在第48行的C:\ xampp ...... \ simpletest3.php中进行字符串转换

所以我使用了print_r()

我试图把(例如) [flightId]放入一个PHP变量中:

$jsondecode1 = json_decode($json1);
$jsondecode2 = $jsondecode1->flightId;
print_r ($jsondecode2);

输出: 注意 :尝试在C中获取非对象的属性'flightId' 48行的......... \ simpletest3.php

我也尝试了其他一些代码,但输出非常相似,我不想让我的问题比它需要的时间更长。 那么,如何将(例如) [flightId]放入PHP变量中。

编辑:

解:

$jsondecode1 = json_decode($json1);
$jsondecode2 = $jsondecode1[0]->flightId;
print_r ($jsondecode2);

要么:

foreach ($jsondecode1 as $data) {
    print_r($data->flightId);
}

I´m trying to put JSON-Information into a PHP variable and I´m getting Errors all the time.

This is my Code to make the JSON readable:

header('Content-Type: application/json');
$jsondecode1 = json_decode($json1);
print_r ($jsondecode1);

This is a snippet of the $jsondecode1:

Array
(
    [0] => stdClass Object
        (
            [flightId] => ******
            [******] => stdClass Object
                (
                    [******] => ******
                    [******] => ******
                    [******] => ******
                    [******] => ******
                    [******] => ******                
                )

Notice: When I echo the jsondecode1 it outputs this:

Array to string conversion in C:\xampp......\simpletest3.php on line 48

So I used print_r().

What I tried to put (for example) [flightId] into a PHP Variable:

$jsondecode1 = json_decode($json1);
$jsondecode2 = $jsondecode1->flightId;
print_r ($jsondecode2);

Output: Notice: Trying to get property 'flightId' of non-object in C:.........\simpletest3.php on line 48

I tried some other codes too, but the Outputs were very similar and I don´t want to make my Question longer than it needs to be. So, how do I put (for example) the [flightId] into a PHP variable.

Edit:

Solution:

$jsondecode1 = json_decode($json1);
$jsondecode2 = $jsondecode1[0]->flightId;
print_r ($jsondecode2);

Or:

foreach ($jsondecode1 as $data) {
    print_r($data->flightId);
}

原文:https://stackoverflow.com/questions/47991040
更新时间:2020-05-08 20:23

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