从数组中获取单个值 - 从URL获取数组

Hi i am trying to fetch single value from below Array example -

嗨，我想从下面的数组示例中获取单个值 -

i want to fetch order_id and name -

我想获取order_id和名称 -

<?php
    $url = 'http://example.com/index.php?route=feed/rest_api/orders&key=test219';

        $json = file_get_contents($url);

    echo '<pre>'; echo $json; echo '</pre>';
?>

Array showing

数组显示

Array
(
    [0] => Array
        (
            [order_id] => 6
            [name] => zoraya panansar
            [status] => Complete
            [date_added] => 2014-01-24 23:06:09
            [products] => 4
            [total] => 14.8000
            [address_1] => 345 goldhawk road
            [address_2] => hammersmith
            [city] => london
            [Postcode] => w6 0wz
            [Country] => United Kingdom
            [Email] => zpanansar@yahoo.com
            [Tele.] => 0786870150
        )

I already tried

我已经试过了

echo $json->order_id;

and

和

echo $json[0]['order_id'];

but its not working... Please help

但它不起作用......请帮助

3 个解决方案

#1

You can not access the data that the server is returning. Instead of returning you a json object, the server is returning you the print_r() output over an array.

您无法访问服务器返回的数据。服务器不是返回一个json对象，而是通过数组返回print_r（）输出。

If its your server, then stop the print, and use json_encode() over the array before printing.

如果是您的服务器，则停止打印，并在打印前在阵列上使用json_encode（）。

If it's not your server, then you have to parse the values from the text. You can use preg_match_all() for this.

如果它不是您的服务器，那么您必须解析文本中的值。您可以使用preg_match_all（）。

preg_match_all("/\[order_id\] => ([^\n\r]+)/", $json, $m);
print_r($m[1]);

There may be built-in php function that can take such print_r() output and revert back into a php array, but I don't know about it!

可能有内置的PHP函数，可以采取这样的print_r（）输出并恢复到PHP数组，但我不知道它！

#2

index.php is returning wrong data, if you look closer then you will see it is trying to send success = false result as json data but probably for testing purposes, it is also returning array content directly before the json data.

index.php返回错误的数据，如果你仔细观察，那么你会发现它正在尝试发送success = false结果作为json数据，但可能是出于测试目的，它也是在json数据之前直接返回数组内容。

Besides, if this is supposed to be a json result then you need to use json_decode before using the returned json object as a class object like $json->order_id.

此外，如果这应该是一个json结果，那么在使用返回的json对象作为类对象（如$ json-> order_id）之前，需要使用json_decode。

This is what I get when I try your current URL With the parameters you provided: Look at the last line:

这是我在使用您提供的参数尝试当前URL时获得的内容：查看最后一行：

Array
(
    [0] => Array
        (
            [order_id] => 6
            [name] => zoraya panansar
            [status] => Complete
            [date_added] => 2014-01-24 23:06:09
            [products] => 4
            [total] => 14.8000
            [address_1] => 345 goldhawk road
            [address_2] => hammersmith
            [city] => london
            [Postcode] => w6 0wz
            [Country] => United Kingdom
            [Email] => zpanansar@yahoo.com
            [Tele.] => 0786870150
        )

    [1] => Array
        (
            [order_id] => 5
            [name] => ciria ann nuqui
            [status] => Complete
            [date_added] => 2014-01-24 17:58:36
            [products] => 7
            [total] => 36.4000
            [address_1] => 345 goldhawk road 
            [address_2] => hammersmith
            [city] => london
            [Postcode] => w6 0wz
            [Country] => United Kingdom
            [Email] => joycernuqui@yahoo.com
            [Tele.] => 07515732291
        )
)
{"success":false}

#3

Change the response from the url to json (Use json_encode function)
The last Row in url script should be

将响应从url更改为json（使用json_encode函数）url脚本中的最后一行应该是

    echo json_encode($some_array);

And than if you want the content:

而且如果你想要内容：

    $array = json_decode(file_get_contents($url),TRUE);
    if(!is_array($array)){
      echo 'bad response';
      exit();
     }
     var_dump($array);
     echo $array[0]['order_id'];

#1