Question

首先，我是PHP的新手，请原谅我。

我想采取JSON回复并将其拆分。这将来自$_POST变量，但我试图测试硬编码变量中的响应。问题是我甚至无法打印它以便看到我正确地开始这个。

$json = 
'({
  "array":
  [
    "Store #: 00608",
    "Phone #: null",
    "Address: 3014 N. SCOTTSDALE RD.",
    "City: SCOTTSDALE",
    "Zip: 85251",
    "State: AZ",
    "Height: 6`4",
    "Weight: 230",
    "Ethnicity: White",
    "Age: 23",
    "Eye Color: Blue",
    "Favorite Food: Thai",
    "Comments: awesome"
   ]
})';

$data = json_decode($json,true);
$pieces = explode(":", $data);
for ($i = 0; $i < count($data['array']); $i++) {
    echo $pieces[$i];
}

当我在浏览器中启动它时，我得到一个空白屏幕，没有错误。最终目标是将这些数据存储为PHP数组'Store #', '00608'等等。

无论如何，我做错了什么？

Answer 1

试试这个JSON。

$json = 
'{
  "array":
  [
    "Store #: 00608",
    "Phone #: null",
    "Address: 3014 N. SCOTTSDALE RD.",
    "City: SCOTTSDALE",
    "Zip: 85251",
    "State: AZ",
    "Height: 6`4",
    "Weight: 230",
    "Ethnicity: White",
    "Age: 23",
    "Eye Color: Blue",
    "Favorite Food: Thai",
    "Comments: awesome"
   ]
}';

请注意我拿出你的（）！

Answer 2

我认为这几乎就是你要找的东西。

$json = 
'{
  "array":
  [
    "Store #: 00608",
    "Phone #: null",
    "Address: 3014 N. SCOTTSDALE RD.",
    "City: SCOTTSDALE",
    "Zip: 85251",
    "State: AZ",
    "Height: 6`4",
    "Weight: 230",
    "Ethnicity: White",
    "Age: 23",
    "Eye Color: Blue",
    "Favorite Food: Thai",
    "Comments: awesome"
   ]
}';
$data = json_decode($json,true);

foreach($data['array'] as $piece) {
  $array = explode(': ', $piece);
  echo 'Key: '.$array[0].'<br />';
  echo 'Value: '.$array[1].'<br />';
  echo '<br />';
}

返回

Key: Store #
Value: 00608

Key: Phone #
Value: null

Key: Address
Value: 3014 N. SCOTTSDALE RD.

Answer 3

您有无效的JSON，因为您在JSON字符串周围有括号。删除它们，它应该正确解析。

循环通过JSONArray但什么都没得到

3 个答案: