0

我正在从 JSON 中提取值，但是当我回显该值时始终得到一个空结果

$json='[[{"transTime":"2013-10-23 17:30:42","Forename":"Ian","Surname":"Graham","Address Line 1":"RG412GX"}]]';

$obj2 = json_decode($json, true);

$displayName = $obj2->Surname;

echo"$displayName";
4

4 回答 4

1

在这个 json 字符串中有一个对象在另一个对象中

$json='[[{"transTime":"2013-10-23 17:30:42","Forename":"Ian","Surname":"Graham","Address Line 1":"RG412GX"}]]';

$obj2 = json_decode($json);

print_r($obj2);
于 2013-10-23T21:58:02.927 回答
0

您提供的代码有几个问题。首先，您的 json 文本存储在 中$json，但您尝试解码$xmlresponse. 不过，我想这只是一个复制/粘贴错误。其次，您尝试使用对象语法访问姓氏，尽管您明确强制json_decode将对象解码为关联数组。第三，提供的json对数组中的对象进行编码。您忽略了响应的嵌套结构。

尝试这个：

$json='[[{"transTime":"2013-10-23 17:30:42","Forename":"Ian","Surname":"Graham","Address Line 1":"RG412GX"}]]';
$response = json_decode($json);
$displayName = $response[0][0]->Surname;

echo $displayName;
于 2013-10-23T22:02:39.193 回答
0

这应该是这样的：

$json='[[{"transTime":"2013-10-23 17:30:42","Forename":"Ian","Surname":"Graham","Address    Line 1":"RG412GX"}]]';

$obj2 = json_decode($json, true);

$displayName = $obj2->Surname;

echo"$displayName";

您正在混淆/编造变量名...

于 2013-10-23T21:59:21.303 回答
0

尝试这个：

<?php
$json='[[{"transTime":"2013-10-23 17:30:42","Forename":"Ian","Surname":"Graham","Address Line 1":"RG412GX"}]]';
$obj2 = json_decode($json, true);
$displayName = $obj2[0][0]['Surname'];
echo "$displayName";
?>
于 2013-10-23T21:57:39.377 回答