Question

我是一个来自远程网址的json

[{"Name":"Abcd","Alias":["Bcde","Cdef","Fghi","Jklm","Load more"]}]

当我尝试按如下方式打印元素别名时，我得到的错误如“试图获取非对象的属性......”

<?php

$json='[{"Name":"Abcd","Alias":["Bcde","Cdef","Fghi","Jklm","Load More"]}]';
$obj=json_decode($json);
foreach($obj->Alias as $val) // Error: Trying to get property of non-object<br/>
echo $val.'<br/>';
?>

解码的json数组如下

Array
(
    [0] => stdClass Object
        (
            [Name] => Abcd
            [Alias] => Array
                (
                    [0] => Bcde
                    [1] => Cdef
                    [2] => Fghi
                    [3] => Jklm
                    [4] => Load More
                )

        )

)

我还想从结果

中排除最后一个“别名”元素（加载更多）

Plz ...提前帮助谢谢

Answer 1

使用array_pop弹出最后一个元素。

<?php

$json='[{"Name":"Abcd","Alias":["Bcde","Cdef","Fghi","Jklm","Load More"]}]';
$obj=json_decode($json);
$aliases = $obj[0]->Alias;
array_pop($aliases);
foreach($aliases as $alias) print $alias;

?>

Answer 2

$str = '[{"Name":"Abcd","Alias":["Bcde","Cdef","Fghi","Jklm","Load more"]}]';
print_r(json_decode($str, true));

请参阅http://php.net/manual/en/function.json-encode.php

上有关函数参数的文档

Answer 3

这是我的解决方案，它没有将对象转换为关联数组

   <?php

    $json='[{"Name":"Abcd","Alias":["Bcde","Cdef","Fghi","Jklm","Load More"]}]';
    $obj=json_decode($json);
    $obj = $obj[0];
    foreach($obj->Alias as $val)
    echo $val.'<br/>';

    ?>

我只能在发布后6小时发布回复，因为我在这里很新：）

使用php处理json字符串中的特定元素

3 个答案: