Question

这里是

的示例

$arr = json_decode('{"people":[
{
  "id": "8080",
  "content": "foo",
  "member": [123, 456],
  "interval": 7
},
{ 
  "id": "8097",
  "content": "bar",
  "member": [1234, 4567],
  "interval": 7
}


]}', true);

$searchId = 123;
$results = array_filter($arr['people'], function($people) use ($searchId) {
    return in_array($searchId, $people['member']);
});

$final = json_encode($results);

echo $final;

这会打印[{“ id”：“ 8080”，“ content”：“ foo”，“ member”：[123,456]，“ interval”：7}]

但是当我尝试获取某个元素的特定值（例如“内容”）时，它显示了一个非法的字符串偏移错误

echo $final["content"];

我应该怎么做才能显示“内容”的价值？（在这种情况下为“ foo”）

Answer 1

将echo $final["content"];更改为echo $results[0]["content"];

Answer 2

实际上，您不需要let newArray = varNames.filter({!reservedWords.contains($0)})

直接输出到

$final = json_encode($results);

数组显示但子数组显示“非法字符串偏移”错误-PHP

2 个答案: