Question

我正在尝试从JSON文件中获取PHP数据。这只是我JSON中的一个数据块...在这个JSON文件中，还有50个以上。这是：

{
"ConsultarNfseResposta": {
"ListaNfse": {
  "CompNfse": [
    {
      "Nfse": {
        "InfNfse": {
          "Numero": "12651",
          "CodigoVerificacao": "ECSV-EJZZ",
          "DataEmissao": "2017-07-25T17:51:12",
          "NaturezaOperacao": "1",
          "OptanteSimplesNacional": "1",
          "IncentivadorCultural": "2",
          "Competencia": "2017-07-25T00:00:00",
          "Servico": {
            "Valores": {
              "ValorServicos": "2350",
              "IssRetido": "2",
              "BaseCalculo": "2350",
              "Aliquota": "0.02",
              "ValorLiquidoNfse": "2350"
            },
            "ItemListaServico": "0107",
            "CodigoTributacaoMunicipio": "6209100",
            "Discriminacao": "TAXA: SERVIÇO DE VOTAÇÃO ELETRÔNICA",
            "CodigoMunicipio": "2611606"
          },
          "PrestadorServico": {
            "IdentificacaoPrestador": {
              "Cnpj": "41069964000173",
              "InscricaoMunicipal": "2427745"
            },
            "RazaoSocial": "INFORMATICA LTDA",
            "Endereco": {
              "Endereco": "RUA 241",
              "Numero": "241",
              "Bairro": "Exemplo",
              "CodigoMunicipio": "2611606",
              "Uf": "PE",
              "Cep": "52030190"
            },
            "Contato": {
              "Telefone": "33254854",
              "Email": "exemplo@exemplo.com.br"
            }
          },
          "TomadorServico": {
            "IdentificacaoTomador": {
              "CpfCnpj": {
                "Cnpj": "00085803000196"
              }
            },
            "RazaoSocial": "EXEMPLO - AMBR",
            "Endereco": {
              "Endereco": "ST 06",
              "Bairro": "Asa Sul",
              "CodigoMunicipio": "5300108",
              "Uf": "DF",
              "Cep": "15425845211"
            },
            "Contato": {
              "Email": "exemplo@gmail.com"
            }
          },
          "OrgaoGerador": {
            "CodigoMunicipio": "2611606",
            "Uf": "PE"
          }
         }
       }
      }
    ]
   }
  }
 }

这是我的PHP代码：

<?php

$json = file_get_contents('arquivo.json');

$json_data = json_decode($json,true);

for ($i=0; $i < count($json_data->ConsultarNfseResposta->ListaNfse->CompNfse), $i++;) {

    echo $json_data->ConsultarNfseResposta->ListaNfse->CompNfse[$i]->Nfse->InfNfse->Numero;
    echo $json_data->ConsultarNfseResposta->ListaNfse->CompNfse[$i]->Nfse->InfNfse->CodigoVerificacao;

}

?>

我在第7行遇到错误： 注意：尝试获取非对象的属性。我已经尝试了所有内容，而且我的代码也没有用。

我该怎么办？谢谢！

Answer 1

json_decode()调用中的第二个参数告诉PHP将对象转换为数组。见the manual。如果您删除true。

，您应该没问题

Answer 2

@FernandoJuriolli，你的for声明不对，你在错误的地方有一个逗号 - 它应该是这样的：

for ($i=0; $i < count($json_data->ConsultarNfseResposta->ListaNfse->CompNfse); $i++) {

来自@ jh1711的答案，这应该有效

Answer 3

使用foreach：

$json_data = json_decode(file_get_contents('arquivo.json'));

foreach ($json_data->ConsultarNfseResposta->ListaNfse->CompNfse as $data) {
    echo $data->Nfse->InfNfse->Numero;
    echo $data->Nfse->InfNfse->CodigoVerificacao;
}

尝试使用PHP从JSON文件获取非对象的属性时出错

3 个答案: