Question

<? if(isset($_POST["submit"]))
{

$f_name = $_FILES["filetoupload"]["name"];
$f_tmp = $_FILES["filetoupload"]["tmp_name"]; 
$store = "uploads/".$f_name;

if(move_uploaded_file($f_tmp,$store))
    echo "file uploaded successfully";
    echo"<br>";
}

$line = fgets($f_open);
echo $line;
$url = "http://maps.googleapis.com/maps/api/geocode/json?address=";
$furl ="$url"."$line";
echo "$furl";
$ch = curl_init("$furl");
$fp = fopen("example4.txt","w");

curl_setopt($ch, CURLOPT_PROXY, '10.10.80.11:3128');
curl_setopt($ch, CURLOPT_FILE, $fp);
curl_setopt($ch, CURLOPT_HEADER, 0);

curl_exec($ch);
curl_close($ch);
fclose($fp);


echo $json['results'][0]['address_components'][0]['types'][0];
echo $json['results'][0]['address_components'][0]['types'][1];

$data=json_decode($jsondata); 
$address=$data->results[0]->address_components; 
?>

bove提到的程序，

我正在尝试从上传的文件中逐行检索值，并将检索值与url连接，

但是我收到了错误消息..

注意：尝试获取非对象的属性第50行的C：\ xampp \ htdocs \ phpprog \ upload_file_add.php ......

我的描述错误在哪里......

Answer 1

我猜你需要在使用json_decode之前检查你的$ jsondata变量是否有效。您可能在$ data中收到null，因此当您尝试访问其属性时，它不再是对象。在json_decode（$ jsondata）之后使用var_dump（$ data）来验证你是否正在接收你期望的内容。

Answer 2

您向我们提供的代码中包含我们不知道其来源的变量，例如$f_open和$json。如果不使用fgets()打开文件，则无法使用fopen()。

$f_name = $_FILES["filetoupload"]["name"];
$f_tmp = $_FILES["filetoupload"]["tmp_name"]; 
$store = "uploads/".$f_name;

if(move_uploaded_file($f_tmp,$store)){
   echo "file uploaded successfully";
   echo"<br>";
}
$f_open = fopen($store,"r");
$line = fgets($f_open);
echo $line;
$url = "http://maps.googleapis.com/maps/api/geocode/json?address=";
$furl ="$url"."$line";
echo "$furl";

阅读包含reading_files_with_PHP的文字文件上面的代码显示

reading_files_with_PHP
http://maps.googleapis.com/maps/api/geocode/json?address=reading_files_with_PHP

从上传的文件中读取值并将其发送到URL地址

2 个答案: