从上传的文件中读取值并将其发送到 url 地址 - To read value from uploaded file and send it to url address

To read value from uploaded file and send it to url address

本文关键字：地址 url 文件读取 | 更新日期: 2024-02-01

<? 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; 
?>

博夫提到的程序，

我正在尝试从上传的文件中逐行检索值，并将检索值与 url 连接起来，

但是我收到了错误消息。

注意：尝试获取非对象的属性 C：''xampp''htdocs''phpprog''upload_file_add.php 在第 50 行...

我的描述错误在哪里...

我想您需要在使用json_decode之前检查您的$jsondata变量是否有效。可能是您在$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(来验证您收到的期望。

您给我们的代码具有我们不知道它们来自何处的变量，例如您的$f_open和$json。如果不使用 fopen() 打开文件，则无法使用fgets()。

$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";

读取包含reading_files_with_PHP的文本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