Question

我对这个字符串有疑问：“NãoVou”。这意味着“我不去”。我试图用php / json文件将其插入数据库。

当我删除空格和〜重音时，它可以正常工作。

我尝试过：

将iso-8859-1更改为utf-8;
将eclipse编码更改为utf-8;
更改了print（json_encode（$ flag））以进行打印（json_encode（'utf8_encode'，$ flag））;

已经阅读了类似的帖子，但其中任何一个都帮助了我。

public class insertNo extends AsyncTask<String, Boolean, Boolean>{

@Override
protected Boolean doInBackground(String... params) {

    ArrayList<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>();
    nameValuePairs.add(new BasicNameValuePair("e_id", subString));
    nameValuePairs.add(new BasicNameValuePair("Não Vou", no));

    try{
        HttpClient httpClient = new DefaultHttpClient();
        HttpPost httpPost = new HttpPost("http://mywebsite.com/includes/insert_counter.php");
        httpPost.setEntity(new UrlEncodedFormEntity(nameValuePairs, "UTF-8"));
        HttpResponse httpResponse = httpClient.execute(httpPost);
        HttpEntity entity = httpResponse.getEntity();
        inputStream = entity.getContent();
        Log.e("pass 1", "connection success");
    }catch(Exception e){
        Log.e("Fail 1", e.toString());
        Log.e("Invalid IP", e.toString());
    }

    try {
        BufferedReader reader = new BufferedReader(new InputStreamReader(inputStream));
        StringBuilder stringBuilder = new StringBuilder();
        while ((line = reader.readLine()) != null) {                
            stringBuilder.append(line + "\n");
        }
        inputStream.close();
        result = stringBuilder.toString();
        Log.e("pass 2", "connection success");
    } catch (Exception e) {
        Log.e("Fail 2", e.toString());
    }

    try {
        JSONObject json_data = new JSONObject(result);
        code = (json_data.getInt("code"));
        if (code==1) {
            Log.d("Insert success", result);
        } else {
            Log.d("Sorry son", result);
        }
    } catch (Exception e) {
        Log.e("Fail 3", e.toString());
    }
    return null;
}

}

php文件

include_once("db_connect.php");

$e_id = $_REQUEST["e_id"];
$yes = $_REQUEST["Vou"];
$no = $_REQUEST["Não Vou"];
$maybe = $_REQUEST["Talvez"];

$flag['code']=0;


if(isset($yes)){
    $r = mysqli_query($mysqli, "UPDATE table1 SET yes = yes + 1 WHERE id='$e_id'");
    $flag['code']=1;
}else if(isset($no)){
    $r = mysqli_query($mysqli, "UPDATE table1 SET no = no + 1 WHERE id='$e_id'");
    $flag['code']=1;
}else if(isset($maybe)){
    $r = mysqli_query($mysqli, "UPDATE table1 SET maybe = maybe + 1 WHERE id='$e_id'");
    $flag['code']=1;
}

print(json_encode($flag));
mysqli_close($mysqli);

Answer 1

From the documentation，PHP变量不应包含特殊字符，空格或

之外的任何内容

a letter is a-z, A-Z, and the bytes from 127 through 255 (0x7f-0xff).

Android无法使用重音和空格插入数据库字符串

1 个答案: