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我正在嘗試使用this文檔的參考進行發佈請求。但問題是另一端的PHP開發人員無法接收參數的值，因此無法發送正確的響應。我在這裏錯過了什麼嗎？URLConnection和POST方法android
//編輯;
我正在發出一個HTTP Post請求。正如你可以看到下面的代碼。我將參數和參數（location_id = 3）寫入輸出流。我也粘貼了我一直在使用的PHP代碼。現在的問題是：
在PHP代碼中沒有收到參數值（它是3），所以我得到一個被else塊包圍的響應。所以，我只是想知道是否有在Android的代碼錯誤或PHP代碼
@覆蓋 保護布爾doInBackground（字符串... PARAMS）{
Log.d(Constants.LOG_TAG,Constants.FETCH_ALL_THEMES_ASYNC_TASK);
Log.d(Constants.LOG_TAG," The url to be fetched "+params[0]);
try {
url = new URL(params[0]);
urlConnection = (HttpURLConnection) url.openConnection();
// /* optional request header */
// urlConnection.setRequestProperty("Content-Type", "application/json");
//
// /* optional request header */
// urlConnection.setRequestProperty("Accept", "application/json");
/* for Get request */
urlConnection.setChunkedStreamingMode(0);
urlConnection.setDoOutput(true);
urlConnection.setDoInput(true);
urlConnection.setRequestMethod("POST");
List<BasicNameValuePair> nameValuePairs = new ArrayList<BasicNameValuePair>();
nameValuePairs.add(new BasicNameValuePair("location_id",params[1]));
outputStream = urlConnection.getOutputStream();
BufferedWriter bufferedWriter = new BufferedWriter(new OutputStreamWriter(outputStream));
bufferedWriter.write(writeToOutputStream(nameValuePairs));
int statusCode = urlConnection.getResponseCode();
/* 200 represents HTTP OK */
if (statusCode == 200) {
inputStream = new BufferedInputStream(urlConnection.getInputStream());
response = convertInputStreamToString(inputStream);
Log.d(Constants.LOG_TAG, " The response is " + response);
return true;
}
else {
return false;
}
} catch (Exception e) {
e.printStackTrace();
} finally {
try {
if(inputStream != null){
inputStream.close();
}
if(outputStream != null){
outputStream.close();
}
} catch (IOException e) {
e.printStackTrace();
}
}
return false;
}
//這裏是代碼對於writeToOutputStream
public String writeToOutputStream(List<BasicNameValuePair> keyValuePair)throws UnsupportedEncodingException{
String result="";
boolean firstTime = true;
for(BasicNameValuePair pair : keyValuePair){
if(firstTime){
firstTime = false;
}
else{
result = result.concat("&");
}
result = result + URLEncoder.encode(pair.getKey(), "UTF-8");
result = result + "=";
result = result+ URLEncoder.encode(pair.getValue(),"UTF-8");
}
Log.d(Constants.LOG_TAG," The result is "+result);
return result;
}
//這裏是convertInputStream爲字符串
public String convertInputStreamToString(InputStream is) throws IOException {
String line="";
String result="";
BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(is));
while((line = bufferedReader.readLine()) != null){
Log.d(Constants.LOG_TAG," The line value is "+line);
result += line;
}
/* Close Stream */
if(null!=inputStream){
inputStream.close();
}
return result;
}
這裏是PHP代碼
<?php
include 'config.php';
header ('Content-Type:application/json');
if(isset($_POST['location_id']))
{
$id=$_POST['location_id'];
$selectThemeQuery = mysql_query("select theme_id from location_theme where location_id='$id'",$conn) or die (mysql_error());
$noRows = mysql_num_rows($selectThemeQuery);
//echo "HI";
if($noRows > 0)
{
$result = array();
while($row = mysql_fetch_assoc($selectThemeQuery))
{
$themeid = $row['theme_id'];
//echo "HI";
$selectNameQuery = mysql_query("select theme_name,theme_image from theme where theme_id='$themeid'",$conn) or die(mysql_error());
$numRows = mysql_num_rows($selectNameQuery);
if($numRows > 0)
{
while($rows = mysql_fetch_assoc($selectNameQuery))
{
$name = $rows['theme_name'];
$image = $rows['theme_image'];
$result[] = array('theme_id'=>$themeid,'theme_name'=>$name, 'theme_image'=>$image);
}
}
}
//echo json_encode($result);
echo json_encode("Hi");
}
else
{
$data2[] = array('Notice'=>false);
echo json_encode($data2);
}
}
else
{
echo "Not Proper Data";
}
?>
只是一個建議：你有沒有考慮過使用可以爲你做95％的東西的東西？例如，POST + JSON聽起來像「REST Service」，爲此Spring-Web提供了一些很棒的工具（但可能有幾十個其他的庫） –
@FlorianSchaetz：請你詳細說明一下，我希望你正在回答android的解決方案。 –
'另一端的PHP開發人員無法接收參數的值。那麼開始告訴他如何接收哪個參數。否則，我們必須猜測，您可以發佈您的PHP腳本。 – greenapps