本文介绍了org.json.JSONException：值＆LT; HTML＆GT;＆LT; HEAD＆GT;＆LT; java.lang.String类型风格不能转换到的JSONObject的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！
问题描述
有人可以请帮我找出什么是错的我试图我的android应用到mysql数据库连接使用PHP和JSON一个简单的登录屏幕。这是我在logcat中越来越能有人告诉我是什么问题的第一个错误。这是我的PHP文件或JSON code
＆LT; PHP
//装载并连接到MySQL数据库的东西
要求（config.inc.php文件）;如果（！空（$ _ POST））{
//获取基于掀起了用户名用户的信息。
$查询=
选择
ID，
用户名，
密码
来自用户的
哪里
用户名=：用户名
$ query_params =阵列（
：用户名=＆GT; $ _ POST [用户名]
）;尝试{
$语句= $ DB-GT＆; prepare（$查询）;
$结果= $ stmt-＆GT;执行（$ query_params）;
}
赶上（PDOException $前）{
//为了测试，你可以使用模具和消息。
//模具（无法运行的问题：$ EX-＆GT;的getMessage（））; //或使用此用这个产品JSON数据：
$响应[成功] = 0;
$响应[消息] =数据库ERROR1请重试！
死亡（json_en code（$响应））;}
$ validated_info = FALSE;//从查询获取所有的行
$行= $ stmt-＆GT;取（）;
如果（$行）{
//如果我们加密的口令，我们就在这里解密，但在我们的例子中，我们只
//比较两个密码
如果（$ _ POST ['密码'] === $行['密码']）{
$ login_ok = TRUE;
}
}
如果（$ login_ok）{
$响应[成功] = 1;
$响应[消息] =登录成功！;
死亡（json_en code（$响应））;
}其他{
$响应[成功] = 0;
$响应[消息] =！无效的凭证;
死亡（json_en code（$响应））;
}
}其他{
？＆GT;
＆LT; H1＆GT;登录和LT; / H1＆GT;
＆LT;形式的行动=login.php中方法=后＆GT;
用户名：其中; BR /＆GT;
＆LT;输入类型=文本名称=用户名占位符=用户名/＆GT;
＆LT; BR /＆GT;＆LT; BR /＆GT;
密码：LT; BR /＆GT;
＆LT;输入类型=密码NAME =密码占位符=密码值='输入code here` /＆GT;
＆LT; BR /＆GT;＆LT; BR /＆GT;
＆LT;输入类型=提交值=登录/＆GT;
＆LT; /表及GT;
＆所述; A HREF =register.php＆GT;寄存器＆下; / A＆GT;
＆LT; PHP
}？＆GT;
包com.example.mysqltest;进口java.io.BufferedReader中;
进口java.io.IOException异常;
进口的java.io.InputStream;
进口java.io.InputStreamReader中;
进口java.io.UnsupportedEncodingException;
进口的java.util.List;进口org.apache.http.HttpEntity;
进口org.apache.http.Htt presponse;
进口org.apache.http.NameValuePair;
进口org.apache.http.client.ClientProtocolException;
进口org.apache.http.client.entity.UrlEn codedFormEntity;
进口org.apache.http.client.methods.HttpGet;
进口org.apache.http.client.methods.HttpPost;
进口org.apache.http.client.utils.URLEn codedUtils;
进口org.apache.http.impl.client.DefaultHttpClient;
进口org.json.JSONException;
进口org.json.JSONObject;进口android.util.Log;
code：
公共类JSONParser {静态InputStream为= NULL;
静态的JSONObject jObj = NULL;
静态JSON字符串=;//构造
公共JSONParser（）{}
公众的JSONObject getJSONFromUrl（最终字符串URL）{ //使HTTP请求
尝试{
//构建客户端和HTTP请求。
DefaultHttpClient的HttpClient =新DefaultHttpClient（）;
HttpPost httpPost =新HttpPost（URL）; //执行POST请求，并在本地存储的响应。
HTT presponse HTT presponse = httpClient.execute（httpPost）;
//从响应中提取数据。
HttpEntity httpEntity = HTT presponse.getEntity（）;
//打开与数据内容的InputStream。
是= httpEntity.getContent（）; }赶上（UnsupportedEncodingException五）{
e.printStackTrace（）;
}赶上（ClientProtocolException E）{
e.printStackTrace（）;
}赶上（IOException异常五）{
e.printStackTrace（）;
} 尝试{
//创建一个BufferedReader通过InputStream的解析。
读者的BufferedReader =新的BufferedReader（新的InputStreamReader（
是，ISO-8859-1），8）;
//声明一个字符串生成器，以帮助解析。
StringBuilder的SB =新的StringBuilder（）;
//声明一个字符串JSON对象数据存储在字符串形式。
串线= NULL; //生成字符串，直到空。
而（（行= reader.readLine（））！= NULL）{
sb.append（行+\\ n）;
} //关闭输入流。
is.close（）;
//字符串生成器的数据转换为实际的字符串。
JSON = sb.toString（）;
}赶上（例外五）{
Log.e（缓冲区错误，错误转换结果+ e.toString（））;
} //尝试将字符串解析到一个JSON对象
尝试{
jObj =新的JSONObject（JSON）;
}赶上（JSONException E）{
Log.e（JSON解析器，错误分析数据+ e.toString（））;
} //返回JSON对象。
返回jObj;}
//函数得到JSON的网址
//通过HTTP POST或GET mehtod
公众的JSONObject makeHtt prequest（URL字符串，字符串的方法，
清单＆LT;＆的NameValuePair GT; PARAMS）{ //使HTTP请求
尝试{ //检查请求的方法
如果（方法==POST）{
//请求方法是POST
// defaultHttpClient
DefaultHttpClient的HttpClient =新DefaultHttpClient（）;
HttpPost httpPost =新HttpPost（URL）;
httpPost.setEntity（新UrlEn codedFormEntity（PARAMS））; HTT presponse HTT presponse = httpClient.execute（httpPost）;
HttpEntity httpEntity = HTT presponse.getEntity（）;
是= httpEntity.getContent（）; }否则如果（方法==GET）{
//请求方法是GET
DefaultHttpClient的HttpClient =新DefaultHttpClient（）;
字符串中的paramString = URLEn codedUtils.format（参数，可以UTF-8）;
网址+ =？ +中的paramString;
HTTPGET HTTPGET =新HTTPGET（URL）; HTT presponse HTT presponse = httpClient.execute（HTTPGET）;
HttpEntity httpEntity = HTT presponse.getEntity（）;
是= httpEntity.getContent（）;
} }赶上（UnsupportedEncodingException五）{
e.printStackTrace（）;
}赶上（ClientProtocolException E）{
e.printStackTrace（）;
}赶上（IOException异常五）{
e.printStackTrace（）;
} 尝试{
读者的BufferedReader =新的BufferedReader（新的InputStreamReader（
是，ISO-8859-1），8）;
StringBuilder的SB =新的StringBuilder（）;
串线= NULL;
而（（行= reader.readLine（））！= NULL）{
sb.append（行+\\ n）;
}
is.close（）;
JSON = sb.toString（）;
}赶上（例外五）{
Log.e（缓冲区错误，错误转换结果+ e.toString（））;
} //尝试分析字符串到一个JSON对象
尝试{
jObj =新的JSONObject（JSON）;
}赶上（JSONException E）{
Log.e（JSON解析器，错误分析数据+ e.toString（））;
} //返回JSON字符串
返回jObj; }
}
解决方案
从PHP脚本，如删除所有HTML：
}其他{
？＆GT;
＆LT; H1＆GT;登录和LT; / H1＆GT;
＆LT;形式的行动=login.php中方法=后＆GT;
用户名：其中; BR /＆GT;
＆LT;输入类型=文本名称=用户名占位符=用户名/＆GT;
＆LT; BR /＆GT;＆LT; BR /＆GT;
密码：LT; BR /＆GT;
＆LT;输入类型=密码NAME =密码占位符=密码值=/＆GT;
＆LT; BR /＆GT;＆LT; BR /＆GT;
＆LT;输入类型=提交值=登录/＆GT;
＆LT; /表及GT;
＆所述; A HREF =register.php＆GT;寄存器＆下; / A＆GT;
＆LT; PHP
}
由于JSON API应该只回声JSON数据。
另外添加以下到PHP的顶部：
标题（内容类型：应用程序/ JSON'）;
can someone please help me figure out what is wrong i am trying to connect my android application to mysql database for a simple login screen using php and json. that is the first error i am getting in the logcat could somebody tell me what the problem is. is it my php file or json code
<?php
//load and connect to MySQL database stuff
require("config.inc.php");
if (!empty($_POST)) {
//gets user's info based off of a username.
$query = "
SELECT
id,
username,
password
FROM users
WHERE
username = :username
";
$query_params = array(
':username' => $_POST['username']
);
try {
$stmt = $db->prepare($query);
$result = $stmt->execute($query_params);
}
catch (PDOException $ex) {
// For testing, you could use a die and message.
//die("Failed to run query: " . $ex->getMessage());
//or just use this use this one to product JSON data:
$response["success"] = 0;
$response["message"] = "Database Error1. Please Try Again!";
die(json_encode($response));
}
$validated_info = false;
//fetching all the rows from the query
$row = $stmt->fetch();
if ($row) {
//if we encrypted the password, we would unencrypt it here, but in our case we just
//compare the two passwords
if ($_POST['password'] === $row['password']) {
$login_ok = true;
}
}
if ($login_ok) {
$response["success"] = 1;
$response["message"] = "Login successful!";
die(json_encode($response));
} else {
$response["success"] = 0;
$response["message"] = "Invalid Credentials!";
die(json_encode($response));
}
} else {
?>
<h1>Login</h1>
<form action="login.php" method="post">
Username:<br />
<input type="text" name="username" placeholder="username" />
<br /><br />
Password:<br />
<input type="password" name="password" placeholder="password" value="" `enter code here`/>
<br /><br />
<input type="submit" value="Login" />
</form>
<a href="register.php">Register</a>
<?php
}
?>
package com.example.mysqltest;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.UnsupportedEncodingException;
import java.util.List;
import org.apache.http.HttpEntity;
import org.apache.http.HttpResponse;
import org.apache.http.NameValuePair;
import org.apache.http.client.ClientProtocolException;
import org.apache.http.client.entity.UrlEncodedFormEntity;
import org.apache.http.client.methods.HttpGet;
import org.apache.http.client.methods.HttpPost;
import org.apache.http.client.utils.URLEncodedUtils;
import org.apache.http.impl.client.DefaultHttpClient;
import org.json.JSONException;
import org.json.JSONObject;
import android.util.Log;
Code:
public class JSONParser {
static InputStream is = null;
static JSONObject jObj = null;
static String json = "";
// constructor
public JSONParser() {
}
public JSONObject getJSONFromUrl(final String url) {
// Making HTTP request
try {
// Construct the client and the HTTP request.
DefaultHttpClient httpClient = new DefaultHttpClient();
HttpPost httpPost = new HttpPost(url);
// Execute the POST request and store the response locally.
HttpResponse httpResponse = httpClient.execute(httpPost);
// Extract data from the response.
HttpEntity httpEntity = httpResponse.getEntity();
// Open an inputStream with the data content.
is = httpEntity.getContent();
} catch (UnsupportedEncodingException e) {
e.printStackTrace();
} catch (ClientProtocolException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}
try {
// Create a BufferedReader to parse through the inputStream.
BufferedReader reader = new BufferedReader(new InputStreamReader(
is, "iso-8859-1"), 8);
// Declare a string builder to help with the parsing.
StringBuilder sb = new StringBuilder();
// Declare a string to store the JSON object data in string form.
String line = null;
// Build the string until null.
while ((line = reader.readLine()) != null) {
sb.append(line + "\n");
}
// Close the input stream.
is.close();
// Convert the string builder data to an actual string.
json = sb.toString();
} catch (Exception e) {
Log.e("Buffer Error", "Error converting result " + e.toString());
}
// Try to parse the string to a JSON object
try {
jObj = new JSONObject(json);
} catch (JSONException e) {
Log.e("JSON Parser", "Error parsing data " + e.toString());
}
// Return the JSON Object.
return jObj;
}
// function get json from url
// by making HTTP POST or GET mehtod
public JSONObject makeHttpRequest(String url, String method,
List<NameValuePair> params) {
// Making HTTP request
try {
// check for request method
if(method == "POST"){
// request method is POST
// defaultHttpClient
DefaultHttpClient httpClient = new DefaultHttpClient();
HttpPost httpPost = new HttpPost(url);
httpPost.setEntity(new UrlEncodedFormEntity(params));
HttpResponse httpResponse = httpClient.execute(httpPost);
HttpEntity httpEntity = httpResponse.getEntity();
is = httpEntity.getContent();
}else if(method == "GET"){
// request method is GET
DefaultHttpClient httpClient = new DefaultHttpClient();
String paramString = URLEncodedUtils.format(params, "utf-8");
url += "?" + paramString;
HttpGet httpGet = new HttpGet(url);
HttpResponse httpResponse = httpClient.execute(httpGet);
HttpEntity httpEntity = httpResponse.getEntity();
is = httpEntity.getContent();
}
} catch (UnsupportedEncodingException e) {
e.printStackTrace();
} catch (ClientProtocolException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}
try {
BufferedReader reader = new BufferedReader(new InputStreamReader(
is, "iso-8859-1"), 8);
StringBuilder sb = new StringBuilder();
String line = null;
while ((line = reader.readLine()) != null) {
sb.append(line + "\n");
}
is.close();
json = sb.toString();
} catch (Exception e) {
Log.e("Buffer Error", "Error converting result " + e.toString());
}
// try parse the string to a JSON object
try {
jObj = new JSONObject(json);
} catch (JSONException e) {
Log.e("JSON Parser", "Error parsing data " + e.toString());
}
// return JSON String
return jObj;
}
}
解决方案
Remove any HTML from the PHP Script such as:
} else {
?>
<h1>Login</h1>
<form action="login.php" method="post">
Username:<br />
<input type="text" name="username" placeholder="username" />
<br /><br />
Password:<br />
<input type="password" name="password" placeholder="password" value="" />
<br /><br />
<input type="submit" value="Login" />
</form>
<a href="register.php">Register</a>
<?php
}
Because JSON API are supposed to only echo JSON data.
Also Add the following to the top of your PHP:
header('Content-type: application/json');
这篇关于org.json.JSONException：值＆LT; HTML＆GT;＆LT; HEAD＆GT;＆LT; java.lang.String类型风格不能转换到的JSONObject的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！