org.json.JSONException：java.lang.String 类型的值 <br 无法转换为 JSONObject
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【中文标题】org.json.JSONException：java.lang.String 类型的值 <br 无法转换为 JSONObject【英文标题】：org.json.JSONException: Value <br of type java.lang.String cannot be converted to JSONObject 【发布时间】：2012-05-09 17:41:37 【问题描述】：我正在尝试从 android 活动将数据连接到我的 sqldatabase 进行注册页面，我收到此错误“org.json.JSONException: Value
首先，谁能建议我在将mysql数据库与php脚本用于android应用程序时如何调试我的程序？因为我通常使用日志猫，但这里的错误不是很清楚：S ...
这是活动代码：
public class Subscribe extends Activity
Button bSubscribe;
EditText etPwdSub, etPwdConf, etLoginSub, etNameSub, etFnSub;
String result = null;
InputStream is = null;
String donnees = "";
protected void onCreate(Bundle savedInstanceState)
super.onCreate(savedInstanceState);
setContentView(R.layout.subscribe);
etLoginSub = (EditText) findViewById(R.id.etLoginSub);
etPwdSub = (EditText) findViewById(R.id.etPwdSub);
etPwdConf = (EditText) findViewById(R.id.etPwdConf);
etNameSub = (EditText) findViewById(R.id.etNameSub);
etFnSub = (EditText) findViewById(R.id.etFnSub);
bSubscribe = (Button) findViewById(R.id.bSubscribe);
bSubscribe.setOnClickListener(new View.OnClickListener()
public void onClick(View v)
Pattern p = Pattern.compile(".+@.+\\.[a-z]+");
Matcher m = p.matcher(etLoginSub.getEditableText());
if (m.matches() == false)
Toast.makeText(
getBaseContext(),
"Le champs email ne correspond pas au format d'une adresse mail",
Toast.LENGTH_SHORT).show();
else
// autre méthode : etPwdSub.equals("")
if (etPwdSub.getEditableText() != null
&& etPwdConf.getEditableText() != null
&& etNameSub.getEditableText() != null
&& etFnSub.getEditableText() != null)
if (etPwdSub.getEditableText().toString().equals(etPwdConf.getEditableText().toString()))
ArrayList<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>();
nameValuePairs.add(new BasicNameValuePair("login", etLoginSub.getText().toString()));
nameValuePairs.add(new BasicNameValuePair("pwd", etPwdConf.getText().toString()));
nameValuePairs.add(new BasicNameValuePair("name", etNameSub.getText().toString()));
nameValuePairs.add(new BasicNameValuePair("firstname", etFnSub.getText().toString()));
try
// commandes httpClient
HttpClient httpclient = new DefaultHttpClient();
HttpPost httppost = new HttpPost(
"http://192.168.1.101/spotnshare/subscribe.php");
httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
HttpResponse response = httpclient.execute(httppost);
HttpEntity entity = response.getEntity();
is = entity.getContent();
catch (Exception e)
Log.i("taghttppost", "" + e.toString());
Toast.makeText(getBaseContext(), e.toString(), Toast.LENGTH_LONG)
.show();
try
BufferedReader reader = new BufferedReader(new InputStreamReader(
is, "UTF-8"));
StringBuilder sb = new StringBuilder();
String line = null;
while ((line = reader.readLine()) != null)
sb.append(line + "\n");
is.close();
result = sb.toString();
catch (Exception e)
Log.i("tagconvertstr", "" + e.toString());
try
JSONObject jObj = new JSONObject(result);
donnees = jObj.getString("message");
Intent ourIntent = new Intent(Subscribe.this,
SubscribeMess.class);
// objet qui vas nous permettre de passe des variables ici la
// variable passInfo
Bundle objetbunble = new Bundle();
objetbunble.putString("message", donnees);
ourIntent.putExtras(objetbunble); // on passe notre objet dans l'intent
// on appelle notre activité
startActivity(ourIntent);
catch (JSONException e)
Log.i("tagjsonexp", "" + e.toString());
catch (ParseException e)
Log.i("tagjsonpars", "" + e.toString());
else
Dialog d = new Dialog(Subscribe.this);
d.setTitle(etPwdSub.getEditableText() +" "+etPwdConf.getEditableText());
d.show();
else
Dialog d = new Dialog(Subscribe.this);
d.setTitle("Fill in all the fields !");
d.show();
);
protected void onPause()
super.onPause();
finish();
这里是 php 脚本：
<?php
if( isset($_POST['login']) && isset($_POST['pwd']) && isset($_POST['name']) && isset($_POST['firstname']))
include("connexion_bdd.php");
if(connexionBDD() == 1)
$login = $_POST['login'];
$pwd = $_POST['pwd'];
$name = $_POST['name'];
$firstname = $_POST['firstname'];
$sql = "SELECT colUserID
FROM userTable
WHERE colUserLogin = '".$login."' ";
$req = mysql_query($sql);
$resultat=mysql_num_rows($req);
if($resultat==0)
$temps = time();
$clef = md5($login . $temps);
$req = mysql_query("INSERT INTO userTable(colUserLogin, colUserPwd, colUserName, colUserFirstname, colUserKey, colUserDate)
VALUES( '$login', '$pwd', '$name', '$firstname', '$clef', '$temps')");
if($req)
$destinataire = $login;
$sujet ="Welcome on SnSR";
$from = "From: SpotnShareReminder@live.com \r\n";
$from .= "Content-Type: text/html; charset=us-ascii\r\n";
$message = ' Clic on the link below :<br/>
<a href="http://localhost/spotnshare/validation_mail.php?usrk='.$clef.' ">
Registration confirmation.
</a> ';
ini_set('SMTP','relay.skynet.be');
if(mail($destinataire,$sujet,$message,$from))
$msg = 'Check your mailbox to activate your account !';
else
$msg = 'Problem sending you the activation mail !';
$req = mysql_query("DELETE FROM userTable WHERE colUserLogin = '".$pseudo."' ");
else
$msg = 'Problem inserting you in our database !';
else
$msg = 'This email has already been used !';
mysql_free_result ($req);
else
$msg = "Connexion problem with de DB"
print(json_encode(array("message" => $msg)));
else
$msg = "Couldn't treat your datas"
print(json_encode(array("message" => $msg)));
?>
【问题讨论】：
【参考方案1】：您对http://192.168.1.101/spotnshare/subscribe.php 的请求失败并返回非 JSON 字符串（可能是 PHP 错误）。您可以使用
打印出该值Log.i("tagconvertstr", "["+result+"]");
在new JSONObject 调用之前，在解析它之前看看你得到了什么。
编辑：如果您使用的是 Eclipse，您可以设置一个断点并逐步查看发生了什么。
【讨论】：
感谢我设法纠正了一些错误 2 ';'丢失了，我可以看到 Log.i("tagconvertstr", "["+result+"]"); 的错误显示的是这样的： 我遇到了同样的情况，谁能帮帮我？【参考方案2】：感谢我设法纠正了一些错误 2 ';'丢失了，我可以看到 Log.i("tagconvertstr", "["+result+"]"); 的错误
它显示的味精是这样的：
[<br/ > font size = 1 table class=''xdebug-erroe' dir='ltr' ... loads of html code that wasn't in my initial code then...."message":"Problem sending you the activation mail !"]
因此，消息“向您发送激活邮件时出现问题”的 json 格式存在问题，但用户注册成功！
所以我第二次尝试该代码时，它会以正确的 json 格式显示：“此电子邮件已被使用”！ （没有任何错误）但我仍然无法在我的 php 代码中找到错误：S
【讨论】：
【参考方案3】：尝试在打印之前阻止任何输出并将print 更改为echo
您也可以删除关闭的?> 以防止在 php 脚本之后进一步输出
<?php
ob_start();
if( isset($_POST['login']) && isset($_POST['pwd']) && isset($_POST['name']) && isset($_POST['firstname']))
include("connexion_bdd.php");
if(connexionBDD() == 1)
$login = $_POST['login'];
$pwd = $_POST['pwd'];
$name = $_POST['name'];
$firstname = $_POST['firstname'];
$sql = "SELECT colUserID
FROM userTable
WHERE colUserLogin = '".$login."' ";
$req = mysql_query($sql);
$resultat=mysql_num_rows($req);
if($resultat==0)
$temps = time();
$clef = md5($login . $temps);
$req = mysql_query("INSERT INTO userTable(colUserLogin, colUserPwd, colUserName, colUserFirstname, colUserKey, colUserDate)
VALUES( '$login', '$pwd', '$name', '$firstname', '$clef', '$temps')");
if($req)
$destinataire = $login;
$sujet ="Welcome on SnSR";
$from = "From: SpotnShareReminder@live.com \r\n";
$from .= "Content-Type: text/html; charset=us-ascii\r\n";
$message = ' Clic on the link below :<br/>
<a href="http://localhost/spotnshare/validation_mail.php?usrk='.$clef.' ">
Registration confirmation.
</a> ';
ini_set('SMTP','relay.skynet.be');
if(mail($destinataire,$sujet,$message,$from))
$msg = 'Check your mailbox to activate your account !';
else
$msg = 'Problem sending you the activation mail !';
$req = mysql_query("DELETE FROM userTable WHERE colUserLogin = '".$pseudo."' ");
else
$msg = 'Problem inserting you in our database !';
else
$msg = 'This email has already been used !';
mysql_free_result ($req);
else
$msg = "Connexion problem with de DB"
print(json_encode(array("message" => $msg)));
else
$msg = "Couldn't treat your datas"
ob_end_clean()
echo(json_encode(array("message" => $msg)));
【讨论】：
【参考方案4】：您应该在发回响应之前设置Content-Type 标头。
header('Content-Type: application/json');
print(json_encode(array("message" => $msg)));
检查这个：Returning JSON from a PHP Script
【讨论】：
【参考方案5】：我也有同样的问题；我找到了简单的方法。
在 Java 代码中只需键入 Log.e("anyText",response);
在 logCat 中它会告诉你什么是问题
【讨论】：
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