JSONException：无法将 java.lang.String 类型的值转换为 JSONArray
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JSONException: Value <br of type java.lang.String cannot be converted to JSONArray
提问by elfoz
I know that this kind of problem has some solutions and applied some solutions for this problem but I can't solve and i'm confused. Please help me. Here is code:
我知道这种问题有一些解决方案，并为这个问题应用了一些解决方案，但我无法解决，我很困惑。请帮我。这是代码：
protected String doInBackground(Boolean... params) {
String result = null;
StringBuilder sb = new StringBuilder();
try {
// http post
HttpClient httpclient = new DefaultHttpClient();
HttpGet httppost = new HttpGet(
"http://192.168.2.245/getProducts.php?login=1&user_name=UserName&password=Password");
HttpResponse response = httpclient.execute(httppost);
if (response.getStatusLine().getStatusCode() != 200) {
Log.d("MyApp", "Server encountered an error");
}
BufferedReader reader = new BufferedReader(new InputStreamReader(
response.getEntity().getContent(), "utf-8"), 8); //old charset iso-8859-1
sb = new StringBuilder();
sb.append(reader.readLine() + "\n");
String line = null;
while ((line = reader.readLine()) != null) {
sb.append(line + "\n");
}
result = sb.toString();
Log.d("test", result);
}
catch (Exception e) {
Log.e("log_tag", "Error converting result " + e.toString());
}
return result;
}
PHP CODE:
代码：
$login=$_GET["login"];
$user_name=$_GET["user_name"];
$password=$_GET["password"];
$output=array();
if ($login) {
$sql=mysql_query("SELECT user_id FROM users WHERE user_name='".$user_name."' AND user_pass='".$password."' ");
while($row=mysql_fetch_array($sql)) {
$user_id=$row["user_id"];
}
$sql=mysql_query("SELECT name,device_id,lat,lon FROM devices WHERE user_id='".$user_id."' LIMIT 100");
while($row=mysql_fetch_assoc($sql)) {
$output[]=$row; }
}
print(json_encode($output));
mysql_close();
The logcat:org.json.JSONException: Value
logcat:org.json.JSONException: Value
What should i do? The Php code runs on web page without any errors but why got an error in this part? In logcat also has: ( ! ) Notice: Undefined variable: user_id in C:\wamp\www\getProducts.php on line 27
我该怎么办？Php 代码在网页上运行没有任何错误，但为什么在这部分出现错误？在 logcat 中也有：（！）注意：未定义变量：user_id in C:\wamp\www\getProducts.php on line 27
Logcat!
逻辑猫！
<br /><font size='1'><table class='xdebug-error xe-notice' dir='ltr' border='1' cellspacing='0' cellpadding='1'><tr><th align='left' bgcolor='#f57900' colspan="5"><span style='background-color: #cc0000; color: #fce94f; font-size: x-large;'>( ! )</span> Notice: Undefined variable: user_id in C:\wamp\www\getProducts.php on line <i>27</i></th></tr><tr><th align='left' bgcolor='#e9b96e' colspan='5'>Call Stack</th></tr>... this kind of code appears in logcat then org.json.JSONException: Value <br of type java.lang.String cannot be converted to JSONArray at org.json.JSON.typeMismatch(JSON.java:107)
采纳答案by A.S.
You do not return a valid JSON, seems that you are returning a HTML snippet with <br>have a look at your returning source of your website.
您没有返回有效的 JSON，似乎您正在返回一个 HTML 片段，<br>并查看您网站的返回源。
What should i do? The Php code runs on web page without any errors but why got an error in this part? In logcat also has: ( ! ) Notice: Undefined variable: user_id in C:\wamp\www\getProducts.php on line 27
我该怎么办？Php 代码在网页上运行没有任何错误，但为什么在这部分出现错误？在 logcat 中也有：( ! ) 注意：未定义变量：C:\wamp\www\getProducts.php 中的 user_id 第 27 行
So the user_idis not defined! Just look at your GET/POST Parameters or show us your relevant php code.
所以user_id没有定义！只需查看您的 GET/POST 参数或向我们展示您的相关 php 代码。
Please use echo json_encode($output);not print.
请echo json_encode($output);不要使用print。
And what are you doing if there is no returning row from your first Query. Try this.
如果您的第一个 Query 没有返回行，您在做什么？尝试这个。
$sql=mysql_query("SELECT user_id FROM users WHERE user_name='".$user_name."' AND user_pass='".$password."' ");
if(mysql_num_rows($sql) == 0){
echo "USERID CANNOT BE FOUND";
}
while($row=mysql_fetch_array($sql)) {
echo "USERID FOUND" .$row["user_id"] ;
$user_id=$row["user_id"];
}
It seems that user_namecould not be found. So just debug it there by trying to echo your user_id after the while. If you do not get the UserID FOUND echo the user_name and password do not exist.
好像user_name找不到。因此，只需在一段时间后尝试回显您的 user_id 来调试它。如果您没有找到 UserID FOUND 回显，则 user_name 和密码不存在。
Edit: As your logact says, there is no Entry for the username you are passing so the $user_id will never be filled.
编辑：正如您的 logact 所说，您传递的用户名没有条目，因此永远不会填充 $user_id。
http://192.168.2.245/getProducts.php?login=1&user_name=UserName&password=Password
http://192.168.2.245/getProducts.php?login=1&user_name=UserName&password=Password
You are passing user_name = UserName and password = Password. Does this entry really exist in your Database?
您正在传递 user_name = UserName 和 password = Password。这个条目真的存在于您的数据库中吗？
Also for better performance have a look at MySQL LETF JOIN and MySQL subqueries.
另外，为了获得更好的性能，请查看 MySQL LETF JOIN 和 MySQL 子查询。
