我是使用PHP和Java的新手。我正在制作一个Android应用程序,但我遇到了SQL语法错误。。。错误:
返回Java:
查看与MySQL服务器对应的手册第1行"@mail.com"附近要使用的正确语法的版本。知道我该怎么解决吗。
我认为这是php脚本的一个问题。我该怎么解决这个问题。非常感谢任何帮助
// Login by email and password if access success setId()
// Saved Email as static string "staticEmail" and used to get CustomerID from customer table
// Get & set CustomerID to "string qr_id" if email=".$email
/* error:
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server
version for the right syntax to use near '@mail.com' at line 1
*/
<?php
//connect to MySQL database
mysql_connect("localhost","name","pass") or die(mysql_error());
mysql_select_db("tls_db");
$output = array();
if (isset($_GET['email'])){
$email = $_GET['email'];
$sql = mysql_query("select CustomerID from customer where email=".$email) or die(mysql_error());
while($row=mysql_fetch_assoc($sql)){
$output[] = $row;
}
mysql_close();
print(json_encode($output));
}
?>
Java:
private void setId() {
InputStream is = null;
ArrayList<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>();
try {
HttpClient httpclient = new DefaultHttpClient();
HttpPost httppost = new HttpPost(
"http://"+URL+"/tls_db/log.php?email=" + staticEmail); //Post email 123@mail.com
httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
HttpResponse response = httpclient.execute(httppost);
HttpEntity entity = response.getEntity();
is = entity.getContent();
} catch (Exception e) {
Log.e("log_tag", "Error in http connection" + e.toString());
}
// Convert response to string
try {
BufferedReader reader = new BufferedReader(new InputStreamReader(
is, "iso-8859-1"), 8);
sb = new StringBuilder();
sb.append(reader.readLine() + "n");
String line = "";
while ((line = reader.readLine()) != null) {
sb.append(line + "n");
}
result = sb.toString();
is.close();
} catch (Exception e) {
Log.e("log_tag", "Error converting result " + e.toString());
}
try {
JSONArray jArray = new JSONArray(result);
for (int i = 0; i < jArray.length(); i++) {
JSONObject json_data = jArray.getJSONObject(i);
// Get CustomerId and set to (static string qr_id)
qr_id = json_data.getString("CustomerID");
}
} catch (JSONException e1) {
//iv.setVisibility(View.GONE);
Toast.makeText(getBaseContext(), "Server Data Error",
Toast.LENGTH_LONG).show();
} catch (ParseException e1) {
e1.printStackTrace();
}
// Open class QRcode
Intent iSuccess = new Intent(Login.this, QRcode.class);
startActivity(iSuccess);
}
您需要在SQL查询中引用$email:
$sql = mysql_query("select CustomerID from customer where email='".$email."'") or die(mysql_error());
顺便说一句,您的代码容易受到SQL注入的攻击。请务必阅读如何保护自己免受这种攻击。