I am developing an Android app. Here in my app an AsyncTask sends a string to PHP. The PHP returns with an array of crop names.
My php file is:
<?php
ini_set('default_charset', 'utf-8');
header('Content-Type: text/html; charset=UTF-8');
mysql_connect("localhost","root","");
mysql_select_db("farm_o_pedia");
mysql_set_charset('utf8');
$result1 = mysql_query("SET NAMES utf8");
$lang=$_POST['LanguageName'];
$query1="select lang_id from lang_selection where lang_name='$lang'";
$lang_id=mysql_query($query1) or die(mysql_error());
$query2="select crop_name from crop_master where lang_id=$lang_id";
$result2=mysql_query($query2) or die(mysql_error());
while($row=mysql_fetch_assoc($result2))
{
$output[]=$row;
}
print(json_encode($output));
mysql_close();
?>
I get this sort of error in Logcat:
02-16 22:08:04.216: I/HTTP ok(1251): org.apache.http.message.BasicHttpResponse@4052f0b8 02-16 22:08:04.216 : I/JsonObj(1251): 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 'id #4' at line 1
I don't know whats wrong with my syntax.
EDIT: Used PDO as suggested and its working fine: This is my new script.
<?php
$db = new PDO('mysql:host=localhost;dbname=farm_o_pedia;charset=utf8', 'root', '');
$db->query("SET NAMES utf8");
$lang=$_POST['LanguageName'];
$query1="select crop_name from crop_master where lang_id=(select lang_id from lang_selection where lang_name='$lang')";
$result2=$db->query($query1);
while(($row=$result2->fetch(PDO::FETCH_ASSOC))!=false)
{
$output[]=$row;
}
print(json_encode($output));
$db=null;
?>
I'm not sure about the syntax for your queries, but I do know that you are better off with one query instead of two. Something like this:
select crop_name
from crop_master cm join lang_selection ls on cm.languageid = ls.languageid
where lang_name = '$lang'
If you are not familiar with joining tables, I've heard good things about the book, Teach Yourself SQL in 10 Minutes
$query2="select crop_name from crop_master where lang_id=$lang_id";
Try replacing it with:
$query2="select crop_name from crop_master where lang_id='$lang_id'";
It seems that you were trying to retrieve crop_name
where the lang_id is equal to $lang_id (which would be an invalid identifier) and not the value of $lang_id itself.
Note you may want to read up on SQL Injection Attacks
:
For instance if a user passed in 1;DROP TABLE USERS
as the LanguageName
parameter the following would be executed in SQL:
$query1="select lang_id from lang_selection where lang_name=1;DROP TABLE USERS;