I am trying to insert data in table in mysql database through php code but I am always getting following error:
Invalid query: Table 'whatsup_wp1.pushDevices' doesn't exist
I am using following code:
$deviceid = $_GET["deviceid"];
$link = mysql_connect('localhost', 'whatsup_wp1', 'XSvUCl0FugzV4');
if (!$link) {
die('Not connected : ' . mysql_error());
}
// make foo the current db
$db_selected = mysql_select_db('whatsup_wp1', $link);
if (!$db_selected) {
echo 'Can\'t use whatsup_wp1 : ' . mysql_error();
}
else
{
//echo 'connect';
}
//$query = "select count(*) from city";
//$query = "insert into devices (pushID) values('".$deviceid."')";
$query = "INSERT INTO pushDevices(device) VALUES ('".$deviceid."')";
echo $query;
$result = mysql_query($query);
if (!$result){
die('Invalid query: ' . mysql_error());
}
echo $result;
?>
This database have more tables and I am able to use them. I am having problem with the tables that I am creating today. They appears in phpmyadmin but somehow I am not able to get use them through my php code.
Any help may be vital for me. I have spent complete day on it.
Thanks
Pankaj
Answer
Its hard to tell by What your saying but i have a suggestion.... It looks like theres no table selected try this
it formatted like this
$query = "INSERT INTO mydb.mytable
(mytablefield)
VALUES
('myfieldvalue')"
$result = mysql_query($query);
if (!$result){
die('Invalid query: ' . mysql_error());
}
My guess is you meant for it to be like this?
$query = "INSERT INTO whatsup_wp1.devices
(device)
VALUES
('".$deviceid."')"
$result = mysql_query($query);
if (!$result){
die('Invalid query: ' . mysql_error());
}
And for security reasons i recommend this...
else
{
//echo 'connect';
$deviceid = mysql_real_escape_string(stripslashes($deviceid));
}
Change to
else
{
//echo 'connect';
$deviceid = mysql_real_escape_string(stripslashes($deviceid));
}
Personally i just use it like this
$result = mysql_query("INSERT INTO mytable
(mytablefield)
VALUES
('myfieldvalue')");
if($result){echo "Works!";}
else{die('Invalid query: ' . mysql_error());exit();}
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