EASY PHP question? - php

Question

HI Guys,
I have following problem. I'm trying to connect from my Iphone app to a php site, to access the mySql db, to get the right information.
This is my code:
<?php
mysql_connect ("localhost", "user", "pass") or die (mysql_error());
echo "Connected to MySql <br /><hr />";
mysql_select_db ("database_com") or die (mysql_error());
echo "Connectted to Database <br /><hr />";
$country = $_POST['country'];// THIS IS THE VALUE I WANT TO LOAD INTO THE SELECT STATEMENT
echo "value <br /><hr />" . $country;
$query = "SELECT * FROM world WHERE land='$country'";
$result = mysql_query($query) or die (mysql_error());
while ($row = mysql_fetch_array($result)) {
$1 = $row['1'];
$2 = $row['2'];
$3 = $row['a3'];
$4 = $row['4'];
$xmltext = "<?xml version=\"1.0\" encoding=\"UTF-8\"?>\n<country></country>";
$xmlobj = simplexml_load_string($xmltext);
$xmlobj->addChild("1", $1);
$xmlobj->addChild("2", $2);
$xmlobj->addChild("a3", $a3);
$xmlobj->addChild("4", $4);
print header("Content-type: text/plain") . $xmlobj->asXML(); // Place future code ABOVE this line
$xml->save(statistic.xml);
}
?>
If I hardcode the value of land = "Germany" i get an answer, but if I do the other thing, nothing comes out of it.
I hope u can help me.

Answer 1

Do you have a form that posts a country to your php page? If so, is the form using method="post"? If you are sticking the country in the url, use $_GET instead of $_POST.
Also, call mysql_real_escape_string($_POST['country']) to avoid SQL injection issues (like someone deleting all your databases).
Also, this line:
print header("Content-type: text/plain") . $xmlobj->asXML();
has a number of issues. It should be:
header("Content-type: text/plain");
print $xmlobj->asXML();
However, you cannot send headers after printing other stuff out - like your echo calls early in the script. It doesn't sound like this is a problem yet, but it will be when your query works.
Also, you're looping through the result set. If you expect only one result, ditch the while loop. If you have more than one result, you will get errors because of the header() call.

Answer 2

What happens if you print_r or var_dump($_POST)? At first glance, it looks like your POST is empty, or at least $_POST['country'].

Answer 3

Have you check the $_POST['country'] actually contains a value?
Try
echo $query;
and update the post so we can see the content.

Answer 4

Try a simple error catch:
<?php
mysql_connect ("localhost", "user", "pass") or die (mysql_error());
echo "Connected to MySql <br /><hr />";
mysql_select_db ("database_com") or die (mysql_error());
echo "Connectted to Database <br /><hr />";
$country = $_POST['country'];// THIS IS THE VALUE I WANT TO LOAD INTO THE SELECT STATEMENT
echo "value <br /><hr />" . $country;
if (!isset($country) || $country == FALSE /* other fail conditions */) {
echo "<p>There was a problem; one, or more, error condition occurred:</p>";
echo "<pre>" . print_r(get_defined_vars(),true) . "</pre>";
}
else {
// database access stuff
}
?>
This way, if the values aren't set (or error-tests are met) you get a display of all the variables currently defined and the database access doesn't occur.
It's crude, but it sometimes helps to see what's happening. Though, personally, I think it far more likely that you're simply experiencing a typo $_POST['country'] instead of $_POST['contry'] or something, or using $_POST instead of $_GET.

Answer 5

A good way to debug this is to put this block at the top of your file:
<?php
if (!isset($_POST)) {
die('A $_POST is required for this page.');
}
...

Answer 6

writing this peace of code can debug your program
<?php
if (!isset($_POST)) {
}
?>

Answer 7

If
$country = $_POST['country'];
is returning proper value than you just need to change
$query = "SELECT * FROM world WHERE land='$country'";
as
$query = "SELECT * FROM world WHERE land='".$country."'";
otherwise just check if you are receiving $_POST['country'] values properly if not then debug it first.