Display mysql data in php function - php

Question

Ok i got a problem now i want to display a data from the database and display it through a function now how do i do that??
like i have fetched a row from the database and its name is $row_field['data']; and it is correct now i have assigned a variable to it like this $data = $row_field['data']; now if i call it in a function it shows undefined variable even after i assigned it global in the function like this
function fun(){
global $data;
echo $data;
}
but if i assign it a value like 1 or 2 or anything it gets displayed without any error why is that so??

Answer 1

If it displays if you assign it a value like 1 or 2 while still in the global scope, then I can only assume that your database did not return the result you thought it did. Does the database value display if you echo it out outside of the function?

Answer 2

Global is evil. I dont know what you are trying to do, but why dont you just do the query in the function itself?

Answer 3

If you have a column named data and your php call was something like
$result = mysql_query("SELECT data FROM mytable");
while ($row_field = mysql_fetch_assoc($result, MYSQL_NUM)) {
...
}
Then you could replace ... with print $row_field['data'].
Else please provide a snippet of your code where you query the db and retrieve the result.

Answer 4

When learning php try to start with simple things. For example in order to get some data from a database follow the examples from php website.
<?php
$conn = mysql_connect("localhost", "mysql_user", "mysql_password");
if (!$conn) {
echo "Unable to connect to DB: " . mysql_error();
exit;
}
if (!mysql_select_db("mydbname")) {
echo "Unable to select mydbname: " . mysql_error();
exit;
}
$sql = "SELECT id as userid, fullname, userstatus
FROM sometable
WHERE userstatus = 1";
$result = mysql_query($sql);
if (!$result) {
echo "Could not successfully run query ($sql) from DB: " . mysql_error();
exit;
}
if (mysql_num_rows($result) == 0) {
echo "No rows found, nothing to print so am exiting";
exit;
}
// While a row of data exists, put that row in $row as an associative array
// Note: If you're expecting just one row, no need to use a loop
// Note: If you put extract($row); inside the following loop, you'll
// then create $userid, $fullname, and $userstatus
while ($row = mysql_fetch_assoc($result)) {
echo $row["userid"];
echo $row["fullname"];
echo $row["userstatus"];
}
mysql_free_result($result);
If all this goes well go a little further change a little the while loop.
$myArray = array();
while ($row = mysql_fetch_assoc($result)) {
$myArray[] = $row;
}
mysql_free_result($result);
// now you can start playing with your data
echo $myArray[0];
Small steps...