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Top Forums Web Development Passing variable from PHP to bash script Post 303009004 by jgt on Friday 8th of December 2017 08:24:11 AM
Old 12-08-2017
php variable $hostname is undefined.
Replace
Code:
$hostname=$argv[1]

with
Code:
$hostname=$_GET["hostname"];

Assuming that you are using Apache as the web server, you can do the following:
Start an additional terminal window, and run "tail -f /var/log/apache2/error.log". You will get al running list of errors from your web development.
 

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GEOIP_REGION_BY_NAME(3) 						 1						   GEOIP_REGION_BY_NAME(3)

geoip_region_by_name - Get the country code and region

SYNOPSIS
array geoip_region_by_name (string $hostname) DESCRIPTION
The geoip_region_by_name(3) function will return the country and region corresponding to a hostname or an IP address. This function is currently only available to users who have bought a commercial GeoIP Region Edition. A warning will be issued if the proper database cannot be located. The names of the different keys of the returning associative array are as follows: o "country_code" -- Two letter country code (see geoip_country_code_by_name(3)) o "region" -- The region code (ex: CA for California) PARAMETERS
o $hostname - The hostname or IP address whose region is to be looked-up. RETURN VALUES
Returns the associative array on success, or FALSE if the address cannot be found in the database. EXAMPLES
Example #1 A geoip_region_by_name(3) example This will print the array containing the country code and region of the host example.com. <?php $region = geoip_region_by_name('www.example.com'); if ($region) { print_r($region); } ?> The above example will output: Array ( [country_code] => US [region] => CA ) PHP Documentation Group GEOIP_REGION_BY_NAME(3)
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