<?php $slc="selected=\"selected\""; function countryselect($country,$select,$change) { print "<select name=\"$select\" class=\"form\" onChange=\"var a=theform.$select.options[theform.$select.selectedIndex].value; document.get
ElementById('$change').src=a\">"; $sql_flags="SELECT * FROM flags ORDER BY name DESC"; </select> else { $country="images/flags/$country.gif"; } function countrylist($country) { $sql_flags2="SELECT * FROM flags ORDER BY name DESC"; if($country=="$flags2->short") { $country2="$flags2->name"; } }
Error jaki wywala to:
Kod
Warning: mysql_query(): supplied argument is not a valid MySQL-Link resource in <b>C:\Xampp\xampp\htdocs\wgk\system\country.php</b> on line 8
Warning: mysql_fetch_object(): supplied argument is not a valid MySQL result resource in <b>C:\Xampp\xampp\htdocs\wgk\system\country.php</b> on line 9
Warning: mysql_fetch_object(): supplied argument is not a valid MySQL result resource in <b>C:\Xampp\xampp\htdocs\wgk\system\country.php</b> on line 9
Plik wywołuję tym:
<?php include ("system/country.php"); countrylist("$users->country"); ?>