I have a testField in DB
testField = NULL
I ran a query:
$query = mysql_query("SELECT testField FROM table");
$result = mysql_fetch_array($query);
I tried different way to no get an error. But i'm still getting "Undefined Variable" error.
if(isset($result['testField'])){$testField = "OK"; echo $testField};if(!empty($result['testField'])){$testField = "OK"; echo $testField};if($result['testField']=='1'){$testField = "OK"; echo $testField};please help. thanks!
mysql_fetch_assoc is the function you intended to use. mysql_fetch_array will give you an array like this by default:
$result[0] = null;
Just use mysql_fetch_assoc and it should be fine.
A good tip when debugging this is to type print_r($result); and see what output you get.
is_null() as well.
mysql_fetch_array() has a second parameter, where you can define in which form you want to get the result. Default is MYSQL_BOTH, what means, that the returned array contains the associative form too.You are getting an error because PHP is case-sensitive. $testField is not the same variable as $testfield.
Variable names are case-sensitive in PHP. The problem is $testField = "OK"; echo $testfield;.
Replace with:
$testField = "OK"; echo $testField;
Reference: http://php.net/manual/en/language.variables.basics.php:
Variable names are case sensitive, so:
$testField != $testfield
Hence why you are getting a variable undefined error.
You are using $testField and $testfield in your if body, these are treated as different variables...
note one has capital F one has lowercase f !
Ok as you've now stated that only the example had this problem and not your original code.
Try splitting the if accross multiple lines to get the actual line producing the error
e.g.
if(isset($result['testField'])){
$testField = "OK";
echo $testField
};
Also you might want to use var_dump to output $result before the if statement to confirm that the query has given you a response (obviously remove again from production code!).
