I have 5 fields in my DB:
test1 = 1, test2 = 1, test3 = NULL, test4 = NULL, test5 = NULL
PHP code:
if(isset($result['test1'])){$test1= "Test1"; echo $test1};
if(isset($result['test2'])){$test2= "Test2"; echo $test2};
if(isset($result['test3'])){$test3= "Test3"; echo $test3};
if(isset($result['test4'])){$test4= "Test4"; echo $test4};
if(isset($result['test5'])){$test5= "Test5"; echo $test5};
$total = implode(", ", array_filter(array($test1, $test2, $test3, $test4, $test5)));
echo $total;
Finaly Output:
Undefined Variable test3 in Line 7
Undefined Variable test4 in Line 7
Undefined Variable test5 in Line开发者_如何学C 7
Test1, Test2
I came up with 3 possible ways to hopefully run the code with NULL values to see if I will get a blank page with no error, unfortunately, they all gave me "Underfined Variable" error:
if(isset($result['test3'])){$test3= "OK"; echo $test3};
if(!empty($result['test3'])){$test3= "OK"; echo $test3};
if($result['test3']=='1'){$test3= "OK"; echo $test3};
Help? Thanks in Advance!
if(isset($result['test1'])){$test1= "Test1"; echo $test1} else { $test1="Abrakadabra"; }
A way to do it without array_filter
:
$result = array();
if(!empty($result['test1'])){$result[] = "Test1"; echo $test1;}
...
$total = implode(", ", $result);
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