我在DB 5个领域:未定义的变量在PHP
test1 = 1, test2 = 1, test3 = NULL, test4 = NULL, test5 = NULL
PHP代码:
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 $total;
Finaly输出:
Undefined Variable test3 in Line 7
Undefined Variable test4 in Line 7
Undefined Variable test5 in Line 7
的Test1,Test2的
我想出了3点可能的方式,希望能够运行带有NULL值的代码,看看我是否会得到一个空白页,没有错误,不幸的是,他们都给我“Underfined变“错误:
if(isset($result['test3'])){$test3= "OK"; echo $test3};
if(!empty($result['test3'])){$test3= "OK"; echo $test3};
if($result['test3']=='1'){$test3= "OK"; echo $test3};
帮助?提前致谢!
请放慢脚步,需要一些时间来正确地格式化代码,使人们可以读取它。 – 2011-07-03 22:03:13
当字段为NULL时,您希望在$ total中看到什么? –
是否有'array_filter'的原因? –