有没有办法根据()中传递的内容返回函数的一部分?举个例子:
function test($wo) {
if function contains $wo and "date" {
//pass $wo through sql query to pull date
return $date
}
if function contains $wo and "otherDate" {
//pass $wo through another sql query to pull another date
return $otherDate
}
if function only contains $wo {
//pass these dates through different methods to get a final output
return $finaldate
}
}
日期:
test($wo, date);
返回:
1/1/2015
otherDate:
test($wo, otherDate);
返回:
10/01/2015
正常输出:
test($wo);
返回:
12/01/2015
答案 0 :(得分:2)
传递一个指定返回内容的参数:
function test($wo, $type='final') {
// pull $date
if($type == 'date') { return $date; }
// pull $otherdate
if($type == 'other') { return $otherdate; }
// construct $finaldate
if($type == 'final') { return $finaldate; }
return false;
}
然后打电话给:
$something = test($a_var, 'other');
// or for final since it is default
$something = test($a_var);
答案 1 :(得分:0)
你的问题很模糊,但如果我理解正确,你需要可选的参数。您可以通过在函数定义中为它们提供默认值来使函数参数可选:
// $a is required
// $b is optional and defaults to 'test' if not specified
// $c is optional and defaults to null if not specified
function test($a, $b = 'test', $c = null)
{
echo "a is $a\n";
echo "b is $b\n";
echo "c is $c\n";
}
现在你可以这样做:
test(1, 'foo', 'bar');
你得到:
a is 1
b is foo
c is bar
或者这个:
test(37);
你得到:
a is 37
b is test
c is