php 从函数返回两个值
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Returning two values from a function
提问by MacMac
Is it possible to return two values when calling a function that would output the values?
调用将输出值的函数时是否可以返回两个值？
For example, I have this:
例如，我有这个：
<?php
function ids($uid = 0, $sid = '')
{
$uid = 1;
$sid = md5(time());
return $uid;
return $sid;
}
echo ids();
?>
Which will output 1. I want to chose what to ouput, e.g. ids($sid), but it will still output 1.
哪个将输出1. 我想选择要输出的内容，例如ids($sid)，但它仍然会输出1.
Is it even possible?
甚至有可能吗？
回答by Gumbo
You can only return one value. But you can use an arraythat itself contains the other two values:
您只能返回一个值。但是您可以使用一个本身包含其他两个值的数组：
return array($uid, $sid);
Then you access the values like:
然后您访问如下值：
$ids = ids();
echo $ids[0]; // uid
echo $ids[1]; // sid
You could also use an associative array:
您还可以使用关联数组：
return array('uid' => $uid, 'sid' => $sid);
And accessing it:
并访问它：
$ids = ids();
echo $ids['uid'];
echo $ids['sid'];
回答by Annika Backstrom
Return an array or an object if you need to return multiple values. For example:
如果需要返回多个值，则返回一个数组或一个对象。例如：
function foo() {
return array(3, 'joe');
}
$data = foo();
$id = $data[0];
$username = $data[1];
// or:
list($id, $username) = foo();
回答by Jerome WAGNER
You can use an array and the list function to get the info easily :
您可以使用数组和列表函数轻松获取信息：
function multi($a,$b) {
return array($a,$b);
}
list($first,$second) = multi(1,2);
I hope this will help you
我希望这能帮到您
Jerome
杰罗姆
回答by Mark Baker
function ids($uid = 0, $sid = '')
{
$uid = 1;
$sid = md5(time());
return array('uid' => $uid,
'sid' => $sid
);
}
$t = ids();
echo $t['uid'],'<br />';
echo $t['sid'],'<br />';
回答by NikiC
Many possibilities:
多种可能性：
// return array
function f() {
return array($uid, $sid);
}
list($uid, $sid) = f();
$uid;
$sid;
// return object
function f() {
$obj = new stdClass;
$obj->uid = $uid;
$obj->sid = $sid;
return $obj;
}
$obj->uid;
$obj->sid;
// assign by reference
function f(&$uid, &$sid) {
$uid = '...';
$sid = '...';
}
f($uid, $sid);
$uid;
$sid;
回答by user2029015
回答by Adrian Cumpanasu
Pass by reference would be a solution. This way you:
通过引用传递将是一个解决方案。这样你：
return a value as you normally would
and modify the second (the one passed by refference to the function)
像往常一样返回一个值
并修改第二个（通过引用函数传递的那个）
An example:
一个例子：
function foo(&$var) // notice the & in front of the parameter passed
{
$var++;
return 1;
}
$a=5;
echo foo($a);
// echoes 1, the returned value
echo $a;
// $a is now 6
