[英]Pass current array as reference [PHP/Drupal6]
I have a variable $doctype
that is an array: 我有一个变量
$doctype
是一个数组:
$doctype = array(
'news' => ('news'),
'documents' => ('documents'),
'forms' => ('forms'),
'other' => ('other'),
);
which is used as options for another array: 它用作另一个数组的选项:
$form['doc_type'] = array(
'#title' => 'Document Type',
'#type' => 'radios',
'#options' => $doctype,
'#size' => '30',
'#required' => TRUE,
'#default_value' => $doctype['news'],
now I'm trying to pass the current value for another function 现在我正试图将当前值传递给另一个函数
I've tried: 我试过了:
form($form_state, &$doctype) {
but it shows up as a missing argument 但它显示为一个缺失的论点
I want to pass it through a reference, not as a return (already occupied/don't want to work around it) 我想通过引用传递它,而不是作为返回(已经占用/不想解决它)
Any help is appreciated 任何帮助表示赞赏
Assuming form()
specifies its second argument as a reference, then you shouldn't pass &$doctype
as a parameter. 假设
form()
指定其第二个参数作为引用,则不应将&$doctype
作为参数传递。 If the function specifies a parameter as a reference, then all you need to do is pass that variable and it will get passed as a reference. 如果函数将参数指定为引用,那么您需要做的就是传递该变量,它将作为引用传递。
Eg 例如
<?php
function form($var1, &$var2) {
$var2[2] = 5;
}
$var2 = array(1,2,3,4);
form('test', $var2);
echo $var2[2]; // Echoes '5';
?>
However, I'm again assuming that form()
is built into drupal, in which case as far as I know there's no way of passing $doctype
as a reference without changing the core code that defines form()
但是,我再次假设
form()
内置于drupal中,在这种情况下,据我所知,没有办法将$doctype
作为引用传递而不更改定义form()
的核心代码
Out of curiosity, why do you need to pass it as a reference? 出于好奇,你为什么要把它作为参考传递? Can you clarify "already occupied/don't want to work around it"
你能澄清一下“已经占据/不想解决它”
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