[英]how can access argv and options together in php through command line arguments
[英]Parsing through command line arguments/argv
所以我必須創建一個解析器來解析傳遞給它的命令行參數。 下面是我到目前為止的代碼。
public function parse($argv = null)
{
$argv = $this->argsUnparsed;
$argvs = array();
array_shift($argvs);
foreach($argv as $arg)
{
// This is supposed to find the -- characters in a string
if(substr($arg,0,2) == '--')
{
$equals = strpos($arg, '=');
// If character in string equals '=' saving anything before it as a key and anything afterward as a value
if($equals)
{
$argvs[substr($arg,2,$equals - 2)] = substr($arg,$equals + 1);
}
else
{
$k = substr($arg,2);
if(!isset($argvs[$k]))
{
$argvs[$k] = true;
}
}
}
else if(substr($arg,0,1) == '-')
{
foreach(str_split(substr($arg,1)) as $k)
{
if(!isset($argvs[$k]))
{
$argvs[$k] = true;
}
elseif($equals == false)
{
$argvs[substr($arg,2,$equals - 2)] = substr($arg,$equals + 1);
}
}
}
else
{
$argvs[] = $arg;
}
}
return $argvs;
}
這是我解析命令行參數的函數。 我的問題是,當我輸入“php testArgs.php -v -T 4 -l val1,val2,val3 --names = Austin,Duncan,Eddie --type = gold”時,它將其輸出為
Array
(
[0] => testArgs.php
[v] => 1
[T] => 1
[1] => 4
[l] => 1
[2] => val1,val2,val3
[names] => Austin,Duncan,Eddie
[type] => gold
)
第二個和最后兩個陣列插槽是完全正確的,但其余部分應該按如下方式打印出來。
[T] => 4
[l] => val1,val2,val3
此外,不應顯示第一個數組鍵值對[0] => testArgs.php。
你的第二個和第三個else if
塊應該更簡單。 嘗試這樣的事情:
<?php
...
// Check if a key is passed that begins with "-"
else if(substr($arg,0,1) == '-')
{
$k = substr($arg,1);
if(!isset($argvs[$k]))
{
$argvs[$k] = true;
$previous_key = $k;
}
}
// Copy the value being parsed to the previous key
else if(!is_empty($previous_key))
{
$argvs[$previous_key] = $arg;
}
$previous_key = "";
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