我到处都找过这样的东西,但我认为"使用"这个词可能太常见了,不适合任何有用的结果:
从PHP代码库中的类文件中删除所有未使用的use
语句的最简单方法是什么?
编辑:为了简单起见,我们可以忽略检测用于注释的use
语句。
检查FriendsOfPHP的PHP CS Fixerhttps://github.com/FriendsOfPHP/PHP-CS-Fixer
编辑
我已经完全重写了它,所以现在它更强大了:
- 忽略特性和匿名/lambda函数
- 现在处理catch块、类扩展和接口
- 缩进和注释无关紧要
- 命名空间别名的多个声明也适用
- 静态和对象类调用被识别为";用法";($u->getUsages())
- 不处理完全和一半合格的用途
测试文件class.php:
use My'Full'Classname as Another, My'Full'NSname, Some'Other'Space;
/* some insane commentary */ use My'Full'NSname1; use ArrayObject;
$obj = new namespaced'Another;
$obj = new Another;
$a = new ArrayObject(array(1));
Space::call();
$a = function($a, $b, $c = 'test') use ($obj) {
/* use */
};
class MyHelloWorld extends Base {
use traits, hello, world;
}
这里的脚本:
class UseStatementSanitizer
{
protected $content;
public function __construct($file)
{
$this->content = token_get_all(file_get_contents($file));
// we don't need and want them while parsing
$this->removeTokens(T_COMMENT);
$this->removeTokens(T_WHITESPACE);
}
public function getUnused()
{
$uses = $this->getUseStatements();
$usages = $this->getUsages();
$unused = array();
foreach($uses as $use) {
if (!in_array($use, $usages)) {
$unused[] = $use;
}
}
return $unused;
}
public function getUsages()
{
$usages = array();
foreach($this->content as $key => $token) {
if (!is_string($token)) {
$t = $this->content;
// for static calls
if ($token[0] == T_DOUBLE_COLON) {
// only if it is NOT full or half qualified namespace
if ($t[$key-2][0] != T_NAMESPACE) {
$usages[] = $t[$key-1][1];
}
}
// for object instanciations
if ($token[0] == T_NEW) {
if ($t[$key+2][0] != T_NAMESPACE) {
$usages[] = $t[$key+1][1];
}
}
// for class extensions
if ($token[0] == T_EXTENDS || $token[0] == T_IMPLEMENTS) {
if ($t[$key+2][0] != T_NAMESPACE) {
$usages[] = $t[$key+1][1];
}
}
// for catch blocks
if ($token[0] == T_CATCH) {
if ($t[$key+3][0] != T_NAMESPACE) {
$usages[] = $t[$key+2][1];
}
}
}
}
return array_values(array_unique($usages));
}
public function getUseStatements()
{
$tokenUses = array();
$level = 0;
foreach($this->content as $key => $token) {
// for traits, only first level uses should be captured
if (is_string($token)) {
if ($token == '{') {
$level++;
}
if ($token == '}') {
$level--;
}
}
// capture all use statements besides trait-uses in class
if (!is_string($token) && $token[0] == T_USE && $level == 0) {
$tokenUses[] = $key;
}
}
$useStatements = array();
// get rid of uses in lambda functions
foreach($tokenUses as $key => $tokenKey) {
$i = $tokenKey;
$char = '';
$useStatements[$key] = '';
while($char != ';') {
++$i;
$char = is_string($this->content[$i]) ? $this->content[$i] : $this->content[$i][1];
if (!is_string($this->content[$i]) && $this->content[$i][0] == T_AS) {
$useStatements[$key] .= ' AS ';
} else {
$useStatements[$key] .= $char;
}
if ($char == '(') {
unset($useStatements[$key]);
break;
}
}
}
$allUses = array();
// get all use statements
foreach($useStatements as $fullStmt) {
$fullStmt = rtrim($fullStmt, ';');
$fullStmt = preg_replace('/^.+ AS /', '', $fullStmt);
$fullStmt = explode(',', $fullStmt);
foreach($fullStmt as $singleStmt) {
// $singleStmt only for full qualified use
$fqUses[] = $singleStmt;
$singleStmt = explode('''', $singleStmt);
$allUses[] = array_pop($singleStmt);
}
}
return $allUses;
}
public function removeTokens($tokenId)
{
foreach($this->content as $key => $token) {
if (isset($token[0]) && $token[0] == $tokenId) {
unset($this->content[$key]);
}
}
// reindex
$this->content = array_values($this->content);
}
}
$unused = new UseStatementSanitizer('class.php');
print_r($unused->getUnused());
退货:
Array
(
[0] => NSname
[1] => NSname1
)
这可能取决于代码的设置方式。如果您的代码使用这样的名称空间:
namespace Foo
{
<one or more classes in namespace Foo>
}
那么,如果你只是单独检查每个文件,你可能就没事了。这仍然意味着您必须解析PHP代码以找到use语句,然后确定使用了哪些语句。
简单的方法是使用已经构建好的工具。我最近开始使用PhpStorm IDE(30天免费试用,或早期访问程序),当你在文件中有未使用的use
语句时,它可以通知你(你甚至可以指定这应该是警告还是错误)。不过,它仍然需要您打开每个文件。但你也可以检查你正在编辑的文件,这样你的代码最终会更干净。