Issues (21)

src/ArrayParser.php (2 issues)

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<?php
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declare(strict_types=1);
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namespace Yiisoft\Db\Pgsql;
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use function in_array;
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/**
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 * Array representation to PHP array parser for PostgreSQL Server.
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 */
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final class ArrayParser
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{
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    /**
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     * Convert an array from PostgresSQL to PHP.
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     *
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     * @param string|null $value String to convert.
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     */
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    public function parse(string|null $value): array|null
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    {
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        return $value !== null && $value[0] === '{'
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            ? $this->parseArray($value)
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            : null;
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    }
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    /**
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     * Parse PostgreSQL array encoded in string.
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     *
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     * @param string $value String to parse.
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     * @param int $i parse starting position.
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     */
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    private function parseArray(string $value, int &$i = 0): array
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    {
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        if ($value[++$i] === '}') {
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            ++$i;
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            return [];
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        }
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        for ($result = [];; ++$i) {
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            $result[] = match ($value[$i]) {
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                '{' => $this->parseArray($value, $i),
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                ',', '}' => null,
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                '"' => $this->parseQuotedString($value, $i),
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                default => $this->parseUnquotedString($value, $i),
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            };
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            if ($value[$i] === '}') {
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                ++$i;
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                return $result;
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            }
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        }
0 ignored issues
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Bug Best Practice introduced by
In this branch, the function will implicitly return null which is incompatible with the type-hinted return array. Consider adding a return statement or allowing null as return value.

For hinted functions/methods where all return statements with the correct type are only reachable via conditions, ?null? gets implicitly returned which may be incompatible with the hinted type. Let?s take a look at an example:

interface ReturnsInt {
    public function returnsIntHinted(): int;
}

class MyClass implements ReturnsInt {
    public function returnsIntHinted(): int
    {
        if (foo()) {
            return 123;
        }
        // here: null is implicitly returned
    }
}
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    }
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    /**
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     * Parses quoted string.
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     */
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    private function parseQuotedString(string $value, int &$i): string
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    {
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        for ($result = '', ++$i;; ++$i) {
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            if ($value[$i] === '\\') {
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                ++$i;
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            } elseif ($value[$i] === '"') {
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                ++$i;
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                return $result;
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            }
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            $result .= $value[$i];
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        }
0 ignored issues
show
Bug Best Practice introduced by
In this branch, the function will implicitly return null which is incompatible with the type-hinted return string. Consider adding a return statement or allowing null as return value.

For hinted functions/methods where all return statements with the correct type are only reachable via conditions, ?null? gets implicitly returned which may be incompatible with the hinted type. Let?s take a look at an example:

interface ReturnsInt {
    public function returnsIntHinted(): int;
}

class MyClass implements ReturnsInt {
    public function returnsIntHinted(): int
    {
        if (foo()) {
            return 123;
        }
        // here: null is implicitly returned
    }
}
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    }
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    /**
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     * Parses unquoted string.
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     */
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    private function parseUnquotedString(string $value, int &$i): string|null
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    {
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        for ($result = '';; ++$i) {
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            if (in_array($value[$i], [',', '}'], true)) {
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                return $result !== 'NULL'
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                    ? $result
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                    : null;
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            }
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            $result .= $value[$i];
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        }
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    }
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}
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