Новая Школа - онлайн-школа подготовки к ЕГЭ
Задача Задача №3623450
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Аноним
18 июня 2026 14:23 · публичный вопрос
А
Аноним

(select concat(0x3a,char(64),char(52),char(51),char(55),char(48),char(49),DATABASE(),char(64),char(49),char(48),char(55),char(51),char(52)))

Ответ нейросети
18 июня 2026 14:23

This is a MySQL expression using CONCAT and character codes to assemble a string. It effectively builds:

  • 0x3a -> ":"
  • char(64) -> "@"
  • char(52) -> "4"
  • char(51) -> "3"
  • char(55) -> "7"
  • char(48) -> "0"
  • char(49) -> "1"
  • DATABASE() -> the current database name
  • char(64) -> "@"
  • char(49) -> "1"
  • char(48) -> "0"
  • char(55) -> "7"
  • char(51) -> "3"
  • char(52) -> "4"

So the final result is:

":@43701" + DATABASE() + "@10734"

Example: if the current database is named mydb, the output would be: ":@43701mydb@10734"

This is a typical SQL injection-style payload to reveal the database name. In exams you might be asked to decode it, identify what it outputs, or discuss defenses such as using parameterized queries, least-privilege accounts, and input validation to prevent such injections.