php - mysqli 查询，里面有 mysql 函数

Question

这个问题已经有答案了:

mysqli - fetch_Array error call to a member function fetch_array() on a non-object mysqli [duplicate] (2 个回答)

已关闭 6 年前。

我正在尝试使用我在某个地方找到的函数在数据库上生成密码。如果我直接在数据库上使用 sql 查询，它会根据需要生成一个字符串，但在我的 php 页面中它会给出错误:

"Fatal error: Uncaught Error: Call to a member function fetch_array() on boolean"

$query_rsCaptchaID = "DELIMITER $$
DROP FUNCTION IF EXISTS `randomPasswordGenerator` $$
CREATE FUNCTION `randomPasswordGenerator`(
  ) RETURNS varchar(64) CHARSET utf8
BEGIN
  DECLARE charCount TINYINT(1) DEFAULT 0;
  DECLARE charDiceRoll TINYINT(2);
  DECLARE randomChar CHAR(1);
  DECLARE randomPassword CHAR(64) DEFAULT '';
  REPEAT
    SET charCount = charCount + 1;
    SET charDiceRoll = 1 + FLOOR(RAND() * 94);
    IF (charDiceRoll <= 32)
    THEN
      SET randomChar = ELT(charDiceRoll,
      '`', '~', '!', '@', '#', '$', '%', '^',
      '&', '*', '(', ')', '-', '=', '_', '+',
      '[', ']', '{', '}', '\\', '/', '|', '?',
      ';', ':', '\'', '\"', ',', '.', '<', '>');
    ELSEIF (charDiceRoll >= 33)
      AND (charDiceRoll <= 68)
    THEN
      SET charDiceRoll = charDiceRoll - 33;
      SET randomChar = CONV(
        charDiceRoll,
        10, 36);
    ELSE
      SET charDiceRoll = charDiceRoll - 59;
      SET randomChar = LOWER(
        CONV(
          charDiceRoll,
          10, 36)
      );
    END IF;
    SET randomPassword = CONCAT(randomPassword, randomChar);
  UNTIL (charCount = 64)
  END REPEAT;
  RETURN randomPassword;
END $$
DELIMITER ;
SELECT randomPasswordGenerator() AS captchaID;";
$rsCaptchaID = mysqli_query($db,$query_rsCaptchaID);
$row_rsCaptchaID = $rsCaptchaID->fetch_array();

有人有什么想法吗？我对 MySQLi 很陌生，不会说英语，抱歉有错误。

Answer 1

发生该错误是因为您的查询失败。在您发布的最后两行之间，您需要进行一些错误检查:

$rsCaptchaID = mysqli_query($db,$query_rsCaptchaID);
if(!$rsCaptchaID) die(mysqli_error($db); // <- will show you the SQL error
$row_rsCaptchaID = $rsCaptchaID->fetch_array();