准备好的语句在不应该的时候失败

Question

我的网站上有一个注册表单，每当新用户注册时，下面的代码应该运行：

$sql = "SELECT * FROM users WHERE uidUsers=?;";
$stmt = mysqli_stmt_init($conn);
if (!mysqli_stmt_prepare($stmt)) {
    header("Location: ../signup.php?error=sqlerror");
    exit();
}
else {
 mysqli_stmt_bind_param($stmt, "s", $username);
 mysqli_stmt_execute($stmt);
 $resultCheck = mysqli_stmt_num_rows($stmt);
if ($resultCheck > 0) {
    header("Location: ../signup.php?error=usertaken");
    exit();
}
else {
    $hashedPwd = password_hash(PASSWORD_DEFAULT, $password);
    $sql = "INSERT INTO users (`uidUsers`, `pwdUsers`, `phraseUsers`) VALUES(?, 
?, ?)";
    $stmt = mysqli_stmt_init($conn);
if (!mysqli_stmt_prepare($stmt)) {
    header("Location: ../signup.php?error=sqlerror");
    exit();
}
else {
    mysqli_stmt_bind_param($stmt, "sss", $username, $hashedPwd, 
    $securityphrase);
    mysqli_stmt_execute($stmt);
    header("Location: ../login.php?signup=success");
    exit();
}

!mysqli_stmt_prepare 错误处理程序在不应该给定数据库环境以及正确的 INSERT 语句时被触发。所以我不明白为什么它会被触发，我想问为什么？

Stack Overflow

上也有类似的问题

Answer 1

谢谢大家的帮助，问题出在哪里password_hash命令，应该是这样的password_hash($var, PASSWORD_DEFAULT);。我犯的第二个错误是 mysqli_stmt_prepare 中没有包含我的 $sql 语句，应该是这样的 !mysqli_stmt_prepare($stmt, $sql).

准备好的语句在不应该的时候失败

Prepared statement fails when it shouldn't

php

mysql

mysqli

prepared-statement