多次查询后PHP stmt prepare失败

警告：mysqli_stmt_bind_param（）要求参数1为mysqli_stmt，第64行给出布尔值

我已经找到了很多其他关于这个问题的帖子，但是我尝试过的解决方案没有起到任何作用。 像

stmt\u close（）

或

stmt\u free（）

这样的事情没有什么区别

下面是我的实际代码片段：

第一，准备并执行：

// Connect to database
$mysqliID = mysqli_connect($mysqliHost, $mysqliUsername, $mysqliPassword);
if (!$mysqliID) throw new Exception("");
if (!mysqli_select_db($mysqliID, $mysqliDB)) throw new Exception("");

// Check if the username or email address aren't already accupied.
$stmt = mysqli_prepare($mysqliID, "SELECT id FROM Account WHERE username=? OR email=?");
mysqli_stmt_bind_param($stmt, "ss", $username, $email);
if (!mysqli_stmt_execute($stmt)) throw new Exception("");
$duplicates = 0;
mysqli_stmt_bind_result($stmt, $duplicateid);
while (mysqli_stmt_fetch($stmt))
{
    die ("error:-gebruikersnaam of email al in gebruik.");
}

// Save account with salted and hashed password.
include "PasswordHash.php";
$hashResult = create_hash($password);
$hashResultSplitted = explode(":", $hashResult);

$salt = $hashResultSplitted[2];
$hash = $hashResultSplitted[3];
$verified = 0;
$stmt = mysqli_prepare($mysqliID, "INSERT INTO Account (username, hash, salt, email, verified) VALUES (?, ?, ?, ?, ?)");
mysqli_stmt_bind_param($stmt, "ssssi", $username, $hash, $salt, $email, $verified);
if (!mysqli_stmt_execute($stmt)) throw new Exception("");

// Save verification token.
$token = bin2hex(random_bytes(16));
$expirationtime = time() + (15 * 60);
$accountid = mysqli_insert_id($mysqliID);
$stmt = mysqli_prepare($mysqliID, "INSERT INTO Verifications (token, expirationtime, accountid) VALUES (?, ?, ?)");
mysqli_stmt_bind_param($stmt, "sii", $token, $expirationtime, $accountid);
if (!mysqli_stmt_execute($stmt)) throw new Exception("");

第二，准备并执行：

// Connect to database
$mysqliID = mysqli_connect($mysqliHost, $mysqliUsername, $mysqliPassword);
if (!$mysqliID) throw new Exception("");
if (!mysqli_select_db($mysqliID, $mysqliDB)) throw new Exception("");

// Check if the username or email address aren't already accupied.
$stmt = mysqli_prepare($mysqliID, "SELECT id FROM Account WHERE username=? OR email=?");
mysqli_stmt_bind_param($stmt, "ss", $username, $email);
if (!mysqli_stmt_execute($stmt)) throw new Exception("");
$duplicates = 0;
mysqli_stmt_bind_result($stmt, $duplicateid);
while (mysqli_stmt_fetch($stmt))
{
    die ("error:-gebruikersnaam of email al in gebruik.");
}

// Save account with salted and hashed password.
include "PasswordHash.php";
$hashResult = create_hash($password);
$hashResultSplitted = explode(":", $hashResult);

$salt = $hashResultSplitted[2];
$hash = $hashResultSplitted[3];
$verified = 0;
$stmt = mysqli_prepare($mysqliID, "INSERT INTO Account (username, hash, salt, email, verified) VALUES (?, ?, ?, ?, ?)");
mysqli_stmt_bind_param($stmt, "ssssi", $username, $hash, $salt, $email, $verified);
if (!mysqli_stmt_execute($stmt)) throw new Exception("");

// Save verification token.
$token = bin2hex(random_bytes(16));
$expirationtime = time() + (15 * 60);
$accountid = mysqli_insert_id($mysqliID);
$stmt = mysqli_prepare($mysqliID, "INSERT INTO Verifications (token, expirationtime, accountid) VALUES (?, ?, ?)");
mysqli_stmt_bind_param($stmt, "sii", $token, $expirationtime, $accountid);
if (!mysqli_stmt_execute($stmt)) throw new Exception("");

第三次也是最后一次准备和执行：

// Connect to database
$mysqliID = mysqli_connect($mysqliHost, $mysqliUsername, $mysqliPassword);
if (!$mysqliID) throw new Exception("");
if (!mysqli_select_db($mysqliID, $mysqliDB)) throw new Exception("");

// Check if the username or email address aren't already accupied.
$stmt = mysqli_prepare($mysqliID, "SELECT id FROM Account WHERE username=? OR email=?");
mysqli_stmt_bind_param($stmt, "ss", $username, $email);
if (!mysqli_stmt_execute($stmt)) throw new Exception("");
$duplicates = 0;
mysqli_stmt_bind_result($stmt, $duplicateid);
while (mysqli_stmt_fetch($stmt))
{
    die ("error:-gebruikersnaam of email al in gebruik.");
}

// Save account with salted and hashed password.
include "PasswordHash.php";
$hashResult = create_hash($password);
$hashResultSplitted = explode(":", $hashResult);

$salt = $hashResultSplitted[2];
$hash = $hashResultSplitted[3];
$verified = 0;
$stmt = mysqli_prepare($mysqliID, "INSERT INTO Account (username, hash, salt, email, verified) VALUES (?, ?, ?, ?, ?)");
mysqli_stmt_bind_param($stmt, "ssssi", $username, $hash, $salt, $email, $verified);
if (!mysqli_stmt_execute($stmt)) throw new Exception("");

// Save verification token.
$token = bin2hex(random_bytes(16));
$expirationtime = time() + (15 * 60);
$accountid = mysqli_insert_id($mysqliID);
$stmt = mysqli_prepare($mysqliID, "INSERT INTO Verifications (token, expirationtime, accountid) VALUES (?, ?, ?)");
mysqli_stmt_bind_param($stmt, "sii", $token, $expirationtime, $accountid);
if (!mysqli_stmt_execute($stmt)) throw new Exception("");

因此，在第三段也是最后一段代码中，我在mysqli_prepare上得到了警告，但前两段都成功地完成了

---

解决方案：

---

原来名为Verifications的表应该是Verifications。通过良好的阅读和睡眠后的护理发现。

你应该检查一下，我试过了。但是因为mysqli_prepare返回一个布尔值。mysqli_stmt_error（）不接受我的$stmt。因为它必须是mysqli_stmt对象。>警告：mysqli_stmt_error（）要求参数1为mysqli_stmt，第58mysqli_prepare（）行中给出的布尔值返回一条语句

object

或

FALSE

，如果发生错误。如果收到布尔值，则准备失败。