預處理語句適用於INSERT，但不適用於SELECT

Question

編輯：解決了，請看下面，但仍然不知道我做了什么才能使其起作用:)

好吧，我現在被困住了。 我有表users ：

ID int PRIMARY AUTO_INCREMENT
EMAIL varchar(60)
NICK varchar(60)
//...

如果我做：

<?php
$email = $_POST["mai"];
$nickname = $_POST["nck"];
$mysqli = new mysqli($db_host, $db_username, $db_password, $database);
$prepared_statement = "INSERT INTO users VALUES(?,?,?)";
if ($stmt = $mysqli->prepare($prepared_statement)) {
 $id = "";
 $stmt->bind_param("iss",$id,$email,$nickname); 
 $stmt->execute();
}
?>

它每一次都有效。 但是，如果我對select執行相同操作：

<?php
 $previous_entries = new mysqli($db_host, $db_username, $db_password, $database);
 $check = $previous_entries->prepare("SELECT ID,NICK FROM users WHERE EMAIL=?;");
 $check->bind_param("s",$email);
 $check->execute();
 $check->bind_result($maybe_id,$maybe_got_something);

  while ($check->fetch()) { //here was typo, but fixed now 
    if ($maybe_got_something==$nickname){
     echo "Hooray!";
    }
  } 

?>

我從未見過“萬歲！”

但是，如果我這樣更改它：

$previous_entries = new mysqli($db_host, $db_username, $db_password, $database);
$prepared_statement = "SELECT ID,NICK FROM users WHERE EMAIL=";
$prepared_statement .=$email;
$result = $previous_entries->query($prepared_statement);
while ($row = $result->fetch_array()){
  if ($row["NICK"]==$nickname){
     echo "Hooray!";
    }
}

那一切都很好。

我在准備好的陳述中犯了一個可怕的錯誤。 但是我真的找不到它...我在這里做錯了什么？

編輯：更新了腳本以糾正錯誤的錯字，並添加了這兩行：

echo "maybeid: ".. $maybe_id;
echo "maybenick:". $maybe_got_something;

頁面與此相呼應：

  maybeid: maybenick: 

編輯：工作代碼嘗試對其進行調試時，我明白了這一點：

$previous_entries = new mysqli($db_host, $db_username, $db_password, $database);
$check = $previous_entries->prepare("SELECT ID,NICK FROM users WHERE EMAIL=?;");
$check->bind_param("s",$email);
$check->execute();
$check->bind_result($maybeid,$maybenick);
echo "maybeid: ".$maybeid;
echo "maybenick:". $maybenick;
// got rid of the if statement
  while ($check->fetch()) {
    echo "maybeid: ".$maybeid;
    echo "maybenick:". $maybenick;
    if ($maybenick==$nickname){

     echo "Hooray!";
    }
  } 

但是...為什么要擔心呢？ :)

Answer 1

准備好的查詢是$check但是在獲取時您正在使用$stmt 。

Answer 2

嘗試以下操作（使用從prepare返回的語句對象）：

<?php
 $previous_entries = new mysqli($db_host, $db_username, $db_password, $database);
 $check = $previous_entries->prepare("SELECT ID,NICK FROM users WHERE EMAIL=?;");
 $check->bind_param("s",$email);
 $check->execute();
 $check->bind_result($maybe_id,$maybe_got_something);
 if($check->num_rows > 0){
  while ($check->fetch()) {
    if ($maybe_got_something==$nickname){
     echo "Hooray!";
    }
  } 
}
?>

Answer 3

嘗試使用此$check->fetch()更改$stmt->fetch() $check->fetch()

您可能需要將error_reporting級別設置為類似

error_reporting = E_ALL和〜E_NOTICE

在您的php.ini開發環境中可以避免這種錯誤。

問候

Answer 4

您必須像這樣更改代碼：

...
$check = $previous_entries->prepare("SELECT ID,NICK FROM users WHERE EMAIL=:email;");
$check->bind_param(":email",$email);
...