使用bindParam與PDO

Question

我一直在抓我的頭在此代碼爲幾個小時.... 沒有道理給我，爲什麼它不工作

$isCorrect =($question->correct_answer == $body->answer) ? 1:0; 
// the values are all there....... 
// echo $body->question . "\n"; //335 
// echo $body->user . "\n";  //51324123 
// echo $question->day . "\n"; //0 
// echo $isCorrect . "\n";  //0 

//but still the below part fails. 
$db = getConnection(); 
$sql = "INSERT INTO `answers` (`id`, `question_id`, `user`, `day`, `is_correct`) VALUES (NULL, ':question', ':user', ':day', :is_correct)"; 
$stmt = $db->prepare($sql); 
$stmt->bindParam(":question_id", $body->question); 
$stmt->bindParam(":user", $body->user); 
$stmt->bindParam(":day", $question->day, PDO::PARAM_INT); 
$stmt->bindParam(":is_correct", $isCorrect, PDO::PARAM_INT); 
$stmt->execute(); 

給出了這樣的錯誤：

SQLSTATE[HY093]: Invalid parameter number: number of bound variables does not match number of tokens 

我在計算4個標記...我錯過了什麼？顯然我做錯了什麼。

Answer 1

變化：

$stmt->bindParam(":question_id", $body->question); 

到：

$stmt->bindParam(":question", $body->question); 

您在查詢:question使用，但用錯了鍵（:question_id）結合。

Answer 2

$stmt->bindParam(":question_id", $body->question); 

應該

$stmt->bindParam(":question", $body->question); 

這只是一個小錯字。

Answer 3

只是不使用bindParam和PDO
以及命名參數。它會爲你節省大量的頭痛

$db = getConnection(); 
$sql = "INSERT INTO `answers` VALUES (NULL, ?,?,?,?)"; 
$data = [$body->question,$body->user,$question->day,$isCorrect]; 
$stmt = $db->prepare($sql)->execute($data); 

Answer 4

試試這樣說：

$sql = "INSERT INTO `answers` (`id`, `question_id`, `user`, `day`, `is_correct`) 
     VALUES 
     --The :variable shouldn't be surrounded by ''-- 
     (NULL, :question, :user, :day, :is_correct)"; 
$stmt = $db->prepare($sql); 
//The values used in $sql should be the same here, so not :question_id but :question 
$stmt->bindParam(":question", $body->question); 
$stmt->bindParam(":user", $body->user); 
$stmt->bindParam(":day", $question->day, PDO::PARAM_INT); 
$stmt->bindParam(":is_correct", $isCorrect, PDO::PARAM_INT);