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我有一個表單，用戶可以輸入成員，學生和/或講師的姓名（姓和名）。他們還輸入與這個問題無關的其他數據。這全部插入名爲games的表格中。名爲members,students和instructors的表格已經存在，其中的字段包括id,first_name和last_name。所有必需的外鍵約束都是按順序排列的。所以當用戶提交表格時，我只想插入成員，學生和/或教師的相應ID，而不是他們的名字。PHP MySQL將名稱轉換爲ID並且插入返回NULL值
我有以下代碼，並且它成功地提交除成員，學生和/或教師的id之外的所有信息，這些信息在記錄中顯示爲NULL。
如果有人可以在這裏強調這個問題，我將不勝感激。
形式：
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<title>Form</title>
</head>
<body>
<form action="action.php" method="post">
<p>
<p>
<label for="instructor_name">Instructor:</label>
<input type="text" name="instructor_name" id="instructor_name">
</p>
<p>
<label for="student_name">Student:</label>
<input type="text" name="student_name" id="student_name">
</p>
<p>
<label for="member_name">Member:</label>
<input type="text" name="member_name" id="member_name">
</p>
<p>
<label for="exercises">Exercises:</label>
<input type="text" name="exercises" id="exercises">
</p>
<p>
<label for="auth_by">Authorised By:</label>
<input type="text" name="auth_by" id="auth_by">
</p>
<p>
<label for="auth_duration">Authorisation Duration:</label>
<input type="text" name="auth_duration" id="auth_duration">
</p>
<input type="submit" value="Submit">
</form>
</body>
</html>
操作：
<?php
// connection information
include 'pdo_config.php';
try {
// new pdo connection
$conn = new PDO($dsn, $user, $pass, $opt);
// post data
$member_name = $_POST['member_name'];
$exercises = $_POST['exercises'];
$auth_by = $_POST['auth_by'];
$auth_duration = $_POST['auth_duration'];
$student_name = $_POST['student_name'];
$instructor_name = $_POST['instructor_name'];
// get corresponding id's of names
$name_m = explode(" ", $member_name);
$first_name_m = $name_m[0];
$last_name_m = $name_m[1];
$statement1 = $conn->prepare("SELECT member_id FROM tbl_members WHERE first_name = :first_name AND last_name = :last_name");
$statement1->execute(array(':first_name' => $first_name_m, ':last_name' => $last_name_m));
$row1 = $statement1->fetch();
$member_id = $row1['member_id'];
$name_i = explode(" ", $instructor_name);
$first_name_i = $name_i[0];
$last_name_i = $name_i[1];
$statement2 = $conn->prepare("SELECT instructor_id FROM tbl_instructors WHERE first_name = :first_name AND last_name = :last_name");
$statement2->execute(array(':first_name' => $first_name_i, ':last_name' => $last_name_i));
$row2 = $statement2->fetch();
$instructor_id = $row2['instructor_id'];
$name_s = explode(" ", $student_name);
$first_name_s = $name_s[0];
$last_name_s = $name_s[1];
$statement3 = $conn->prepare("SELECT student_id FROM tbl_students WHERE first_name = :first_name AND last_name = :last_name");
$statement3->execute(array(':first_name' => $first_name_s, ':last_name' => $last_name_s));
$row3 = $statement3->fetch();
$student_id = $row3['student_id'];
// prepare statements and bind parameters
$stmt = $conn->prepare("INSERT INTO tbl_games (member_id, exercises, auth_by, auth_duration, student_id, instructor_id) VALUES (:member_id, :exercises, :auth_by, :auth_duration, :student_id, :instructor_id)");
$stmt->bindParam(':member_id', $member_id);
$stmt->bindParam(':exercises', $exercises);
$stmt->bindParam(':auth_by', $auth_by);
$stmt->bindParam(':auth_duration', $auth_duration);
$stmt->bindParam(':student_id', $student_id);
$stmt->bindParam(':instructor_id', $instructor_id);
// execute statements
$stmt->execute();
// success or error message
echo "New record created successfully";
}
catch(PDOException $e)
{
echo "Error: " . $e->getMessage();
}
$conn = null;
?>
如果您想要將POST數組用作多維，則需要將輸入視爲數組。 I.e .:'name =「xxx []」'這裏似乎就是這種情況。否則，不要。 –
當你調試這個時，你插入的ID的變量值是多少？是否有任何由'SELECT'查詢返回的記錄？你在這些*查詢中使用的值是什麼，並且任何記錄都完全匹配那些值*？ （另外，在三個獨立表中存儲非常相似的數據似乎有點奇怪，並且首先要求所有這些不同的查詢。） – David
顯然只回答「回答」而不是「評論」。好吧，我離開這裏。我花了足夠的時間看這個。 –