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我在使用PDO更新查詢代碼時遇到問題。當我通過更改信息並單擊保存按鈕保存信息來編輯人員的單個記錄時,它會影響只應更新單個記錄的所有記錄。所有記錄現在都有相同的信息。如何在不影響其他人的情況下更新單個記錄?先謝謝你。PHP PDO更新查詢影響執行時的所有行
update.php
<?php
include ('includes/connection.php');
$id = isset($_GET['id']) ? $_GET['id']: die('Error: Record ID not found.');
try {
$query_select = "SELECT id, profile_picture, first_name, last_name, gender, age, date_birth FROM tbl_records WHERE id = ? LIMIT 0,1";
$query_statement = $db_connection->prepare($query_select);
$query_statement->bindParam(1, $id);
$query_statement->execute();
$row = $query_statement->fetch();
$profilePicture = $row['profile_picture'];
$firstName = $row['first_name'];
$lastName = $row['last_name'];
$gender = $row['gender'];
$age = $row['age'];
$dateBirth = $row['date_birth'];
}
catch(PDOException $e) {
die('Error 1: '. $e->getMessage());
}
if($_POST) {
try {
$query_update = "UPDATE tbl_records SET
profile_picture = :t_profile_picture,
first_name = :t_first_name,
last_name = :t_last_name,
gender = :t_gender,
age = :t_age;
date_birth = :t_date_birth
WHERE id = :t_id";
$query_statement = $db_connection->prepare($query_update);
$profilePicture = htmlspecialchars(strip_tags($_POST['profile-picture']));
$firstName = htmlspecialchars(strip_tags($_POST['first-name']));
$lastName = htmlspecialchars(strip_tags($_POST['last-name']));
$gender = htmlspecialchars(strip_tags($_POST['gender']));
$age = htmlspecialchars(strip_tags($_POST['age']));
$dateBirth = htmlspecialchars(strip_tags($_POST['date-birth']));
$query_statement->bindParam(':t_profile_picture', $profilePicture);
$query_statement->bindParam(':t_first_name', $firstName);
$query_statement->bindParam(':t_last_name', $lastName);
$query_statement->bindParam(':t_gender', $gender);
$query_statement->bindParam(':t_age', $age);
$query_statement->bindParam(':t_date_birth', $dateBirth);
$query_statement->bindParam(':t_id', $id);
if($query_statement->execute()) {
echo "<div class='alert alert-success' role='start'>Record was updated</div>";
}
else {
echo "<div class='alert alert-danger' role='start'>Unable to update the record.</div>";
}
echo var_dump($query_statement->rowCount());
}
catch(PDOException $e) {
die('ERROR 2: ' . $e->getMessage());
}
}
?>
<html>
<body>
<form action="update.php?id=<?php echo htmlspecialchars($id); ?>" method="post">
<input type="hidden" name="id" value="<?php echo htmlspecialchars($id, ENT_QUOTES); ?>" />
<input type="file" name="profile-picture" value="<?php echo htmlspecialchars($profilePicture, ENT_QUOTES); ?>" />
<label for="first-name">First name:</label> <br />
<input type="text" name="first-name" value="<?php echo htmlspecialchars($firstName, ENT_QUOTES); ?>" /> <br />
<label for="last-name">Last name:</label> <br />
<input type="text" name="last-name" value="<?php echo htmlspecialchars($lastName, ENT_QUOTES); ?>" /> <br />
<label for="gender">Gender:</label> <br />
<input type="text" name="gender" value="<?php echo htmlspecialchars($gender); ?>" /> <br />
<label for="age">Age:</label> <br />
<input type="text" name="age" value="<?php echo htmlspecialchars($age); ?>" /> <br />
<label for="date-birth">Date of Birth:</label> <br />
<input type="date" name="date-birth" value="<?php echo htmlspecialchars($dateBirth); ?>" /> <br />
<input class="button-style" type="submit" value="SAVE" />
</form>
</body>
</html>
'age =:t_age;'<<<語句結束。這應該是一個逗號。這就是爲什麼它正在更新一切。分號實際上是一個有效的字符,不會爲其輸出錯誤。它也不會更新'date_birth'列。 –
投票結束爲一個錯字。 –
@ Fred-ii-非常感謝。我正在尋找花了幾個小時的問題,原因只是一個錯字><非常感謝! – lvlzero