我一直盯着这个页面半小时试图弄清楚我哪里出错了。找到前两个变量并将其插入到数据库中,但最后两个,'email'和'password'未找到,未插入数据库但仍传递if语句。任何帮助将不胜感激。
form.php的
<form name="signup" method="POST" action="signup.php">
<label for="signupFirstName">First Name</label>
<input type="text" id="signupFirstName" name="signupFirstName" />
<label for="signupLastName">Last Name</label>
<input type="text" id="signupLastName" name="signupLastName"/>
<label for="signupEmail">Email</label>
<input type="text" id="signupEmail" name="signupEmail" />
<label for="signupConfirmEmail">Confirm Email</label>
<input type="text" id="signupConfirmEmail" name="signupConfirmEmail"/>
<label for="signupPassword">Password</label>
<input type="text" id="signupPassword" name="signupPassword"/>
<label for="signupConfirmPassword">Confirm Password</label>
<input type="text" id="signupConfirmPassword" name="signupConfirmPassword"/>
<button name="submit" type="submit" >Submit Form</button>
</form>
signup.php
<?php
if (isset($_POST['signupFirstName']) || isset($_POST['signupLastName']) || isset($_POST['signupEmail']) || isset($_POST['signupPassword']) ) {
echo $_POST['signupEmail'];
$mysqli = new mysqli('localhost', 'user1', 'password', 'db2');
/* check connection */
if (mysqli_connect_errno()) {
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}
$stmt = $mysqli->prepare("INSERT INTO members (First_Name, Last_Name, Email, Password) VALUES (?,?,?,?)");
$stmt->bind_param('ssss',$sample,$lastName,$email,$password);
// escape the POST data for added protection
$sample = isset($_POST['signupFirstName'])
? $mysqli->real_escape_string($_POST['signupFirstName'])
: '';
$lastName = isset($_POST['signupLastName'])
? $mysqli->real_escape_string($_POST['signupLastName'])
: '';
$email = isset($_POST['signupEmail'])
? $mysqli->real_escape_string($_POST['signupEmail'])
: '';
$password = isset($_POST['signupPassword'])
? $mysqli->real_escape_string($_POST['signupPassword'])
: '';
/* execute prepared statement */
$stmt->execute();
printf("%d Row inserted.\n", $stmt->affected_rows);
/* close statement and connection */
$stmt->close();
/* close connection */
$mysqli->close();
}
else{
echo "broken";
}
?>
答案 0 :(得分:0)
在实际设置变量之前,您似乎将params绑定到查询。将bind_param()
来电移至execute()
来电之上。
您还可以重构代码以获取大量垃圾。示例如下:
<?php
function arr_get($array, $key) {
if (isset($array[$key])) {
return $array[$key];
}
return '';
}
if (isset($_POST['signupFirstName']) || isset($_POST['signupLastName']) || isset($_POST['signupEmail']) || isset($_POST['signupPassword']) ) {
echo $_POST['signupEmail'];
$mysqli = new mysqli('localhost', 'user1', 'password', 'db2');
/* check connection */
if (mysqli_connect_errno()) {
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}
$stmt = $mysqli->prepare("INSERT INTO members (First_Name, Last_Name, Email, Password) VALUES (?,?,?,?)");
$stmt->bind_param('ssss', arr_get($_POST, 'signupFirstName'), arr_get($_POST, 'signupLastName'), arr_get($_POST, 'signupEmail'), arr_get($_POST, 'signupPassword'));
/* execute prepared statement */
$stmt->execute();
printf("%d Row inserted.\n", $stmt->affected_rows);
/* close statement and connection */
$stmt->close();
/* close connection */
$mysqli->close();
}
else {
echo "broken";
}
?>