PHP服务器和mysql表单
PHP服务器和mysql表单
提问于 2018-05-22 04:39:26
我不能让这个起作用。我碰到了一个路障。请有人看一下，如果你发现我遗漏了什么，请告诉我。已验证与mysql的连接是否正常。当我提交表单时，它会显示"Record inserted successfully“。但是在数据库中没有创建实际的记录。我尝试过不同的方法，但都没有成功。
<?php
if ($_SERVER["REQUEST_METHOD"] == "POST"){
$usrID = $_POST["usrID"];
$Fname = $_POST["Fname"];
$Lname = $_POST["Lname"];
$short_title = $_POST["short_title"];
$email = $_POST["email"];
$phone = $_POST["phone"];
$link = mysqli_connect("localhost", "username", "topsecret", "DBname");
if($link === false){
die("ERROR: Could not connect. " . mysqli_connect_error());
}
$sql = "INSERT INTO tableName (usrID, Fname, Lname, short_title, email, phone) VALUES (?, ?, ?, ?, ?, ?)";
if($stmt = mysqli_prepare($link, $sql)){
mysqli_stmt_bind_param($stmt, "sss", $usrID, $Fname, $Lname, $short_title, $email, $phone);
mysqli_stmt_execute($stmt);
echo "Record inserted successfully.";
} else{
echo "ERROR: Could not prepare query: $sql. " . mysqli_error($link);
}
mysqli_stmt_close($stmt);
mysqli_close($link);
}
?>
<!doctype html>
<html>
<head>
<meta charset="utf-8">
<title>Need Help!</title>
</head>
<body>
<form action="<?php echo htmlspecialchars($_SERVER["PHP_SELF"]);?>" method="post">
<table align="center" width="650" border="1" cellspacing="0" cellpadding="3">
<tbody>
<tr>
<td colspan="4" align="center">
<span>Input New User Information: </span></td>
</tr>
<tr>
<td colspan="4" align="center">
<label for="usrID">User ID:</label>
<input type="text" name="usrID" id="usrID" required></td>
</tr>
<tr>
<td colspan="2"><label for="Fname">First Name: </label><input type="text" name="Fname" id="Fname" required></td>
<td colspan="2"><label for="Lname">Last Name: </label><input type="text" name="Lname" id="Lname" required></td>
</tr>
<tr>
<td colspan="2"><label for="short_title">Abreviated Title(s): </label><input type="text" name="short_title" id="short_title" placeholder="Generally goes after Last Name and a comma"></td>
<td colspan="2"><label for="email">Agent Email: </label><input type="email" name="email" id="email"></td>
</tr>
<tr>
<td colspan="4"><label for="phone">Agent Phone: </label><input type="tel" name="phone" id="phone" placeholder="XXX.XXX.XXXX"></td>
</tr>
<tr>
<td colspan="2"><input type="reset"></td>
<td colspan="2"><input type="submit" name="submit" value="Submit"></td>
</tr>
</tbody>
</table>
</form>
</body>
</html>回答 1
Stack Overflow用户
发布于 2018-05-22 04:43:19
看起来这个纹身返回了真的：
if($stmt = mysqli_prepare($link, $sql)){无论执行的语句返回什么，如果上述条件为真，它都会打印您的代码行：
mysqli_stmt_execute($stmt);
echo "Record inserted successfully.";$stmt中的语句是什么？检查sql并验证其是否正确。
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/50456332
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