嗨，我似乎找不到有关此主题的任何帮助，因为没有抛出错误。我试图使用 mysqli_query 通过 php 将记录插入数据库，但在重定向后没有进行任何更改。我正在使用三个文件:index.php、conn.php 和 new.php。 index.php 和 new.php 位于同一文件夹中，但 conn.php 位于下一个目录。 索引.php:
<!DOCTYPE html>
<html>
<head>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.2.1/jquery.min.js"></script>
<link rel="stylesheet" href="style.css" type="text/css" >
<link href="https://fonts.googleapis.com/css?family=Raleway" rel="stylesheet">
<link href="https://fonts.googleapis.com/css?family=Khula" rel="stylesheet">
</head>
<script>
$(function()
{
$('.error').fadeOut(10000);
});
</script>
<body>
<header>
<img src="images/logo.png">
<p>The reliable bus company</p>
</header>
<div class="wrapper">
<div class="container">
<div class="titletxt">
<h4>Drivers</h4>
</div>
<?php
include '../conn.php';
mysqli_query($conn, "SET NAMES utf8");
$result = mysqli_query($conn, "SELECT * FROM tbl_employee");
echo "
<div class='table_content'>
<table align='center'>
<tr>
<th>Employee ID</th>
<th>Title</th>
<th>Name</th>
<th>Address</th>
<th>Contact Number</th>
<th>Job Position</th>
<th>Gender</th>
<th>DOB</th>
</tr>
";
while($row = mysqli_fetch_array($result))
{
echo "<tr>";
echo "<td>" . $row['employeeID'] . "</td>";
echo "<td>" . $row['title'] . "</td>";
echo "<td>" . $row['name'] . "</td>";
echo "<td>" . $row['address'] . "</td>";
echo "<td>" . $row['contactNum'] . "</td>";
echo "<td>" . $row['position'] . "</td>";
echo "<td>" . $row['gender'] . "</td>";
echo "<td>" . $row['DOB'] . "</td>";
echo "</tr>";
}
echo "</table></div>";
?>
<!-- Record Insert -->
<br>
<div class="titletxt">
<h4>Insert a Record</h4>
</div>
<h3 style="font-weight: 400; margin-left: 5px;">New Employee</h3>
<form class="insert_form" action="new.php" method="post" name="insert_form">
<label>Title: </label>
<input type="text" name="title" required><br>
<span class="error"><?php echo $titleErr ?></span>
<br>
<label>Name: </label>
<input type="text" name="name" required> <br>
<span class="error"><?php echo $nameErr ?></span>
<br>
<label>Address:</label>
<input type="text" name="address" required><br>
<span class="error"><?php echo $addressErr ?></span>
<br>
<label>Contact Number</label>
<input type="text" name="contactNum" required><br>
<span class="error"><?php echo $contactErr ?></span>
<br>
<label>Job Position</label>
<input type="text" name="position" required><br>
<span class="error"><?php echo $positionErr ?></span>
<br>
<label>Gender: </label>
<input type="radio" name="gender" value="male" required> Male
<input type="radio" name="gender" value="female" required> Female<br>
<span class="error"><?php echo $genderErr ?></span>
<br>
<label>DOB: </label>
<input style="width: 60px;" type="text" name="DOB_year" required>YYYY
<input style="width: 30px;" type="text" name="DOB_months" required>MM
<input type="text" name="DOB_day" style="width: 30px" required>DD<br>
<span class="error"><?php echo $DOBErr ?></span>
<br>
<input type="submit" Value="Insert Entry">
</form>
</div>
</div>
</body>
</html>
conn.php:
<?php
$server = "localhost";
$user = "root";
$password = "";
$db = "bus_db";
global $conn;
$conn = mysqli_connect($server, $user, $password, $db);
if(mysqli_connect_errno())
{
echo "Mysql Error has occured" . mysqli_connect_error;
}
else if(!mysqli_connect_errno())
{
echo "<connection>Connection Established</connection>";
}
function close_connection()
{
global $conn;
mysqli_close($conn);
}
$title = $name = $address = $contact = $position = $gender = $DOB = "";
$titleErr = $nameErr = $addressErr = $contactErr = $positionErr = $genderErr = $DOBErr = "";
mysqli_query($conn, "SET NAMES utf8");
if ($_SERVER["REQUEST_METHOD"] == "POST")
{
if (empty($_POST["title"]))
{
$titleErr = "Title is Required";
}else{
$title = input($_POST["title"]);
}
if (empty($_POST["name"]))
{
$nameErr = "Name is Required";
}else
{
$name = input($_POST["name"]);
if (!preg_match("/^[a-zA-Z ]*$/",$name)) {
$nameErr = "Invalid Name";
}
}
if (empty($_POST["address"]))
{
$addressErr = "Address is Required";
}else{
$address = input($_POST["address"]);
}
if (empty($_POST["contactNum"]))
{
$contactErr = "Contact Number is required ";
}else{
$contact = input($_POST["contactNum"]);
$regex = "^([0-9]{10,11})$^";
if (!preg_match($regex, $contact)) {
$contactErr = "Invalid Phone Number";
}
}
if(empty($_POST["position"]))
{
$positionErr = "Position is required";
}else{
$position = input($_POST["position"]);
}
if (empty($_POST["gender"]))
{
$genderErr = "Gender is Required";
}else{
$gender = input($_POST["gender"]);
}
if (empty($_POST["DOB_year"]) || empty($_POST["DOB_months"]) || empty($_POST["DOB_day"]))
{
$DOBErr = "Invalid entry for date of birth";
}else
{
$DOB = input($_POST["DOB_year"] + "/" + $_POST["DOB_months"] + "/" + $_POST["DOB_day"]);
}
}
function input($data) {
$data = trim($data);
$data = stripslashes($data);
$data = htmlspecialchars($data);
return $data;
}
function insert_records($p_title, $p_name, $p_address, $p_contact, $p_position, $p_gender, $p_DOB)
{
global $conn;
mysqli_query($conn, "INSERT INTO tbl_employee VALUES(null, '" .$p_title."', '".$p_name."', '".$p_address."', '".$p_contact."', '".$p_position."', '".$p_gender."', '".$p_DOB."')");
}
?>
新的.php:
<?php
include '../conn.php';
insert_records($title, $name, $address, $contact, $position, $gender, $DOB);
header( 'Location:index.php');
close_connection();
?>
我将不胜感激，谢谢
最佳答案
您应该编辑 insert_records()，以便在 mysqli_query 失败时向您提供反馈。
function insert_records($p_title, $p_name, $p_address, $p_contact, $p_position, $p_gender, $p_DOB)
{
global $conn;
$result = mysqli_query($conn, 'some query') or die('Query failed: ' . mysqli_error($conn));
return $result;
}
并在此处阅读有关如何防止 MySQL 注入(inject)的信息:How can I prevent SQL injection in PHP?
<小时/>编辑:
$DOB = input($_POST["DOB_year"] + "/" + $_POST["DOB_months"] + "/" + $_POST["DOB_day"]);
在php中'+'用于进行计算。如果你想连接字符串使用'.'
$DOB = input($_POST["DOB_year"] . "/" . $_POST["DOB_months"] . "/" . $_POST["DOB_day"]);
关于PHP 不会将数据插入 Mysql 数据库 - 不会引发错误，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/44101191/