无法通过PHP将新记录插入到mysql数据库中
[英]Can't insert new records through PHP into mysql database
问题描述
When I press the submit button I get error. 
当我按下提交按钮时，我收到错误。
object not found error.
And the page automatically adds empty entries with auto incremented primary key (without pressing the submit button). 页面会自动添加带有自动递增主键的空条目（无需按提交按钮）。
I am still a beginner in PHP, I searched thoroughly but I can't find out what's wrong in code. 我仍然是PHP的初学者，我彻底搜索，但我无法找到代码中的错误。
<html>
<head>
<title>Add New Record in MySQL Database</title>
</head>
<body>
<form action="insert.php" method="post">
<p>
<label for="Name">Full Name:</label>
<input type="text" name="Name" id="Name">
</p>
<p>
<label for="Code">Code:</label>
<input type="text" name="Code" id="Code">
</p>
<p>
<label for="GPA">GPA:</label>
<input type="text" name="GPA" id="GPA">
</p>
<input type="submit" value="Submit">
</form>
<?php
/* Attempt MySQL server connection. Assuming you are running MySQL
server with default setting (user 'root' with no password) */
$link = mysqli_connect("localhost", "username", "password", "students");
// Check connection
if ($link === false) {
die("ERROR: Could not connect. " . mysqli_connect_error());
}
// Escape user inputs for security
$full_name = filter_input(INPUT_POST, 'full_name');
$code = filter_input(INPUT_POST, 'code');
$gpa = filter_input(INPUT_POST, 'gpa');
// attempt insert query execution
$sql = "INSERT INTO info VALUES ('$full_name', '$code', '$gpa')";
if (mysqli_query($link, $sql)) {
echo "Records added successfully. $full_name";
} else {
echo "ERROR: Could not able to execute $sql. " . mysqli_error($link);
}
// close connection
mysqli_close($link);
?>
</body>
</html>
3 个解决方案
解决方案1
13 已采纳 2016-10-02 16:29:55
Try this: 试试这个：
$full_name = filter_input(INPUT_POST, 'Name');
$code = filter_input(INPUT_POST, 'Code');
$gpa = filter_input(INPUT_POST, 'GPA');
The reason why I wrote that is because your input names contain Name , Code and GPA so you need to write this exactly as your input names ( case-sensitive ). 我写这个的原因是因为你的输入名称包含Name ， Code和GPA所以你需要把它写成输入名称（ 区分大小写 ）。
解决方案2
6 2016-10-02 16:39:51
Do with isset() . 用isset() 。 when the submit button clicks only the code runs. 当提交按钮单击时，只运行代码。
Inside the php you should use the form input name field. 在php中你应该使用表单输入名称字段。
<?php
if(isset($_POST['submit'])){
$link = mysqli_connect("localhost", "username", "password", "students");
if ($link === false) {
die("ERROR: Could not connect. " . mysqli_connect_error());
}
// Escape user inputs for security
$full_name = filter_input(INPUT_POST, 'full_name');
$code = filter_input(INPUT_POST, 'code');
$gpa = filter_input(INPUT_POST, 'gpa');
//to prevent sql injection attack
$full_name = mysqli_real_escape_string($link, $full_name);
$code = mysqli_real_escape_string($link, $code);
$gpa = mysqli_real_escape_string($link, $gpa);
// attempt insert query execution
$sql = "INSERT INTO info (Name,Code,GPA) VALUES ('$full_name', '$code', '$gpa')";
if (mysqli_query($link, $sql)) {
echo "Records added successfully. $full_name";
} else {
echo "ERROR: Could not able to execute $sql. " . mysqli_error($link);
}
// close connection
mysqli_close($link);
}
?>
<html>
<head>
<title>Add New Record in MySQL Database</title>
</head>
<body>
<form action="insert.php" method="post">
<p>
<label for="Name">Full Name:</label>
<input type="text" name="full_name" id="Name">
</p>
<p>
<label for="Code">Code:</label>
<input type="text" name="code" id="Code">
</p>
<p>
<label for="GPA">GPA:</label>
<input type="text" name="gpa" id="GPA">
</p>
<input type="submit" name="submit" value="submit">
</form>
</body>
</html>
解决方案3
2 2016-10-03 01:43:41
The problem is the input name. 问题是输入名称。 You named Full Name input with name="Name" , but you declare $full_name = filter_input(INPUT_POST, 'full_name'); 您使用name="Name"命名了Full Name输入，但是声明$full_name = filter_input(INPUT_POST, 'full_name'); in php section. 在PHP部分。 you must change full_name to Name . 您必须将full_name更改为Name 。 As well as the Code and GPA input. 以及代码和GPA输入。
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