我从终端创建了一个users表，mysql然后尝试创建简单的任务：从表单中插入值。这是我的dbConfig file<?php$mysqli = new mysqli("localhost", "root", "pass", "testDB");/* check connection */if (mysqli_connect_errno()) { printf("Connect failed: %s\n", mysqli_connect_error()); exit();}?>这是我的Index.php。<!doctype html><html> <head> <meta charset="utf-8"> <meta http-equiv="X-UA-Compatible" content="IE=edge,chrome=1"> <meta name="description" content="$1"> <meta name="viewport" content="width=device-width, initial-scale=1"> <link rel="stylesheet" type="text/css" href="style.css"> <title>test</title> <?php include_once 'dbConfig.php'; ?></head><body> <?php if(isset($_POST['save'])){ $sql = "INSERT INTO users (username, password, email) VALUES ('".$_POST["username"]."','".$_POST["password"]."','".$_POST["email"]."')"; } ?> <form method="post"> <label id="first"> First name:</label><br/> <input type="text" name="username"><br/> <label id="first">Password</label><br/> <input type="password" name="password"><br/> <label id="first">Email</label><br/> <input type="text" name="email"><br/> <button type="submit" name="save">save</button> <button type="submit" name="get">get</button> </form></body></html>按下我的保存按钮后，什么也没有发生，数据库仍然为空。我尝试echo'ing了INSERT查询，当它被认为它从形式的所有值。在尝试从终端检查是否可以正常工作后，我登录到sql尝试从用户表返回所有数据的操作，但得到了空集。
3 回答
侃侃无极
TA贡献2051条经验 获得超10个赞
您的代码中有两个问题。
表单中找不到任何动作。
您尚未执行查询mysqli_query（）
dbConfig.php
<?php
$conn=mysqli_connect("localhost","root","password","testDB");
if(!$conn)
{
die("Connection failed: " . mysqli_connect_error());
}
?>
的index.php
include('dbConfig.php');
<!Doctype html>
<html>
<head>
<meta charset="utf-8">
<meta http-equiv="X-UA-Compatible" content="IE=edge,chrome=1">
<meta name="description" content="$1">
<meta name="viewport" content="width=device-width, initial-scale=1">
<link rel="stylesheet" type="text/css" href="style.css">
<title>test</title>
</head>
<body>
<?php
if(isset($_POST['save']))
{
$sql = "INSERT INTO users (username, password, email)
VALUES ('".$_POST["username"]."','".$_POST["password"]."','".$_POST["email"]."')";
$result = mysqli_query($conn,$sql);
}
?>
<form action="index.php" method="post">
<label id="first"> First name:</label><br/>
<input type="text" name="username"><br/>
<label id="first">Password</label><br/>
<input type="password" name="password"><br/>
<label id="first">Email</label><br/>
<input type="text" name="email"><br/>
<button type="submit" name="save">save</button>
</form>
</body>
</html>
慕尼黑8549860
TA贡献1818条经验 获得超11个赞
尝试这个：
dbConfig.php
<?php
$mysqli = new mysqli('localhost', 'root', 'pwd', 'yr db name');
if($mysqli->connect_error)
{
echo $mysqli->connect_error;
}
?>
的index.php
<html>
<head><title>Inserting data in database table </title>
</head>
<body>
<form action="control_table.php" method="post">
<table border="1" background="red" align="center">
<tr>
<td>Login Name</td>
<td><input type="text" name="txtname" /></td>
</tr>
<br>
<tr>
<td>Password</td>
<td><input type="text" name="txtpwd" /></td>
</tr>
<tr>
<td> </td>
<td><input type="submit" name="txtbutton" value="SUBMIT" /></td>
</tr>
</table>
control_table.php
<?php include 'config.php'; ?>
<?php
$name=$pwd="";
if(isset($_POST['txtbutton']))
{
$name = $_POST['txtname'];
$pwd = $_POST['txtpwd'];
$mysqli->query("insert into users(name,pwd) values('$name', '$pwd')");
if(!$mysqli)
{ echo mysqli_error(); }
else
{
echo "Successfully Inserted <br />";
echo "<a href='show.php'>View Result</a>";
}
}
?>
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