PHP：将值从窗体插入MySQL
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。
GCT10153回答
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侃侃无极
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,$sql);}?><form action="index.php" method="post"> <label id="first"> First name:</label><br/><input type="text" name="username"><br/><label id="first">Password</label><br/><input type="password" name="password"><br/><label id="first">Email</label><br/><input type="text" name="email"><br/><button type="submit" name="save">save</button></form></body></html> -
慕尼黑8549860
尝试这个：dbConfig.php<?php$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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