PHP：将表单中的值插入 MySQL
[英]PHP: Inserting Values from the Form into MySQL
问题描述
I created a users table in mysql from the terminal and I am trying to create simple task: insert values from the form.
我从终端在mysql创建了一个users表，我正在尝试创建简单的任务：从表单插入值。 This is my dbConfig file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();
}
?>
and This is my Index.php .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>
After hitting my save button, nothing happens, database is still empty.点击我的保存按钮后，没有任何反应，数据库仍然是空的。 I tried echo'ing the INSERT query and it takes all values from the form as it is supposed to.我尝试echo'ing显INSERT查询，它按照预期从表单中获取所有值。 After I try to check if this worked from terminal, I login into my sql try to return all data from users table and I get empty set.在我尝试从终端检查这是否有效后，我登录到我的sql尝试从用户表中返回所有数据，但我得到了空集。
2 个解决方案
解决方案1
47 已采纳 2016-05-21 21:48:27
The following code just declares a string variable that contains a MySQL query:以下代码仅声明了一个包含 MySQL 查询的字符串变量：
$sql = "INSERT INTO users (username, password, email)
VALUES ('".$_POST["username"]."','".$_POST["password"]."','".$_POST["email"]."')";
It does not execute the query.它不执行查询。 In order to do that you need to use some functions but let me explain something else first.为了做到这一点，你需要使用一些函数，但让我先解释一下其他的东西。
NEVER TRUST USER INPUT : You should never append user input (such as form input from $_GET or $_POST ) directly to your query.永远不要信任用户输入：您永远不应该将用户输入（例如来自$_GET或$_POST表单输入）直接附加到您的查询中。 Someone can carefully manipulate the input in such a way so that it can cause great damage to your database.有人可以以这种方式小心地操纵输入，从而可能对您的数据库造成很大的损害。 That's called SQL Injection.这就是所谓的 SQL 注入。 You can read more about it here你可以在这里阅读更多关于它的信息
To protect your script from such an attack you must use Prepared Statements.为了保护您的脚本免受此类攻击，您必须使用准备好的语句。 More on prepared statements here更多关于准备好的声明在这里
Include prepared statements to your code like this:在您的代码中包含准备好的语句，如下所示：
$sql = "INSERT INTO users (username, password, email)
VALUES (?,?,?)";
Notice how the ?注意如何? are used as placeholders for the values.用作值的占位符。 Next you should prepare the statement using mysqli_prepare :接下来，您应该使用mysqli_prepare准备语句：
$stmt = $mysqli->prepare($sql);
Then start binding the input variables to the prepared statement:然后开始将输入变量绑定到准备好的语句：
$stmt->bind_param("sss", $_POST['username'], $_POST['email'], $_POST['password']);
And finally execute the prepared statements.最后执行准备好的语句。 (This is where the actual insertion takes place) （这是实际插入发生的地方）
$stmt->execute();
NOTE Although not part of the question, I strongly advice you to never store passwords in clear text.注意虽然不是问题的一部分，但我强烈建议您永远不要以明文形式存储密码。 Instead you should use password_hash to store a hash of the password相反，您应该使用password_hash来存储password_hash的哈希值
解决方案2
6 2016-09-23 07:04:51
There are two problems in your code.您的代码中有两个问题。
- No action found in form.未在表单中找到任何操作。
- You have not executed the query mysqli_query()您还没有执行查询mysqli_query()
dbConfig.php配置文件
<?php
$conn=mysqli_connect("localhost","root","password","testDB");
if(!$conn)
{
die("Connection failed: " . mysqli_connect_error());
}
?>
index.php索引.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>
解决方案3
-1 2017-03-01 18:07:18
Try this: 尝试这个：
dbConfig.php dbConfig.php
<?php
$mysqli = new mysqli('localhost', 'root', 'pwd', 'yr db name');
if($mysqli->connect_error)
{
echo $mysqli->connect_error;
}
?>
Index.php 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>";
}
}
?>
解决方案4
-2 2017-06-22 03:57:08
<!DOCTYPE html>
<?php
$con = new mysqli("localhost","root","","form");
?>
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.2.1/jquery.min.js"></script>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
<title>Untitled Document</title>
<script type="text/javascript">
$(document).ready(function(){
//$("form").submit(function(e){
$("#btn1").click(function(e){
e.preventDefault();
// alert('here');
$(".apnew").append('<input type="text" placeholder="Enter youy Name" name="e1[]"/><br>');
});
//}
});
</script>
</head>
<body>
<h2><b>Register Form<b></h2>
<form method="post" enctype="multipart/form-data">
<table>
<tr><td>Name:</td><td><input type="text" placeholder="Enter youy Name" name="e1[]"/>
<div class="apnew"></div><button id="btn1">Add</button></td></tr>
<tr><td>Image:</td><td><input type="file" name="e5[]" multiple="" accept="image/jpeg,image/gif,image/png,image/jpg"/></td></tr>
<tr><td>Address:</td><td><textarea cols="20" rows="4" name="e2"></textarea></td></tr>
<tr><td>Contact:</td><td><div id="textnew"><input type="number" maxlength="10" name="e3"/></div></td></tr>
<tr><td>Gender:</td><td><input type="radio" name="r1" value="Male" checked="checked"/>Male<input type="radio" name="r1" value="feale"/>Female</td></tr>
<tr><td><input id="submit" type="submit" name="t1" value="save" /></td></tr>
</table>
<?php
//echo '<pre>';print_r($_FILES);exit();
if(isset($_POST['t1']))
{
$values = implode(", ", $_POST['e1']);
$imgarryimp=array();
foreach($_FILES["e5"]["tmp_name"] as $key=>$val){
move_uploaded_file($_FILES["e5"]["tmp_name"][$key],"images/".$_FILES["e5"]["name"][$key]);
$fname = $_FILES['e5']['name'][$key];
$imgarryimp[]=$fname;
//echo $fname;
if(strlen($fname)>0)
{
$img = $fname;
}
$d="insert into form(name,address,contact,gender,image)values('$values','$_POST[e2]','$_POST[e3]','$_POST[r1]','$img')";
if($con->query($d)==TRUE)
{
echo "Yoy Data Save Successfully!!!";
}
}
exit;
// echo $values;exit;
//foreach($_POST['e1'] as $row)
//{
$d="insert into form(name,address,contact,gender,image)values('$values','$_POST[e2]','$_POST[e3]','$_POST[r1]','$img')";
if($con->query($d)==TRUE)
{
echo "Yoy Data Save Successfully!!!";
}
//}
//exit;
}
?>
</form>
<table>
<?php
$t="select * from form";
$y=$con->query($t);
foreach ($y as $q);
{
?>
<tr>
<td>Name:<?php echo $q['name'];?></td>
<td>Address:<?php echo $q['address'];?></td>
<td>Contact:<?php echo $q['contact'];?></td>
<td>Gender:<?php echo $q['gender'];?></td>
</tr>
<?php }?>
</table>
</body>
</html>
解决方案5
-3 2017-06-19 05:46:34
<?php
$username="root";
$password="";
$database="test";
#get the data from form fields
$Id=$_POST['Id'];
$P_name=$_POST['P_name'];
$address1=$_POST['address1'];
$address2=$_POST['address2'];
$email=$_POST['email'];
mysql_connect(localhost,$username,$password);
@mysql_select_db($database) or die("unable to select database");
if($_POST['insertrecord']=="insert"){
$query="insert into person values('$Id','$P_name','$address1','$address2','$email')";
echo "inside";
mysql_query($query);
$query1="select * from person";
$result=mysql_query($query1);
$num= mysql_numrows($result);
#echo"<b>output</b>";
print"<table border size=1 >
<tr><th>Id</th>
<th>P_name</th>
<th>address1</th>
<th>address2</th>
<th>email</th>
</tr>";
$i=0;
while($i<$num)
{
$Id=mysql_result($result,$i,"Id");
$P_name=mysql_result($result,$i,"P_name");
$address1=mysql_result($result,$i,"address1");
$address2=mysql_result($result,$i,"address2");
$email=mysql_result($result,$i,"email");
echo"<tr><td>$Id</td>
<td>$P_name</td>
<td>$address1</td>
<td>$address2</td>
<td>$email</td>
</tr>";
$i++;
}
print"</table>";
}
if($_POST['searchdata']=="Search")
{
$P_name=$_POST['name'];
$query="select * from person where P_name='$P_name'";
$result=mysql_query($query);
print"<table border size=1><tr><th>Id</th>
<th>P_name</th>
<th>address1</th>
<th>address2</th>
<th>email</th>
</tr>";
while($row=mysql_fetch_array($result))
{
$Id=$row[Id];
$P_name=$row[P_name];
$address1=$row[address1];
$address2=$row[address2];
$email=$row[email];
echo"<tr><td>$Id</td>
<td>$P_name</td>
<td>$address1</td>
<td>$address2</td>
<td>$email</td>
</tr>";
}
echo"</table>";
}
echo"<a href=lab2.html> Back </a>";
?>
解决方案6
-4 2017-10-27 22:10:49
When you click the Button 当您单击按钮时
if(isset($_POST['save'])){
$sql = "INSERT INTO `members`(`id`, `membership_id`, `email`, `first_name`)
VALUES ('".$_POST["id"]."','".$_POST["membership_id"]."','".$_POST["email"]."','".$_POST["firstname"]."')";
**if ($conn->query($sql) === TRUE) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}**
}
This will execute the Query in the variable $sql 这将在变量$ sql中执行查询
if ($conn->query($sql) === TRUE) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
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