PHP：将表单中的值插入 MySQL

Question

提问by notrealme

I created a userstable in mysqlfrom the terminal and I am trying to create simple task: insert values from the form. This is my dbConfig file

我从终端创建了一个users表，mysql我正在尝试创建简单的任务：从表单插入值。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();
}
?>

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="">
    <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. I tried echo'ingthe INSERTquery and it takes all values from the form as it is supposed to. After I try to check if this worked from terminal, I login into my sqltry to return all data from users table and I get empty set.

点击我的保存按钮后，没有任何反应，数据库仍然是空的。我尝试echo'ing了INSERT查询，它按照预期从表单中获取所有值。在我尝试从终端检查这是否有效后，我登录sql尝试从用户表中返回所有数据，但我得到了空集。

Answer 1

回答by dimlucas

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. In order to do that you need to use some functions but let me explain something else first.

它不执行查询。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 $_GETor $_POST) directlyto your query. Someone can carefully manipulate the input in such a way so that it can cause great damage to your database. That's called SQL Injection. You can read more about it here

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$_GET或的表单输入$_POST）直接附加到您的查询中。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。这就是所谓的 SQL 注入。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 mustuse Prepared Statements. More on prepared statements here

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。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 ?are used as placeholders for the values. Next you should prepare the statement using mysqli_prepare:

请注意如何将?用作值的占位符。接下来，您应该使用mysqli_prepare以下方法准备语句：

$stmt = mysqli_prepare($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. (This is where the actual insertion takes place)

最后执行准备好的语句。（这是实际插入发生的地方）

$stmt->execute();

NOTEAlthough not part of the question, I strongly advice you to never store passwords in clear text. Instead you should use password_hashto store a hash of the password

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。相反，您应该使用password_hash来存储密码的哈希值

Answer 2

回答by Syed Muneeb Ul Hasan

There are two problems in your code.

您的代码中有两个问题。

No action found in form.
You have not executed the query mysqli_query()

未在表单中找到任何操作。
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="">
<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>

Answer 3

回答by Dewesh Sinha

Try this:

尝试这个：

dbConfig.php

配置文件

<?php
$mysqli = new mysqli('localhost', 'root', 'pwd', 'yr db name');
    if($mysqli->connect_error)
        {
        echo $mysqli->connect_error;
        }
    ?>

Index.php

索引.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>&nbsp;</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>";
    }

         }  

    ?>

Answer 4

回答by mittal bhatt

<!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>

Answer 5

回答by Jalak Nimavat

<?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>";
?>

Answer 6

回答by ntthushara

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;
    }

PHP：将表单中的值插入 MySQL

提问by notrealme

回答by dimlucas

回答by Syed Muneeb Ul Hasan

回答by Dewesh Sinha

回答by mittal bhatt

回答by Jalak Nimavat

回答by ntthushara

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PHP：将表单中的值插入 MySQL

提问by notrealme

回答by dimlucas

回答by Syed Muneeb Ul Hasan

回答by Dewesh Sinha

回答by mittal bhatt

回答by Jalak Nimavat

回答by ntthushara

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