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php - 无法在XAMPP中使用php更新MYSQL数据库表字段

转载 作者：行者123 更新时间：2023-11-29 19:52:04 26 4
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所以，我正在尝试设计一个 php 网站，到目前为止，它在向列表添加条目方面效果良好。

问题是，它无法使用edit.php更新表。单击编辑链接时，它会显示一条消息:

"There is no data to be edited."

但是如果我尝试手动放置 localhost/edit.php**?id=1** 它会显示 id 编号列表并且工作正常。请帮忙。

home.php

<html>
    <head>
        <title>My first PHP Website</title>
    </head>
   <?php
   session_start(); //starts the session
   if($_SESSION['user']){ // checks if the user is logged in  
   }
   else{
      header("location: index.php"); // redirects if user is not logged in
   }
   $user = $_SESSION['user']; //assigns user value
   ?>
    <body>
        <h2>Home Page</h2>

<hello>! 
 <!--Display's user name-->
        <a href="logout.php">Click here to go logout</a><br/><br/>
        <form action="add.php" method="POST">
           Add more to list: <input type="text" name="details" /> <br/>
           Public post? <input type="checkbox" name="public[]" value="yes" /> <br/>
           <input type="submit" value="Add to list"/>
        </form>
    <h2 align="center">My list</h2>
    <table border="1px" width="100%">
    <tr>
      <th>Id</th>
      <th>Details</th>
      <th>Post Time</th>
      <th>Edit Time</th>
      <th>Edit</th>
      <th>Delete</th>
      <th>Public Post</th>
    </tr>
    <?php
      mysql_connect("localhost","root","") or die(mysql_error());
      mysql_select_db("first_db") or die("Cannot connect to database");
      $query = mysql_query("select * from list");
      while($row = mysql_fetch_array($query))
      {
        print "<tr>";
          print '<td align="center">'. $row['id'] . "</td>";
          print '<td align="center">'. $row['details'] . "</td>";
          print '<td align="center">'. $row['date_posted'] . " - " . $row['time_posted'] . "</td>";
          print '<td align="center">'. $row['date_edited'] . " - " . $row['time_edited'] . "</td>";
          print '<td align="center"><a href="edit.php">edit</a></td>';
          print '<td align="center"><a href="delete.php">delete</a></td>';
          print '<td align="center">'. $row['public'] . "</td>";
        print "</tr>";
      }
    ?>
    </table>
  </body>
</html>

image )

编辑.php

<html>
    <head>
        <title>My first PHP website</title>
    </head>
    <?php
    session_start(); //starts the session
    if($_SESSION['user']){ //checks if user is logged in
    }
    else{
        header("location:index.php"); // redirects if user is not logged in
    }
    $user = $_SESSION['user']; //assigns user value
    $id_exists = false;
    ?>
    <body>
        <h2>Home Page</h2>
        <p>Hello <?php Print "$user"?>!</p> <!--Displays user's name-->
        <a href="logout.php">Click here to logout</a><br/><br/>
        <a href="home.php">Return to Home page</a>
        <h2 align="center">Currently Selected</h2>
        <table border="1px" width="100%">
            <tr>
                <th>Id</th>
                <th>Details</th>
                <th>Post Time</th>
                <th>Edit Time</th>
                <th>Public Post</th>
            </tr>
            <?php
                if(!empty($_GET['id']))
                {
                    $id = $_GET['id'];
                    $_SESSION['id'] = $id;
                    $id_exists = true;
                    mysql_connect("localhost", "root","") or die(mysql_error()); //Connect to server
                    mysql_select_db("first_db") or die("Cannot connect to database"); //connect to database
                    $query = mysql_query("Select * from list Where id='$id'"); // SQL Query
                    $count = mysql_num_rows($query);
                    if($count > 0)
                    {
                        while($row = mysql_fetch_array($query))
                        {
                            Print "<tr>";
                                Print '<td align="center">'. $row['id'] . "</td>";
                                Print '<td align="center">'. $row['details'] . "</td>";
                                Print '<td align="center">'. $row['date_posted']. " - ". $row['time_posted']."</td>";
                                Print '<td align="center">'. $row['date_edited']. " - ". $row['time_edited']. "</td>";
                                Print '<td align="center">'. $row['public']. "</td>";
                            Print "</tr>";
                        }
                    }
                    else
                    {
                        $id_exists = false;
                    }
                }
            ?>
        </table>
        <br/>
        <?php
        if($id_exists)
        {
        Print '
        <form action="edit.php" method="POST">
            Enter new detail: <input type="text" name="details"/><br/>
            public post? <input type="checkbox" name="public[]" value="yes"/><br/>
            <input type="submit" value="Update List"/>
        </form>
        ';
        }
        else
        {
            Print '<h2 align="center">There is no data to be edited.</h2>';
        }
        ?>
    </body>
</html>

<?php
    if($_SERVER['REQUEST_METHOD'] == "POST")
    {
        mysql_connect("localhost", "root","") or die(mysql_error()); //Connect to server
        mysql_select_db("first_db") or die("Cannot connect to database"); //Connect to database
        $details = mysql_real_escape_string($_POST['details']);
        $public = "no";
        $id = $_SESSION['id'];
        $time = strftime("%X");//time
        $date = strftime("%B %d, %Y");//date
        foreach($_POST['public'] as $list)
        {
            if($list != null)
            {
                $public = "yes";
            }
        }
        mysql_query("UPDATE list SET details='$details', public='$public', date_edited='$date', time_edited='$time' WHERE id='$id'") ;
        header("location: home.php");
    }
?>

http://s15.postimg.org/m8dloz7d7/screenshot_20.png

这是网址中带有 ?id=1 的http;//s15,postimg,org/yoabiq0p7/screenshot_21,png(用句点更改逗号)。

最佳答案

您只打印 edit.php，您需要打印整个编辑链接。

print '<td align="center"><a href="edit.php">edit</a></td>';

将此行替换为:

print '<td align="center"><a href="edit.php?id=' . $row["id"] . '">edit</a></td>';

这将解决问题。

P.S:小心，你的代码是开放的 SQL 注入(inject)!确保在这个地方使用mysql_real_escape_string():

$id = mysql_real_escape_string($_GET['id']);

如果id只是数字，您也可以执行以下操作以避免SQL注入(inject):

$id = intval($_GET["id"]);

SQL 注入(inject)问题非常严重，您需要过滤来自外部的内容。我也建议使用准备好的语句 PDO。

关于php - 无法在XAMPP中使用php更新MYSQL数据库表字段，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/40794996/

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