Question

我写了一段PHP代码。我无法从数据库中检索结果并删除。在提交时它只是给出一个空白页而不会抛出任何错误。我是新手，所以请回答，即使你认为这是一个愚蠢的问题。请参阅代码并向我建议一些可以使我的代码正常工作的更改。

 <!DOCTYPE html>
    <html>

    <head>
    	<link rel="stylesheet" type="text/css" href="style.css">
    </head>

    <body>

    	<title>EDIT SCREEN</title>


    	<form action="test4.php" method="post">
    		<ul>
    			
    			<li>
    				Employee ID:</br>
    				<input type="text" name="eid">
    			</li>

    			<li>
    				<input type="submit" value="SUBMIT">
    			</li>
    		</ul>

    	</form>

    </body>

    </html>

//test.php

    <?php



    define('DB_NAME', 'test');
    define('DB_USER', '**');
    define("DB_PASSWORD", '**');
    define('DB_HOST', 'localhost');

    $link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD);


    $db_selected = mysql_select_db(DB_NAME, $link);


    if(isset($_POST['ok'])){
    $value1 = $_POST['eid'];

    $res = mysql_query("SELECT * from 'add' WHERE empid = '".$value1."'");
    echo "<table border='1'>
          <tr><th>Name</th>
          <th>EmployeeID</th><th>Address</th></tr>";

             while($row = mysql_fetch_array($res))
             {
                echo "<tr>";
                echo "<td>" . $row['name'] . "</td>";
                echo "<td>" . $row['empid'] . "</td>";
                echo "<td>" . $row['desig'] . "</td>";
                echo "<td><a href='test5.php?del=$row[empid]'>Delete</a></td>";
                echo "</tr>";
             }
             echo "</table>";
          }

    if (!mysql_query($sql))  
    {
           die('Error: ' . mysql_error());
    }



    mysql_close();

    ?>

//test5.php

<?php



define('DB_NAME', 'test');
define('DB_USER', '');
define("DB_PASSWORD", '');
define('DB_HOST', 'localhost');

$link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD);


$db_selected = mysql_select_db(DB_NAME, $link);


$value1 = $_POST['del'];

mysql_query("DELETE FROM add WHERE empid = '$value1'")


?>

Answer 1

<?php
if(isset($_POST['ok'])){
$value1 = $_POST['eid'];

$res = mysql_query("SELECT * from `add` WHERE empid = '".$value1."'");
echo "<table border='1'>
      <tr><th>Name</th>
      <th>EmployeeID</th><th>Address</th></tr>";

         while($row = mysql_fetch_array($res))
         {
            echo "<tr>";
            echo "<td>" . $row['name'] . "</td>";
            echo "<td>" . $row['empid'] . "</td>";
            echo "<td>" . $row['add'] . "</td>";
            echo "</tr>";
         }
         echo "</table>";
      }   
?>




New Code

<!DOCTYPE html>
    <html>

    <head>
        <link rel="stylesheet" type="text/css" href="style.css">
    </head>

    <body>

        <title>EDIT SCREEN</title>


        <form action="result.php" method="post">
            <ul>

                <li>
                    Employee ID:</br>
                    <input type="text" name="eid">
                </li>

                <li>
                    <input type="submit" value="SUBMIT" name="ok">
                </li>
            </ul>

        </form>

    </body>

    </html>



                <?php
                mysql_connect("localhost","root","");
                mysql_select_db("fdd");
                if(isset($_POST['ok'])){
                $value1 = $_POST['eid'];

                $res = mysql_query("SELECT * from `jobs` WHERE id = '".$value1."'");
                echo "<table border='1'>
                      <tr><th>Name</th>
                      <th>EmployeeID</th><th>Address</th></tr>";

                         while($row = mysql_fetch_array($res))
                         {
                            echo "<tr>";
                            echo "<td>" . $row['job_date'] . "</td>";
                            echo "<td>" . $row['client_code'] . "</td>";
                            echo "<td>" . $row['department'] . "</td>";
                            echo "</tr>";
                         }
                         echo "</table>";
                      }   
                ?>

     output

enter image description here

Answer 2

在查询数据库之前，您必须先建立连接。请按照以下代码段示例首先创建与mysql的连接，然后查询数据库。

<?php
$con=mysqli_connect("localhost","my_user","my_password","my_db");
// Check connection
if (mysqli_connect_errno())
  {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
  }

// Perform queries 
mysqli_query($con,"SELECT * FROM Persons");
mysqli_query($con,"INSERT INTO Persons (FirstName,LastName,Age) 
VALUES ('Glenn','Quagmire',33)");

mysqli_close($con);
?>

Answer 3

尝试这段代码，不要直接在mysql查询中使用id，在使用你的id直接将mysql查询从 mysql_real_escape_string 传递给它之前，它会从sql注入中清除id。请参阅SQL注入，REF，

<!DOCTYPE html>
<html>
<head>
    <link rel="stylesheet" type="text/css" href="style.css">
</head>
<body>
    <title>EDIT SCREEN</title>
        <form action="test4.php" method="post">
        <ul>
        <li>Employee ID:</br><input type="text" name="eid"></li>
        <li><input type="submit" value="SUBMIT"></li>
        </ul>
    </form>
</body>
</html>

********* test4.php file ***********

<?php

$value1 =   mysql_real_escape_string($_POST['eid']);
if($value1)
{
    $Connection     =   mysql_connect('localhost','root','') or die(mysql_error());
    $ConnectionDB   =   mysql_select_db($Connection) or die(mysql_error());
    $res = mysql_query("SELECT * from `add` WHERE empid = '".$value1."'") or die(mysql_error());
    echo "<table border='1'>
           <tr>
            <th>Name</th>
            <th>EmployeeID</th>
            <th>Address</th>
           </tr>";
         while($row = mysql_fetch_array($res))
         {
            echo "<tr>";
            echo "<td>" . $row['name'] . "</td>";
            echo "<td>" . $row['empid'] . "</td>";
            echo "<td>" . $row['add'] . "</td>";
            echo "</tr>";
         }
         echo "</table>";
}

注意：当您的代码有效时，请使用＆＃34;删除此＆＃34;或者删除（mysql_error（））＆＃34; ; ＆＃34; 它仅用于生产模式。

列表＆amp;使用php错误删除mysql中的记录

3 个答案: