Question

   <html>

   <head>
   <title>Add New Record in MySQL Database</title>
   </head>

    <body>
     <?php
     if(isset($_POST['add'])) {
        $dbhost = 'localhost';
        $dbuser = 'root';
        $dbpass = 'password';
        $conn = mysqli_connect($dbhost, $dbuser, $dbpass);

        if(! $conn ) {
           die('Could not connect: ' . mysql_error());
        }

        if(! get_magic_quotes_gpc() ) {
           $emp_name = addslashes ($_POST['emp_name']);
           $emp_address = addslashes ($_POST['emp_address']);
        }else {
           $emp_name = $_POST['emp_name'];
           $emp_address = $_POST['emp_address'];
        }

        $emp_salary = $_POST['emp_salary'];

        $sql = "insert into employee(emp_name,emp_address, emp_salary)values('$emp_name','$emp_address','$emp_salary')";

        mysqli_select_db($conn,"test_db");
        $retval = mysqli_query($conn,$sql);

        if(!$retval) {
           die('Could not enter data: ' . mysql_error());
        }

        echo "Entered data successfully\n";

        mysql_close($conn);
     }else {
        ?>

           <form method = "post" action = "<?php $_PHP_SELF ?>">
              <table width = "400" border = "0" cellspacing = "1" 
                 cellpadding = "2">

                 <tr>
                    <td width = "100">Employee Name</td>
                    <td><input name = "emp_name" type = "text" 
                       id = "emp_name"></td>
                 </tr>

                 <tr>
                    <td width = "100">Employee Address</td>
                    <td><input name = "emp_address" type = "text" 
                       id = "emp_address"></td>
                 </tr>

                 <tr>
                    <td width = "100">Employee Salary</td>
                    <td><input name = "emp_salary" type = "text" 
                       id = "emp_salary"></td>
                 </tr>

                 <tr>
                    <td width = "100"> </td>
                    <td> </td>
                 </tr>

                 <tr>
                    <td width = "100"> </td>
                    <td>
                       <input name = "add" type = "submit" id = "add" 
                          value = "Add Employee">
                    </td>
                 </tr>

              </table>
           </form>

        <?php
     }
  ?>

当我尝试输入值并在此时按下提交按钮时，我没有收到任何错误，但我无法在数据库中输入值。   问题是我得到的文字是＆＃34;无法输入数据：表＆＃39;员工＆＃39;是只读的＃34;。有人可以帮我解决这个问题吗？

我在wamp服务器中创建了数据库（test_db）和table（employee）。

Answer 1

我确定您未授权用户在您的表格中输入数据请编辑schema_name，并在DB上执行查询：

GRANT ALL ON TABLE schema_name.employee TO root;

您也可以在没有架构的情况下尝试： GRANT ALL ON TABLE employee to root;

Answer 2

你的问题解决了。尽管如此，我强烈建议您使用Prepared Statements，否则您的代码将打开SQL注入和可能的引用问题。

你正在混合mysql和mysqli。停下来。既然你正在使用mysqli，利用准备好的语句和bind_param，否则你就是打开SQL注入和可能的引用问题。 - @ aynber

<强> 更改

将die('Could not connect: ' . mysql_error());更改为die('Could not connect: ' . mysqli_connect_error());
将mysql_close($conn);更改为mysqli_close($conn);
将action = "<?php $_PHP_SELF ?>"更改为action = "<?php echo $_SERVER['PHP_SELF']; ?>"
使用Prepared Statements。

更新代码

<html>

   <head>
    <title>Add New Record in MySQL Database</title>
   </head>

    <body>
     <?php
     if(isset($_POST['add'])) {
        $dbhost = 'localhost';
        $dbuser = 'root';
        $dbpass = 'password';
        $db = "test_db";

        $conn = mysqli_connect($dbhost, $dbuser, $dbpass, $db);

        if(! $conn ) {
           die('Could not connect: ' . mysqli_connect_error());
        }

        $stmt = mysqli_prepare($conn, "INSERT INTO employee(emp_name,emp_address, emp_salary) VALUES (?, ?, ?)");
        mysqli_stmt_bind_param($stmt, 'sss', $_POST['emp_name'], $_POST['emp_address'], $_POST['emp_salary']);

        if(!mysqli_stmt_execute($stmt)) {
           die('Could not enter data: ' . mysqli_error($conn));
        }

        echo "Entered data successfully\n";

        mysqli_close($conn);
     } else {
        ?>
           <form method = "post" action = "<?php echo $_SERVER['PHP_SELF']; ?>">
                <table width = "400" border = "0" cellspacing = "1"  cellpadding = "2">
                    <tr>
                        <td width = "100">Employee Name</td>
                        <td><input name = "emp_name" type = "text" id = "emp_name"></td>
                    </tr>
                     <tr>
                        <td width = "100">Employee Address</td>
                        <td><input name = "emp_address" type = "text" id = "emp_address"></td>
                     </tr>
                     <tr>
                        <td width = "100">Employee Salary</td>
                        <td><input name = "emp_salary" type = "text" id = "emp_salary"></td>
                     </tr>
                     <tr>
                        <td width = "100"> </td>
                        <td> </td>
                     </tr>
                     <tr>
                        <td width = "100"> </td>
                        <td><input name = "add" type = "submit" id = "add" value = "Add Employee"></td>
                     </tr>
                </table>
           </form>

        <?php
     }
  ?>

快速查看

无法在数据库中输入数据

2 个答案: