Question

我是stackoverflow的新手。我想创建这个表单，将信息上传到数据库。当我点击提交时，它不会上传。我检查了我的连接文件，这是正确的。代码如下。请帮忙。

<?php
 include("/connections/db_conx.php");

 if ( isset($_POST['submit']) ) {


$title = mysqli_real_escape_string($_POST['title']);
 $text = mysqli_real_escape_string($_POST['text']);
   $picture = mysqli_real_escape_string($_POST['picture']);


  $sql = "INSERT INTO news  (title, text, picture) VALUES('$title','$text','$picture', now(),now(),now())";

    $query = mysqli_query ($db_conx, $sql);


     echo 'Entered into the news table'; 
     }
     ?>


  <html>


  <head>



 </head>


 <body>

  <table border="0"> 
  <tr>
  <form method="post" action="index.php" id="tech">
  <td>Title</td><td> <input type="text" name="title"></td> </tr>
  <tr> <td>Text</td><td> <textarea rows="4" name="text" cols="50"       name="comment"       form="tech"> </textarea>
       </td> </tr>

 <tr> <td>Picture</td><td> <input type="varchar" name="picture"></td> </tr>

 <tr> <td><input id="button" type="submit" name="submit" value="Submit"></td>

 </tr>
 </form>


 </table> 



</body>




</html>

Answer 1

以下是您需要使用的代码：

<?php
    include("/connections/db_conx.php");
    if(isset($_POST['submit'])) {
        $title   = mysqli_real_escape_string($db_conx, $_POST['title']);
        $text    = mysqli_real_escape_string($db_conx, $_POST['text']);
        $picture = mysqli_real_escape_string($db_conx, $_POST['picture']);
        $sql     = "INSERT INTO news (`title`, `text`, `picture`) VALUES('$title','$text','$picture');";
        if(!$result = $db_conx->query($sql)){
            die('There was an error running the query [' . $db_conx->error . ']');
        }
        echo 'Entered into the news table';
    }
?>


<html>
    <head>
    </head>
    <body>
        <form method="post" action="index.php" id="tech">
            <table border="0">
                <tr>
                    <td>Title</td>
                    <td> <input type="text" name="title"></td>
                </tr>
                <tr> 
                    <td>Text</td>
                    <td><textarea rows="4" name="text" cols="50" name="comment" form="tech"> </textarea></td> 
                </tr>
                <tr> 
                    <td>Picture</td>
                    <td> <input type="varchar" name="picture"></td> 
                </tr>
                <tr> 
                    <td><input id="button" type="submit" name="submit" value="Submit"></td>
                </tr>
            </table>
        </form>
    </body>
</html>

您的问题是mysqli_real_escape_string()函数需要2个参数：数据库连接和要转义的字符串。

我还包括完全重新格式化的代码和错误检查。

Answer 2

试试这个：

$title = mysqli_real_escape_string($db_conx, $_POST['title']);
$text = mysqli_real_escape_string($db_conx, $_POST['text']);
$picture = mysqli_real_escape_string($db_conx, $_POST['picture']);

$sql = "INSERT INTO news  (title, text, picture) VALUES('$title','$text','$picture')";

$query = mysqli_query ($db_conx, $sql);

if($query){
echo 'Entered into the news table'; // Your Success Message
}
else { 
    echo mysqli_error($db_conx); 
}

php注册到数据库

2 个答案: