1

我有一个连接到 mysql 数据库的 php 页面。我知道与数据库的连接很好，因为我有一个 php 代码，可以将数据库中的信息显示到网页上。当我尝试将新数据插入数据库时​​，页面刷新并且数据未插入。我已检查以确保 insert into 命令具有正确的值。

      <?php 

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

            include "connect_to_mysql.php";

            $name = mysql_real_escape_string($_POST["Name"]);

            $sql = mysql_query("SELECT TestID FROM test WHERE Name='$name' LIMIT 1")or die (mysql_error());
            $productMatch = mysql_num_rows($sql); 

            if ($productMatch > 0) 
                {
                    echo 'Sorry you tried to place a duplicate "User Account" into the system, <a href="index.php">click here</a>';
                    exit();
                }
                else
                {

                    $sql = mysql_query("INSERT INTO test (TestID,Name) 
                        VALUES('', '$name')") or die (mysql_error());
                     $uid = mysql_insert_id();

                    header("location: index.php"); 
                    exit();
                }
        }
?>
<?php 

    include "connect_to_mysql.php";
    $User_list = "";
    $sql = mysql_query("SELECT * FROM test");
    $UserCount = mysql_num_rows($sql); 
    if ($UserCount > 0) 
        {
            while($row = mysql_fetch_array($sql))
                { 
                         $id = $row["TestID"];
                         $name = $row["Name"];

                         $User_list .= "Users ID: $id - <strong>$name</strong>&nbsp; &nbsp; &nbsp;<br />";
                }
        } 
        else 
        {
            $User_list = "You have no users listed in the database.";
        }
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>

<body>
<div align="center" id="mainWrapper">
  <div id="pageContent"><br />
    <div align="right" style="margin-right:32px;"><a href="index.php#UserForm">+ Add New User</a></div>
<div align="left" style="margin-left:24px;">
      <h2>User list</h2>
      <?php echo $User_list; ?>
    </div>
    <hr />
    <a name="UserForm" id="UserForm"></a>
    <h3>
    &darr; Add New User Form &darr;
    </h3>
    <form action="index.php" enctype="multipart/form-data" name="myForm" id="myform" method="post">
    <table width="90%" border="0" cellspacing="0" cellpadding="6">
      <tr>
        <td width="20%" align="right">Name</td>
        <td width="80%"><label>
          <input name="name" type="text" id="name" size="50" />
        </label></td>
      </tr> 
      <tr>
        <td>&nbsp;</td>
        <td><label>
          <input type="submit" name="button" id="button" value="Add This Name Now" />
        </label></td>
      </tr>
    </table>
    </form>
    <br />
  <br />
  </div>
</div>
</body>
</html>
4

4 回答 4

0

我马上看到了两个问题（可能还有更多）。首先，PHP 数组键区分大小写。您正在访问$_POST['Name']，但您的表单输入是name. 其次，您正在测试$_POST['User_Name']似乎在任何地方都不存在的：

// Look for name in the $_POST
if (isset($_POST['name'])) 
{
    include "connect_to_mysql.php";
    // name is case-sensitive
    $name = mysql_real_escape_string($_POST["name"]);

稍后，如果您的表有一个 AUTO_INCREMENT id on TestID，您应该省略它或在插入语句中插入 NULL：

// Don't include TestID if it is AUTO_INCREMENT. That will happen automatically
$sql = mysql_query("INSERT INTO test (Name) 
                    VALUES('$name')") or die (mysql_error());
于 2012-08-17T01:54:16.873 回答
0

我认为如果你改变它会起作用

 if (isset($_POST['User_Name']))

到

if (isset($_POST['Name']))

您检查表单中不存在的东西的存在。

于 2012-08-17T01:54:27.823 回答
0

添加：

如果TestID是自动增量，则更改以下内容

"INSERT INTO test (TestID,Name) VALUES('', '$name')"

到

"INSERT INTO test (Name) VALUES('$name')"
于 2012-08-17T01:56:31.977 回答
0

如果您没有收到任何错误，这意味着您的 MySQL 语法有错误，有两种测试方法是将语法复制到 PHPMyAdmin 或您的本机 MySQL 命令行中，然后查看是否收到输出错误。或者您可以做的另一件事是修改所有mysql_query(); 通过添加功能mysql_query()or die(mysql_error());

于 2012-08-17T01:56:42.670 回答