Question

有谁可以解释为什么这个PHP / MySQL无效？基本上我需要从表单一次插入行的负载，因此会有多个名称字段，多个短，中，长字段等。我得到这个错误：

Notice: Undefined variable: Short1 in C:\xampp\htdocs\process.php on line 95
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'Long, VLong, Extreme, LJump, HJump, Shotputt, Discuss, Javelin, Date, Year) VAL' at line 2

这是我的PHP

<?php


$host = "localhost";
$databasename = "pe_results";
$databaseusername = "root";
$databasepassword = "";

$conn = mysql_connect("$host", "$databaseusername", "$databasepassword"); 
mysql_select_db("$databasename", $conn); 

        if (isset($_POST['Name1'])) { 
        $Name1 = $_POST['Name1'];
        }
        if (isset($_POST['Short1'])) { 
        $Short1 = $_POST['Short1'];
        }
        if (isset($_POST['Med1'])) { 
        $Med1 = $_POST['Med1'];
        }
        if (isset($_POST['Long1'])) { 
        $Long1 = $_POST['Long1'];
        }
        if (isset($_POST['VLong1'])) { 
        $VLong1 = $_POST['VLong1'];
        }
        if (isset($_POST['Extreme1'])) { 
        $Extreme1 = $_POST['Extreme1'];
        }
        if (isset($_POST['LJump1'])) { 
        $LJump1 = $_POST['LJump1'];
        }
        if (isset($_POST['HJump1'])) { 
        $HJump1 = $_POST['HJump1'];
        }
        if (isset($_POST['Shotputt1'])) { 
        $Shotputt1 = $_POST['Shotputt1'];
        }
        if (isset($_POST['Discuss1'])) { 
        $Discuss1 = $_POST['Discuss1'];
        }
        if (isset($_POST['Javelin1'])) { 
        $Javelin1 = $_POST['Javelin1'];
        }
        if (isset($_POST['Date'])) { 
        $Date = $_POST['Date'];
        }
        if (isset($_POST['Year'])) { 
        $Year = $_POST['Year'];
        }
        // Sector 2 */
            if (isset($_POST['Name2'])) { 
        $Name2 = $_POST['Name2'];
        }
        if (isset($_POST['Short2'])) { 
        $Short2 = $_POST['Short2'];
        }
        if (isset($_POST['Med2'])) { 
        $Med2 = $_POST['Med2'];
        }
        if (isset($_POST['Long2'])) { 
        $Long2 = $_POST['Long2'];
        }
        if (isset($_POST['VLong2'])) { 
        $VLong2 = $_POST['VLong2'];
        }
        if (isset($_POST['Extreme2'])) { 
        $Extreme2 = $_POST['Extreme2'];
        }
        if (isset($_POST['LJump2'])) { 
        $LJump2 = $_POST['LJump2'];
        }
        if (isset($_POST['HJump2'])) { 
        $HJump2 = $_POST['HJump2'];
        }
        if (isset($_POST['Shotputt2'])) { 
        $Shotputt2 = $_POST['Shotputt2'];
        }
        if (isset($_POST['Discuss2'])) { 
        $Discuss2 = $_POST['Discuss2'];
        }
        if (isset($_POST['Javelin2'])) { 
        $Javelin2 = $_POST['Javelin2'];
        }
        if (isset($_POST['Date'])) { 
        $Date = $_POST['Date'];
        }
        if (isset($_POST['Year'])) { 
        $Year = $_POST['Year'];
        }

        $sql="INSERT INTO results_main
  (Name, Short, Med, Long, VLong, Extreme, LJump, HJump, Shotputt, Discuss, Javelin, Date, Year)
VALUES
  ('$Name1', '$Short1', '$Med1', '$Long1', '$VLong1', '$Extreme1', '$LJump1', '$HJump1', '$Shotputt1', '$Discuss1', '$Javelin1', '$Date', '$Year'),
  ('$Name2', '$Short2', '$Med2', '$Long2', '$VLong2', '$Extreme2', '$LJump2', '$HJump2', '$Shotputt2', '$Discuss2', '$Javelin2', '$Date', '$Year');
";

$result = mysql_query($sql) or die(mysql_error());

// close connection 
mysql_close($conn);

?>

JW的新错误消息

Notice: Undefined variable: Short1 in C:\xampp\htdocs\process.php on line 95
INSERT INTO results_main (`Name`, `Short`, `Med`, `Long`, `VLong`, `Extreme`, `LJump`, `HJump`, `Shotputt`, `Discuss`, `Javelin`, `Date`, `Year`) VALUES (`1`, ``, `1`, `1`, `1`, `1`, `1`, `1`, `1`, `1`, `1`, `2013-04-26`, `10`), (`2`, `2`, `2`, `2`, `2`, `2`, `2`, `2`, `2`, `2`, `2`, `2013-04-26`, `10`); Unknown column '1' in 'field list'

Answer 1

LONG是保留关键字，恰好是列的名称。为了避免语法错误，应使用反引号转义列名。

INSERT INTO results_main(Name, Short, Med, `Long`, VLong, ...) VALUES (....)

MySQL Reserved Keywords List

如果您有权更改列，请将名称更改为非保留关键字，以避免以后出现问题。

作为旁注，如果变量的值（ s ）来自外部，则查询易受SQL Injection攻击。请查看下面的文章，了解如何防止它。通过使用PreparedStatements，您可以摆脱在值周围使用单引号。

How to prevent SQL injection in PHP?

Answer 2

检查你的帖子数组，你有错误的$ _POST ['Short1'];

另外不要使用MYSQL保留关键字，Long是保留关键字。如果你使用，你应该通过

逃避它

`Long`

Answer 3

我建议通过转储你的$ _POST数组并查看它。根据您的代码变量仅在$ _POST数组中存在值时才设置。

一次将多个值插入MySQL

3 个答案: