Inserting multiple values into a MySQL at once [duplicate]

问题

could anyone shed some light as to why this PHP / MySQL is not working? Basically I need to insert loads of rows at once from a form, so there will be multiple name fields, multiple short, med, long fields etc.. Im getting this error:

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

Here is my 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);

?>

New error message for JW

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'

回答1:

LONG is a reserved keyword and happens to be the name of your column. In order to avoid syntax error, the column name should be escape with backticks.

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

MySQL Reserved Keywords List

If you have the privilege to alter the column, change the name to a non-reserved keyword to avoid problem getting back on the future.

As a sidenote, the query is vulnerable with SQL Injection if the value(s) of the variables came from the outside. Please take a look at the article below to learn how to prevent from it. By using PreparedStatements you can get rid of using single quotes around values.

How to prevent SQL injection in PHP?

回答2:

Check your post array, you have something wrong with $_POST['Short1'];

Also don't use MYSQL reserved keywords, Long is a reserved keyword. If you use you should escape it by

`Long`

回答3:

I suggest by dumping your $_POST array and looking at it. According to your code variables are only set if there is a value in the $_POST array.

来源：https://stackoverflow.com/questions/16015818/inserting-multiple-values-into-a-mysql-at-once

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