任何人都可以闡明爲什麼這個PHP/MySQL不工作?基本上,我需要一次從表格中插入行的負載,所以將有多個名稱字段,多個短,中,長字段等即時得到這個錯誤:一次向MySQL中插入多個值
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
這是我的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);
?>
爲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'
謝謝,我用反引號排序了這個問題,但現在我得到這個錯誤:注意:未定義的變量:Short1在95行的C:\ xampp \ htdocs \ process.php中 '字段列表中的未知列'1' ' –
在調用'mysql_query()'之前,你能發佈這個結果嗎:'echo $ sql'? –
更新了問題的結果,我已經解決了調用$ Short1的問題,但是我仍然收到未知列1錯誤 –