2013-08-17 267 views
0

我試圖tomake形式將數據添加到我的資料庫,並即時得到一個錯誤將數據添加到MySQL數據庫

「無法輸入數據:您的SQL語法錯誤;檢查手冊這對應於您的MySQL服務器版本的正確語法使用附近')VALUES('s','d','f','g','h')'在第1行「

這裏是使用

<html> 
<head> 
<title>Add New Record in MySQL Database</title> 
</head> 
<body> 
<?php 
if(isset($_POST['add'])) 
{ 
$dbhost = 'localhost'; 
$dbuser = 'root'; 
$dbpass = ''; 
$conn = mysql_connect($dbhost, $dbuser, $dbpass); 
if(! $conn) 
{ 
    die('Could not connect: ' . mysql_error()); 
} 

if(! get_magic_quotes_gpc()) 
{ 
    $make = addslashes ($_POST['make']); 
    $model = addslashes ($_POST['model']); 
    $value = addslashes ($_POST['value']); 
    $bcosts = addslashes ($_POST['bcosts']); 
} 
else 
{ 
    $make = $_POST['make']; 
    $model = $_POST['model']; 
    $value = $_POST['value']; 
    $bcosts = $_POST['bcosts']; 
} 
$rcosts = $_POST['rcosts']; 

$sql="INSERT INTO referb (make,model,rcosts,bcosts,value,) VALUES ('$make','$model','$rcosts','$bcosts','$value')"; 

mysql_select_db('ely'); 
$retval = mysql_query($sql, $conn); 
if(! $retval) 
{ 
    die('Could not enter data: ' . mysql_error()); 
} 
echo "Entered data successfully\n"; 
mysql_close($conn); 
} 
else 
{ 
?> 
<form method="post" action="<?php $_PHP_SELF ?>"> 
<table width="400" border="0" cellspacing="1" cellpadding="2"> 
<tr> 
<td width="100">make</td> 
<td><input name="make" type="text" id="make"></td> 
</tr> 
<tr> 
<td width="100">model</td> 
<td><input name="model" type="text" id="model"></td> 
</tr> 
<tr> 
<td width="100">repair costs</td> 
<td><input name="rcosts" type="text" id="rcosts"></td> 
</tr> 
<tr> 
<td width="100">paid costs</td> 
<td><input name="bcosts" type="text" id="bcosts"></td> 
</tr> 
<tr> 
<td width="100">value</td> 
<td><input name="value" type="text" id="value"></td> 
</tr> 
<tr> 
<td width="100"> </td> 
<td> </td> 
</tr> 
<tr> 
<td width="100"> </td> 
<td> 
<input name="add" type="submit" id="add" value="Add"> 
</td> 
</tr> 
</table> 
</form> 
<?php 
} 
?> 
</body> 
</html> 

我非常新手因此爲什麼我問笑碼IM不要擔心,這會不會成爲一個活的網站:)

+0

只是爲了確保你知道:http://stackoverflow.com/questions/12859942/why-shouldnt -i-use-mysql-functions-in-php –

+0

非常感謝您的支持 - 我是極度新手,並試圖與之交手,我更喜歡手中的方法。似乎還有更多關於這種方式的樣本和教程,然後是新的方法。你提到的方式有沒有簡單的跟隨教程? –

回答

1

更換value,value這裏:

$sql="INSERT INTO referb (make,model,rcosts,bcosts,value,) VALUES ('$make','$model','$rcosts','$bcosts','$value')"; 

所以就變成:

$sql="INSERT INTO referb (make,model,rcosts,bcosts,value) VALUES ('$make','$model','$rcosts','$bcosts','$value')"; 
+0

感謝工作:) –

+0

@RickNash填補免費點擊「接受」按鈕,如果答案幫助你。 –