2014-09-20 47 views
0

我有几行代码:UPDATE MySQL错误

$sql = mysql_query("UPDATE user SET txnid='" . mysql_real_escape_string($data['txn_id']) . "', 
    payment_amount='" . mysql_real_escape_string($data['payment_amount']) . "' , 
    payment_status='" . mysql_real_escape_string($data['payment_status']) . "', 
    itemid='" . mysql_real_escape_string($data['item_number']) . "', 
    createdtime='" . mysql_real_escape_string(date("Y-m-d H:i:s")) . "' WHERE id='" . mysql_real_escape_string($_COOKIE["buy_user"]) . "'"; 

    $result = mysql_query($sql); 

我试图更新ID 117

我知道我必须使用UPDATE,但我不能这样做。

帮助我

+2

哪里是你的更新? – rybo111 2014-09-20 17:52:52

+0

所以你想要'更新''用户''WHERE \'ID \'= 117'而不是'INSERT INTO \'用户\'VALUES()'? – Sean 2014-09-20 17:55:46

+2

你可以请张贴错误,以及?它是否在日志? – tandy 2014-09-20 18:00:33

回答

0

奇怪的是,你打电话mysql_query功能的两倍。

我建议你遵循这个模式:

# set a variable to contain just a string, the SQL text you want to execute 
$sql = "UPDATE ... "; 

# the SQL text is conveniently available for debugging/logging 
echo $sql; 

# execute the SQL 
$result = mysql_query($sql);