[英]sql syntax error when creating column
I am trying to run the following query, but am getting a sql syntax error, but dreamweaver doesnt seem to be highlighting any error so im not sure where it is ?. 我试图运行以下查询,但我得到一个SQL语法错误,但Dreamweaver似乎没有突出显示任何错误,所以我不知道它在哪里? thanks :-)
谢谢 :-)
<?php
$form_id = $_POST[form_id];
$query = mysql_query("
IF NOT EXISTS (SELECT * FROM INFORMATION_SCHEMA.COLUMNS
WHERE
TABLE_NAME = 'email_history'
AND COLUMN_NAME = `$form_id`)
BEGIN
ALTER TABLE 'email_history' ADD `$form_id` VARCHAR( 255 ) NOT NULL
END;
") or die(mysql_error());
?>
Have a look at 看一下
‘$form_id`
You probably mean 你可能意味着
`$form_id`
Update: Once more wrong quotes 更新:再一次错误的报价
'email_history'
should be 应该
`email_history`
Remind: Backticks ` are for qualifiers (tablenames and such), where the apostroph '
is for values. 提醒:Backticks`用于限定符(tablenames等),其中apostroph
'
用于值。
Here is the correct quotation format. 这是正确的报价格式。
$query = mysql_query("
IF NOT EXISTS (SELECT * FROM INFORMATION_SCHEMA.COLUMNS
WHERE
TABLE_NAME = 'email_history'
AND COLUMN_NAME = '$form_id')
BEGIN
ALTER TABLE `email_history` ADD `$form_id` VARCHAR( 255 ) NOT NULL
END;
") or die(mysql_error());
Try this. 尝试这个。
$form_id = $_POST[form_id];
$query = mysql_query("
IF NOT EXISTS (SELECT * FROM INFORMATION_SCHEMA.COLUMNS
WHERE
TABLE_NAME = 'email_history'
AND COLUMN_NAME = '$form_id')
BEGIN
ALTER TABLE `email_history` ADD `$form_id` VARCHAR( 255 ) NOT NULL
END;
") or die(mysql_error());
Also, just on side note, you should never directly use the values from User input into your SQL. 另外,请注意,您不应该直接使用用户输入中的值到SQL中。 This is to prevent SQL injection.
这是为了防止SQL注入。 Here are a few links on MySQL Injection which I found:
以下是MySQL注入的一些链接,我发现:
http://www.php.net/manual/en/security.database.sql-injection.php http://25yearsofprogramming.com/blog/2011/20110205.htm https://www.owasp.org/index.php/SQL_Injection_Prevention_Cheat_Sheet http://www.php.net/manual/en/security.database.sql-injection.php http://25yearsofprogramming.com/blog/2011/20110205.htm https://www.owasp.org/index。 PHP / SQL_Injection_Prevention_Cheat_Sheet
<?php
$form_id = $_POST[form_id];
$query = mysql_query("
IF NOT EXISTS (SELECT * FROM INFORMATION_SCHEMA.COLUMNS
WHERE
TABLE_NAME = 'email_history'
AND COLUMN_NAME = '$form_id')
BEGIN
ALTER TABLE email_history ADD `$form_id` VARCHAR( 255 ) NOT NULL
END;
") or die(mysql_error());
?>
Everyone has focused on bad quoting, but is there any chance of this query to run? 每个人都专注于糟糕的引用,但这个查询是否有可能运行? Does MySQL allow such conditional statements?
MySQL是否允许这样的条件语句? I know this is possible in stored routines but I have never heard of such direct usage, so I would split this into two queries: column existence check and table alteration.
我知道这在存储例程中是可能的,但我从未听说过这样的直接用法,所以我将它分成两个查询:列存在检查和表更改。
This will look like this: 这将是这样的:
<?php
$form_id = some_function_to_check_if_input_is_valid_etc($_POST[form_id]);
$exist_result = mysql_query("SELECT IF(NOT EXISTS(
SELECT *
FROM INFORMATION_SCHEMA.COLUMNS
WHERE TABLE_NAME = 'email_history'
AND COLUMN_NAME = '{$form_id}'
),1,0) AS should_alter_table
");
$row = mysql_fetch_assoc($exist_result);
if($row['should_alter_table']){
mysql_query("ALTER TABLE email_history ADD `{$form_id}` VARCHAR( 255 ) NOT NULL");
}
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