MySQL语法错误——第1行接近“1”。

When running my PHP script It keeps giving me the error

在运行我的PHP脚本时，它总是给我错误。

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 '1' at line 1

在SQL语法中有一个错误;请检查与您的MySQL服务器版本对应的手册，以便在第1行中使用接近“1”的语法。

This is my sql code I have other than selecting from the table. I have commented out all of this and not gotten an error, so I'm assuming its occuring in this block of code.

这是我从表中选择的sql代码。我已经注释掉了所有这些并没有出错，所以我假设它发生在这段代码中。

    if($status === 1){
        $sqlQ = mysql_query("UPDATE tablename SET status=1 WHERE steam64='$id'");

        if(!mysql_query($sqlQ, $con)){
            die('Error: ' . mysql_error());
        }
    }else if($status !== 1){
        $sqlQ = mysql_query("UPDATE tablename SET status=2 WHERE steam64='$id'");

        if(!mysql_query($sqlQ, $con)){
            die('Error: ' . mysql_error());
        }
    }

What is really confusing me is the line 1 part.

真正让我困惑的是第一部分。

5 个解决方案

#1

You're violating the DRY principle big time. Why not something like...

你违反了干原则。为什么不像……

$statusValue = ($status === 1) ? 1 : 2;
$sqlQuery = mysql_query("UPDATE `14d2_group` SET `status` = $statusValue WHERE `steam64` = '$id'"):

UPDATE 2: It looks like there's a need for additional clarification.

更新2:看起来需要进一步的澄清。

mysql_query function doesn't only create a query: it actually sends in to MySQL - and returns the result. In case of UPDATE it will return FALSE if query has failed. That's why you shouldn't call mysql_query twice, as you did in the original example.

mysql_query函数不仅创建一个查询:它实际上发送到MySQL——并返回结果。如果发生更新，如果查询失败，它将返回FALSE。这就是为什么不应该像在原示例中那样调用mysql_query两次。

You can check how many lines were actually updated with mysql_affected_rows function.

您可以通过mysql_affected_rows函数检查实际更新了多少行。

UPDATE 3: Finally get it. ) That was the reason error appeared: you tried to call mysql_query with result of the last update query. Which was, as TRUE converted to String, just '1'. )

更新3:最终得到它。这就是出现错误的原因:您试图用上次更新查询的结果调用mysql_query。也就是说，它被转换为字符串，只有'1'。

#2

You're using the result from one query as a query itself.

您将一个查询的结果作为查询本身使用。

What you probably wanted to do is:

你可能想做的是:

if($status === 1){
    $sqlQ = mysql_query("UPDATE tablename SET status=1 WHERE steam64='$id'");

    if (!$sqlQ) {
        die('Error: ' . mysql_error());
    }
}
else {// no need for your if-statement here because it would always be true
    $sqlQ = mysql_query("UPDATE tablename SET status=2 WHERE steam64='$id'");

    if(!$sqlQ){
        die('Error: ' . mysql_error());
    }
}

#3

"Line 1" corresponds to line 1 of the query, not the script invoking it. To add the line of the script invoking it, use:

“第1行”对应的是查询的第1行，而不是调用它的脚本。要添加调用它的脚本的行，请使用:

die('Error: ' . mysql_error() . ' in ' . $_SERVER['PHP_SELF'] . ' on line ' . __LINE__ );

As for the query, I don't really see anything jumping out at me. The only suggestion I have right now is to always enclose field names in backticks, just in case they're keywords (it also makes them clearer to read)

至于查询，我真的看不到有什么东西跳出来。我现在唯一的建议是，总是在backticks中包含字段名，以防它们是关键字(它也使阅读变得更清晰)

Also, your else if is redundant. If $status === 1 doesn't run, then clearly $status !== 1 must be true.

另外，如果是多余的。如果$status == 1不运行，那么显然$status !== 1必须为真。

#4

Because of type casting, status=1 is not a problem. I'm assuming $id has some probrem. Once change $id to other safe value (1, 'foo'...) then check it works or not.

由于类型转换，状态=1不是问题。假设$id有一些probrem。一旦将$id更改为其他安全值(1，'foo'…)，检查它是否有效。

#5

-1

The 'line 1' part is SQL saying that hte message it recieved had an error on line 1 -- the first line of the command that SQL tried to process.

“第1行”部分是SQL，说它收到的hte消息在第1行有一个错误——SQL试图处理的命令的第一行。

If I had to make a guess, status isn't set to a number type, so you need to put quotes around it so that SQL knows it's being passed a variable.

如果我必须猜测，状态不是设置为数字类型，所以你需要在它周围加上引号，这样SQL就知道它被传递了一个变量。

Edit: OK, the other solution might be right too. We both made different assumptions about your data structure, and I think his is better. Try it first.

编辑:好的，另一个解决方案也可能是正确的。我们对你的数据结构做了不同的假设，我认为他的更好。先试一试。

#1