用mysqli编写PHP程序代码

问题描述：

I am less familiar with mysqli, and I write Procedure style, not Object orientated style. I know of some of the benefits of object orientated style but it trips my mind so I am slow to adjust with it. My policy is, if I cannot support it, I don't write it.

So... I have a simple function which contains the following:

mysqli_query( $DB, "CREATE TABLE $TABLE ( $SCHEMHA )" );

if (mysqli_connect_errno()) {
    printf("Connect failed: %s
", mysqli_connect_error());
    exit();
}

The query works when creating the table, so long as I don't have a bug in my schema. If my schema fails, my conditional statement does not trap the failure (thus the connect failed message never gets called).

Where am I going wrong?

My goal is to be able to exit/stop/end on selective errors. So for example, if a table already exists (errno 1050), I am comfortable ignoring that error, however if there is an error 1072 (Key column 'whatever' doesn't exist in table) I would like to stop processing.

All help appreciated...

我对mysqli不熟悉，而且我编写过程样式，而不是面向对象的样式。我知道面向对象风格的一些好处，但它让我心烦意乱，所以我很难调整它。我的政策是，如果我不能支持它，我就不会写它。 p>

所以...我有一个简单的函数，其中包含以下内容： p>

  mysqli_query（$ DB，“CREATE TABLE $ TABLE（$ SCHEMHA）”）; 
 
if（mysqli_connect_errno（））{
 printf（“连接失败：％s 
”，mysqli_connect_error（）  ）; 
 exit（）; 
} 
  code>  pre> 
 
 只要我的架构中没有错误，查询就会在创建表时起作用。 如果我的模式失败，我的条件语句不会捕获失败（因此连接失败的消息永远不会被调用）。 p> 
 
 
我哪里出错？ p> 
 
 我的目标是能够退出/停止/结束选择性错误。 因此，例如，如果一个表已经存在（错误1050），我很乐意忽略该错误，但是如果存在错误1072（表中不存在关键列'无论'）我想停止处理。  p> 
 
 
所有帮助表示赞赏......  p> 
  div>

答

You'll want to examine the return value of mysqli_query to know whether the query succeeded or failed.
You can get the exact error code using mysqli_errno.
The _connect_ functions you're using are irrelevant in this context, they give you information about errors which happened during a connection attempt.

if (!mysqli_query(..)) {
    switch (mysqli_errno($DB)) {
        case 1072:
            ...
    }
}

答

I believe something like this would do:

$result = mysqli_query( $DB, "CREATE TABLE $TABLE ( $SCHEMHA )" );

if (!$result) {
    switch(mysqli_connect_error()) {
        // Key column 'whatever' doesn't exist in table
        case 1072:
            printf("Connect failed: %s
", mysqli_connect_error());
            exit();
            break;
        // and so on... 
    }
}

答

The mysqli_connect_errno returns the error from the connection. Your error is resulting in the query though so use mysqli_error.

if (mysqli_error($DB)) {
    printf("Errormessage: %s
", mysqli_error($link));
}

If you just want to check the error code use http://php.net/manual/en/mysqli.errno.php.

printf("Errorcode: %d
", mysqli_errno($DB));

To check for specific codes you could do:

if(in_array(mysqli_errno($DB), array(1072, other codes)) {
    exit('Bad Code');
} else {
    echo 'Still processing...';
}

答

So for example, if a table already exists (errno 1050), I am comfortable ignoring that error,

CREATE TABLE IF NOT EXISTS $TABLE ...

用mysqli编写PHP程序代码

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