错误报告?从PHP导入.sql文件并显示错误。

I am building a way of importing .SQL files into a MySQL database from PHP. This is used for executing batches of queries. The issue I am having is error reporting.

我正在构建一种从PHP导入. sql文件到MySQL数据库的方法。这用于执行多批查询。我遇到的问题是错误报告。

$command = "mysql -u $dbuser --password='$dbpassword' --host='$sqlhost' $dbname < $file";
exec($command, $output);

This is essentially how I am importing my .sql file into my database. The issue is that I have no way of knowing if any errors occurred within the PHP script executing this command. Successful imports are entirely indistinguishable from a failure.

这本质上是我将.sql文件导入数据库的方式。问题是，我无法知道在执行此命令的PHP脚本中是否发生了任何错误。成功的进口与失败是完全不可区分的。

I have tried:

我有尝试:

Using PHP's sql error reporting functions.
使用PHP的sql错误报告函数。
Adding the verbose argument to the command and examining the output. It simply returns the contents of the .sql file and that is all.
向命令添加详细参数并检查输出。它只返回.sql文件的内容，仅此而已。
Setting errors to a user variable within the .sql file and querying it from the PHP script.
将错误设置为.sql文件中的用户变量，并从PHP脚本查询它。

I hope I am not forced to write the errors into a temporary table. Is there a better way?

我希望我不会*将错误写入临时表中。有更好的方法吗?

UPDATE: If possible, it would be very preferable if I could determine WHAT errors occurred, not simply IF one occurred.

更新:如果可能的话，如果我可以确定发生了什么错误，而不仅仅是发生了错误，那就更好了。

4 个解决方案

#1

Try using shell_exec

尝试使用shell_exec

$output = shell_exec( "mysql -u $dbuser --password='$dbpassword' --host='$sqlhost' $dbname < $file" );
// parse $output here for errors

From the manual:

从手册:

shell_exec — Execute command via shell and return the complete output as a string

shell_exec -通过shell执行命令，并以字符串形式返回完整的输出

Note:

注意:

This function is disabled when PHP is running in safe mode.

当PHP在安全模式下运行时，此函数将被禁用。

EDIT: Full solution:

编辑:完整的解决方案:

what you need to do is grab STDERR and discard STDOUT. Do this by adding '2>&1 1> /dev/null' to the end of your command.

您需要做的是抓取STDERR并丢弃STDOUT。通过在命令的末尾添加“2>&1 > /dev/null”来实现这一点。

$output = shell_exec( "mysql -u $dbuser --password='$dbpassword' --host='$sqlhost' $dbname < $file 2>&1 1> /dev/null" );
$lines = explode( PHP_EOL, $output );

$errors = array();

foreach( $lines as $line )
{
    if ( strtolower( substr( $line, 0, 5 ) ) == 'error' )
    {
        $errors[] = $line;
    }
}

if ( count( $errors ) )
{
    echo PHP_EOL . 'Errors occurred during import.';
    echo implode( PHP_EOL, $errors );
}
else
{
    echo 'No Errors' . PHP_EOL;
}

#2

$command = "mysql -u $dbuser --password='$dbpassword' --host='$sqlhost' $dbname"
  . " < $file 2>&1";
exec($command, $output);

The error message you're looking for is probably printed to stderr rather than stdout. 2>&1 causes stderr to be included in stdout, and as a result, also included in $output.

您正在查找的错误消息可能打印到stderr而不是stdout。2>和1使stderr包含在stdout中，因此，也包含在$output中。

Even better, use proc_open instead of exec, which gives you far more control over the process, including separate stdout and stderr pipes.

更好的是，使用proc_open而不是exec，这将使您对流程有更多的控制，包括单独的stdout和stderr管道。

#3

When issuing a exec, the shell will return a 0 on succes, or a number indicating a failure.

当发出exec时，shell将在succes上返回一个0，或者一个表示失败的数字。

$result = exec( $command, $output );

$result = exec($命令，$output);

should do the trick. Check result and handle appropiate.

应该足够了。检查结果并处理批准。

#4

-1

You have done everything but look at the PHP manual! There is an additional parameter for the exec to return a result

除了PHP手册之外，您已经完成了所有工作!exec还有一个额外的参数来返回结果

http://php.net/manual/en/function.exec.php

"If the return_var argument is present along with the output argument, then the return status of the executed command will be written to this variable."

“如果return_var参数与输出参数一起出现，那么被执行的命令的返回状态将被写到这个变量中。”

exec($command,$output,$result);
if ($result === 0) {
 // success
} else {
 // failure
}

#1