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我有生成UID代码:PHP从MS SQL读取二进制UID值作为十六进制
$time_low = str_pad(dechex(mt_rand(0, 65535)), 4, '0', STR_PAD_LEFT) . str_pad(dechex(mt_rand(0, 65535)), 4, '0', STR_PAD_LEFT);
$time_mid = str_pad(dechex(mt_rand(0, 65535)), 4, '0', STR_PAD_LEFT);
$time_high_and_version = mt_rand(0, 255);
$time_high_and_version = $time_high_and_version & hexdec('0f');
$time_high_and_version = $time_high_and_version^hexdec('40'); // Sets the version number to 4 in the high byte
$time_high_and_version = str_pad(dechex($time_high_and_version), 2, '0', STR_PAD_LEFT);
$clock_seq_hi_and_reserved = mt_rand(0, 255);
$clock_seq_hi_and_reserved = $clock_seq_hi_and_reserved & hexdec('3f');
$clock_seq_hi_and_reserved = $clock_seq_hi_and_reserved^hexdec('80'); // Sets the variant for this GUID type to '10x'
$clock_seq_hi_and_reserved = str_pad(dechex($clock_seq_hi_and_reserved), 2, '0', STR_PAD_LEFT);
$clock_seq_low = str_pad(dechex(mt_rand(0, 65535)), 4, '0', STR_PAD_LEFT);
$node = str_pad(dechex(mt_rand(0, 65535)), 4, '0', STR_PAD_LEFT) . str_pad(dechex(mt_rand(0, 65535)), 4, '0', STR_PAD_LEFT) . str_pad(dechex(mt_rand(0, 65535)), 4, '0', STR_PAD_LEFT);
$guid = $time_low . '-' . $time_mid . '-' . $time_high_and_version . $clock_seq_hi_and_reserved . '-' . $clock_seq_low . '-' . $node;
它生成类似的字符串:011FFF33-CA4A-44E8-8CD5-7344D8E94344。当我从MS SQL 2008数据库中读取它时,我得到的二进制字符串如下:3ÿJÊèDŒ'sDØéCD。我如何读取它作为十六进制字符串而不是二进制字符串?谢谢!
将其转换为select调用中的字符串 - 存储的uid是原始位,而不是您看到的很好的十六进制字符串。 mssql驱动程序可能会将这些原始位返回,而不是转换为可读的十六进制字符串本身,因此请在查询中进行。 – 2012-08-13 03:35:46
非常感谢,解决了我的问题!我做了“CAST(Id as varchar(20))”请更改您的评论以回答,我会接受它。 – Kristian 2012-08-13 09:22:08