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我将如何做到（这只是示例）不返回错误？

$field_in = this, that

$result = mysql_query("SHOW COLUMNS FROM " . $table. " WHERE Field IN ( ".$field_in." )")
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$result = mysql_query( 
   "select *
    from INFORMATION_SCHEMA.COLUMNS c
    where c.TABLE_SCHEMA = '" . $table. "' 
    and c.COLUMN_NAME IN ( ".$field_in." )")
于 2012-06-29T14:03:54.313 回答