php - mysql_field_table() 的错误结果

Question

我认为 mysql_field_table 有问题

我的查询:

 SELECT partners.buyer,
        calls_nc_sell_daily.market,
        partners.name,
        calls_nc_sell_daily.partner_id_fk as 'partners_id_fk',
        partners.seller
   FROM partners, calls_nc_sell_daily
  WHERE calls_nc_sell_daily.date >= '$date_start'
    AND calls_nc_sell_daily.date <= '$date_stop'
    AND partners.id = calls_nc_sell_daily.partner_id_fk

WHEN 结果字段由以下 PHP 片段解析:

$tables = array();
for($i=0;$i<$elements;$i++) {
   $name =  mysql_field_name($result,$i);
   $tables[$name] = mysql_field_table($result,$name);
}
print_r($tables);

预期:

Array
(
    [buyer] => partners
    [market] => calls_nc_sell_daily
    [name] => partners
    [partners_id_fk] => calls_nc_sell_daily
    [seller] => partners
)

实际:

Array
(
    [buyer] => partners
    [market] => partners
    [name] => partners
    [partners_id_fk] => partners
    [seller] => partners
)

Answer 1

mysql_field_table()根据 PHP 文档，将字段 offset 而不是其名称作为第二个参数。因此，您的姓名将被转换为数字，这会导致所有迭代的结果为零。

$tables = array();
for($i=0;$i<$elements;$i++) {
   $name =  mysql_field_name($result,$i);
   $tables[$name] = mysql_field_table($result, $i);
}
print_r($tables);

string mysql_field_table ( resource $result , int $field_offset )

field_offset

The numerical field offset. The field_offset starts at 0. If field_offset does not exist, an error of level E_WARNING is also issued.