perl json is needed by mysql_为什么我无法通过Perl通过JSON::XS通过Perl正确编码来自PostgreSQL的布尔值？...

我在PostgreSQL系统上查询返回一个布尔值：

my $sth = $dbh->prepare("select 'f'::boolean");

$sth->execute;

my @vals = $sth->fetchrow_array;根据the DBD::Pg docs，

The current implementation of

PostgreSQL returns 't' for true and

'f' for false. From the Perl point of

view, this is a rather unfortunate

choice. DBD::Pg therefore translates

the result for the BOOL data type in a

Perlish manner: 'f' becomes the number

0 and 't' becomes the number 1. This

way the application does not have to

check the database-specific returned

values for the data-type BOOL because

Perl treats 0 as false and 1 as true.

You may set the pg_bool_tf attribute

to a true value to change the values

back to 't' and 'f' if you wish.

因此，只要pg_bool_tf返回0，那么该语句应该返回一个0，它会这样做。然而，在JSON :: XS(和纯JSON)方式的某处，将返回的0解释为一个字符串：

use JSON::XS qw(encode_json);

my $options =

{

layout => 0,

show_widget_help => $vals[0] // 1,

};

die encode_json($options);......与......一起死去

{ “布局”：0 “show_widget_help”： “0”}

...这将是好的，除了我的JavaScript期望在那里布尔值，并且非空字符串“0”被评估为真。为什么后者被引用而前者不是？

根据the JSON::XS docs，这是一个主要特点：

round-trip integrity

When you serialise a perl data

structure using only data types

supported by JSON, the deserialised

data structure is identical on the

Perl level. (e.g. the string "2.0"

doesn't suddenly become "2" just

because it looks like a number). There

minor are exceptions to this, read the

MAPPING section below to learn about

those.

说：

Simple Perl scalars (any scalar that

is not a reference) are the most

difficult objects to encode: JSON::XS

will encode undefined scalars as JSON

null values, scalars that have last

been used in a string context before

encoding as JSON strings, and anything

else as number value.

但是我从来没有在字符串上下文中使用@vals [0]。也许DBD :: Pg在返回之前将其布尔0作为字符串使用？