Question

（或如何使用perl DBI（DBD :: PG）和占位符迭代信息模式？）

Windows 7，ActiveState Perl 5.20.2，PostgreSQL 9.4.1。

使用占位符作为COLUMN VALUE时，下面的案例A，B和C成功。

没有使用占位符
传递了一个文字
传递了一个变量（填充了相同的文字）

将它提升到数据库对象......（表，视图等）

是很棒的

这里是案例D的错误输出：

Z:\CTAM\data_threat_mapping\DB Stats\perl scripts>test_placeholder.pl

A Row Count: 1
B Row Count: 1
C Row Count: 1

DBD::Pg::st execute failed: ERROR:  syntax error at or near "$1"

LINE 1: SELECT COUNT(*) FROM $1 WHERE status = 'Draft';
                             ^ at Z:\CTAM\data_threat_mapping\DB     Stats\perl 
scripts\test_placeholder.pl line 34.

对任何方向都有责任！

#!/usr/bin/perl -w
use strict;
use diagnostics;
use DBI;

my $num_rows = 0;

# connect
my $dbh = DBI->connect("DBI:Pg:dbname=CTAM;host=localhost",
                       "postgres", "xxxxx",
                       { 'RaiseError' => 1, pg_server_prepare => 1 });

#---------------------
# A - success
my $sthA = $dbh->prepare(
    "SELECT COUNT(*) FROM cwe_compound_element WHERE status = 'Draft';"
);
$sthA->execute(); # no placeholders

#---------------------
# B -  success
my $sthB = $dbh->prepare (
    "SELECT COUNT(*) FROM cwe_compound_element WHERE status = ?;"
);
$sthB->execute('Draft'); # pass 'Draft' to placeholder

#---------------------
# C -  success
my $status_value = 'Draft';
my $sthC = $dbh->prepare(
    "SELECT COUNT(*) FROM cwe_compound_element WHERE status = ?;"
);
$sthC->execute($status_value); # pass variable (column value) to placeholder

#---------------------
# D - failure
my $sthD = $dbh->prepare(
    "SELECT COUNT(*) FROM ? WHERE status = 'Draft';"
);
$sthD->execute('cwe_compound_element'); # pass tablename to placeholder

我尝试过单/双/无引号（q，qq）......

Answer 1

如果

SELECT * FROM Foo WHERE field = ?

表示

SELECT * FROM Foo WHERE field = 'val'

然后

SELECT * FROM ?

装置

SELECT * FROM 'Table'

这显然是错的。占位符只能在表达式中使用。修正：

my $sthD = $dbh->prepare("
   SELECT COUNT(*)
    FROM ".$dbh->quote_identifier($table)."
   WHERE status = 'Draft'
");
$sthD->execute();

如何使用postgres dbd占位符作为DB对象名称

1 个答案: