Perl 如何将postgres dbd占位符用于DB对象名称
(或者如何使用perl DBI(DBD::PG)和占位符迭代信息模式?) Windows7、ActiveState Perl 5.20.2、PostgreSQL 9.4.1 下面的案例A、B和C在使用占位符作为列值时成功。为了Perl 如何将postgres dbd占位符用于DB对象名称,perl,postgresql,Perl,Postgresql,(或者如何使用perl DBI(DBD::PG)和占位符迭代信息模式?) Windows7、ActiveState Perl 5.20.2、PostgreSQL 9.4.1 下面的案例A、B和C在使用占位符作为列值时成功。为了 没有使用占位符 传递一个文本 传递了一个变量(用相同的文本填充) 将其提升到DB对象级别将是非常好的。。(表格、视图等) 下面是案例D的错误输出: Z:\CTAM\data_threat_mapping\DB Stats\perl scripts>test_pl
- 没有使用占位符
- 传递一个文本
- 传递了一个变量(用相同的文本填充)
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.
多谢指点
#!/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
我试过单引号/双引号/无引号(q,qq).如果
SELECT * FROM Foo WHERE field = ?
意味着
SELECT * FROM Foo WHERE field = 'val'
然后
意味着
这显然是错误的。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();
如果
意味着
SELECT * FROM Foo WHERE field = 'val'
然后
意味着
这显然是错误的。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();