我试图表名传递给获取该表的所有字段名,并将其存储到一个数组,然后子使用该阵列中的与另一个SQL查询的fetchrow一起显示在这些数据领域。下面的代码我现在有:搜索结果

使用表名作为参数子呼叫的例子:

shamoo("reqhead_rec");
shamoo("approv_rec");
shamoo("denial_rec");

shamoo子:

sub shamoo
{
    my $table = shift;
    print uc($table)."\n=====================================\n";

    #takes arg (table name) and stores all the field names into an array
    $STMT = <<EOF;
    select first 1 * from $table
    EOF

    my $sth = $db1->prepare($STMT);$sth->execute;

    my ($i, @field);
    my $columns = $sth->{NAME_lc};
    while (my $row = $sth->fetch){for $i (0 .. $#$row){$field[$i] = $columns->[$i];}}

    $STMT = <<EOF;
    select * from $table where frm = '$frm' and req_no = $req_no
    EOF
    $sth = $db1->prepare($STMT);$sth->execute;
    $i=0;
    while ($i!=scalar(@field))
    {
    #need code for in here...
    }
}

我要寻找一种方法来把这个n要的东西,不必明确定义.... 结果

my ($frm, $req_no, $auth_id, $alt_auth_id, $id_acct, $seq_no, $id, $appr_stat, $add_date, $approve_date, $approve_time, $prim);
while(($frm, $req_no, $auth_id, $alt_auth_id, $id_acct, $seq_no, $id, $appr_stat, $add_date, $approve_date, $approve_time, $prim) = $sth->fetchrow_array())
有帮助吗?

解决方案

使用fetchrow_hashref:

sub shamoo {
    my ($dbh, $frm, $req_no, $table) = @_;

    print uc($table), "\n", "=" x 36, "\n";

    #takes arg (table name) and stores all the field names into an array
    my $sth = $dbh->prepare(
        "select * from $table where frm = ? and req_no = ?"
    );

    $sth->execute($frm, $req_no);

    my $i = 1;
    while (my $row = $sth->fetchrow_hashref) {
        print "row ", $i++, "\n";
        for my $col (keys %$row) {
            print "\t$col is $row->{$col}\n";
        }
    }
}

您可能还需要在创建数据库手柄设置FetchHashKeyName"NAME_lc""NAME_uc"

my $dbh = DBI->connect(
    $dsn,
    $user,
    $pass,
    {
        ChopBlanks       => 1,
        AutoCommit       => 1,
        PrintError       => 0,
        RaiseError       => 1,
        FetchHashKeyName => "NAME_lc",
    }
) or die DBI->errstr;

其他提示

不知这种方法会为空表工作。

要获得列元数据的最安全的方法是不看返回hashref的键(这可能不存在),但按规则宁愿玩,并使用DBI提供的$属性某物本身:

$sth->{NAME}->[i]
$sth->{NAME_uc}->[i]
$sth->{NAME_lc}->[i]

请参阅所述DBI手册页的元数据部分的详细说明。

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