I would like to know is there any way to set Header with 60 char in SALV? Since there is a CHAR limit of 40 when I try using SET_LONG_TEXT method. I am using this SALV to both display and download it as an excel too.
My Prod Code looks like this below.
TRY.
" Call SALV Factory Method
cl_salv_table=>factory(
EXPORTING
list_display = abap_false
IMPORTING
r_salv_table = DATA(lo_salv)
CHANGING
t_table = <fs_global_factor_data> ).
DATA(lo_columns) = lo_salv->get_columns( ).
lo_columns->set_optimize( ).
DATA(lo_col_ref) = lo_columns->get( ).
" Get the Field Description in the table
lr_strucdesc ?= cl_abap_typedescr=>describe_by_data( ls_struct ).
/stl/cl_gf_validations=>at_gf_table_struct = lr_strucdesc->get_ddic_field_list( ).
Here, I am replacing the long text with my own texts with a length of 60 Char from the AT_FIELD_HEADER table.
" Remove Short and Medium texts and Replace long texts from AT_FIELD_HEADER Table
LOOP AT lo_col_ref ASSIGNING FIELD-SYMBOL(<fs_col_ref>).
<fs_col_ref>-r_column->set_short_text( space ).
<fs_col_ref>-r_column->set_medium_text( space ).
<fs_col_ref>-r_column->set_fixed_header_text( lc_char_l ) .
<fs_col_ref>-r_column->set_long_text( SWITCH #( <fs_col_ref>-columnname+0(5) WHEN /stl/cl_gf_validations=>ac_desc_field_prefix THEN /stl/cl_gf_validations=>ac_fd_description
ELSE CONV #( LET line = VALUE /stl/cl_gf_validations=>ty_field_header( /stl/cl_gf_validations=>at_field_header[ fieldname = <fs_col_ref>-columnname ] OPTIONAL )
IN COND #( WHEN line-title IS NOT INITIAL THEN line-title ELSE line-description )
) ) ).
ENDLOOP.
DATA(lv_xml_type) = if_salv_bs_xml=>c_type_xlsx.
DATA(lv_data_in_xstring) = lo_salv->to_xml( xml_type = lv_xml_type ).
CATCH cx_root.
MESSAGE e033(/stl/bw_gf_msg).
ENDTRY.
" Convert XSTRING TO BIN
CALL FUNCTION 'SCMS_XSTRING_TO_BINARY'
EXPORTING
buffer = lv_data_in_xstring
IMPORTING
output_length = lv_file_size
TABLES
binary_tab = lt_data_in_bin.
Here, GUI_DOWNLOAD is used to download it as an excel file.
" Download the Excel
cl_gui_frontend_services=>gui_download(
EXPORTING
bin_filesize = lv_file_size
filename = lv_fullpath
filetype = lc_filetype_bin
CHANGING
data_tab = lt_data_in_bin
EXCEPTIONS
file_write_error = 1
no_batch = 2
gui_refuse_filetransfer = 3
invalid_type = 4
no_authority = 5
unknown_error = 6
header_not_allowed = 7
separator_not_allowed = 8
filesize_not_allowed = 9
header_too_long = 10
dp_error_create = 11
dp_error_send = 12
dp_error_write = 13
unknown_dp_error = 14
access_denied = 15
dp_out_of_memory = 16
disk_full = 17
dp_timeout = 18
file_not_found = 19
dataprovider_exception = 20
control_flush_error = 21
not_supported_by_gui = 22
error_no_gui = 23
OTHERS = 24
).
IF sy-subrc <> 0.
MESSAGE ID sy-msgid TYPE sy-msgty NUMBER sy-msgno
WITH sy-msgv1 sy-msgv2 sy-msgv3 sy-msgv4.
ENDIF.
Also, I am using the same piece of code to display the SALV where header gets truncated to 40 char.
Making a game in GameMaker and I'm having a problem where the variable is not set and I'm not sure why. When pressing down on either W,A,S, or D and clicking LMB it works but crashes if I pressing down on W,A,S, or D then let go and afterwards click LMB.
Code:
var look
var bullet
if (keyboard_check(ord("A"))){
x = x - 5;
sprite_index = spr_west_hiro
image_xscale = 3
image_yscale = 3
look = 180
} else if (keyboard_check(ord("D"))){
x = x + 5;
sprite_index = spr_east_hiro
image_xscale = 3
image_yscale = 3
look = 0
} else if (keyboard_check(ord("S"))){
y = y + 5;
sprite_index = spr_south_hiro
image_xscale = 3
image_yscale = 3
look = 270
} else if (keyboard_check(ord("W"))){
y = y - 5;
sprite_index = spr_north_hiro
image_xscale = 3
image_yscale = 3
look = 90
}
//------------------------------------------------------------------------------
if (keyboard_check_released(ord("A"))){
sprite_index = spr_idlewest_hiro
image_xscale = 3
image_yscale = 3
look = 180
} else if (keyboard_check_released(ord("D"))){
sprite_index = spr_idleast_hiro
image_xscale = 3
image_yscale = 3
look = 0
} else if (keyboard_check_released(ord("S"))){
sprite_index = spr_idlesouth_hiro
image_xscale = 3
image_yscale = 3
look = 270
} else if (keyboard_check_released(ord("W"))){
sprite_index = spr_idlenorth_hiro
image_xscale = 3
image_yscale = 3
look = 90
}
//--------------------------------------------------------------------------------
if (mouse_check_button_pressed(mb_left)){
var bullet = instance_create_layer(x,y, "Instances", obj_bullet)
bullet.direction = look
}
Error:
ERROR!!! :: ############################################################################################
FATAL ERROR in
action number 1
of Step Event0
for object obj_hiro:
local variable look(100001, -2147483648) not set before reading it.
at gml_Object_obj_hiro_Step_0 (line 61) - bullet.direction = look
############################################################################################
--------------------------------------------------------------------------------------------
stack frame is
gml_Object_obj_hiro_Step_0 (line 61)
I have reviewed the code multiple times and I am still stumped. Particularly because of the fact that it seems as though the variable doesn't save the coefficient set to it despite the fact that it should when W,A,S or D is pressed down then released.
Firstly, you would want to assign a value to your look variable, as otherwise it will remain not set to anything if no buttons had been pressed.
Secondly, you may want to do so in Create event, as you probably want your character to shoot in last pressed direction, not just "somewhere"
I'm processing through Telegram history (txt file) reading it line by line when I when there is a pattern
Free_Trade_Calls__AltSignals:* / * (*)
example
Free_Trade_Calls__AltSignals:GAS / BTC (binance)
then store "following lines" in a separate variables each (and also store that that traiding pair RDN/ BTC and exchange bittrex) in separate variables) until it hits timestamp (today's date), then search for the pattern again
but when I see pattern
Free_Trade_Calls__AltSignals:TARGET
proceed to next line
whole example
Free_Trade_Calls__AltSignals:IOC/ BTC (bittrex)
BUY : 0.00164
SELL :
TARGET 1 : 0.00180
TARGET 2 : 0.00205
TARGET 3 : 0.00240
STOP LOS : 0.000120
2018-04-19 15:46:57 Free_Trade_Calls__AltSignals:TARGET
I've got some incomplete pseudocode but it is mostly BS anyway - quite new to parsing text aynway here is
let now = new Date();
// var diff = (now.getTime() - date.getTime()) / 60000; // diff in min
const fs = require('fs'),
path = require('path'),
filePath = path.join(__dirname, 'chats.txt'),
// myString = "Free_Trade_Calls__AltSignals";
buy,
sell = [3],
sl;
let lines = [];
const buf = fs.readFileSync(filePath)
lines [] = buf.toString().split('\n'){
for (line < lines.length){
if (!err) {
if(line.indexOf("Trade_Calls__AltSignals: * / * (*)") > -1) {
currency = line.indexOf(2);
exchange = line.indexOf(4);
nextline();
while (line.indexOf($new.Date) < - 1){ // no containing Date
if (line.indexOf(buy) > -1 ){
line = buy
}
if (line.indexOf(buy) > -1 ){
}
}else{
nextline();
}
}
}
}
}
I have a string S which consists of a's and b's. Perform the below operation once. Objective is to obtain the lexicographically smallest string.
Operation: Reverse exactly one substring of S
e.g.
if S = abab then Output = aabb (reverse ba of string S)
if S = abba then Output = aabb (reverse bba of string S)
My approach
Case 1: If all characters of the input string are same then output will be the string itself.
Case 2: if S is of the form aaaaaaa....bbbbbb.... then answer will be S itself.
otherwise: Find the first occurence of b in S say the position is i. String S will look like
aa...bbb...aaaa...bbbb....aaaa....bbbb....aaaaa...
|
i
In order to obtain the lexicographically smallest string the substring that will be reversed starts from index i. See below for possible ending j.
aa...bbb...aaaa...bbbb....aaaa....bbbb....aaaaa...
| | | |
i j j j
Reverse substring S[i:j] for every j and find the smallest string.
The complexity of the algorithm will be O(|S|*|S|) where |S| is the length of the string.
Is there a better way to solve this problem? Probably O(|S|) solution.
What I am thinking if we can pick the correct j in linear time then we are done. We will pick that j where number of a's is maximum. If there is one maximum then we solved the problem but what if it's not the case? I have tried a lot. Please help.
So, I came up with an algorithm, that seems to be more efficient that O(|S|^2), but I'm not quite sure of it's complexity. Here's a rough outline:
Strip of the leading a's, storing in variable start.
Group the rest of the string into letter chunks.
Find the indices of the groups with the longest sequences of a's.
If only one index remains, proceed to 10.
Filter these indices so that the length of the [first] group of b's after reversal is at a minimum.
If only one index remains, proceed to 10.
Filter these indices so that the length of the [first] group of a's (not including the leading a's) after reversal is at a minimum.
If only one index remains, proceed to 10.
Go back to 5, except inspect the [second/third/...] groups of a's and b's this time.
Return start, plus the reversed groups up to index, plus the remaining groups.
Since any substring that is being reversed begins with a b and ends in an a, no two hypothesized reversals are palindromes and thus two reversals will not result in the same output, guaranteeing that there is a unique optimal solution and that the algorithm will terminate.
My intuition says this approach of probably O(log(|S|)*|S|), but I'm not too sure. An example implementation (not a very good one albeit) in Python is provided below.
from itertools import groupby
def get_next_bs(i, groups, off):
d = 1 + 2*off
before_bs = len(groups[i-d]) if i >= d else 0
after_bs = len(groups[i+d]) if i <= d and len(groups) > i + d else 0
return before_bs + after_bs
def get_next_as(i, groups, off):
d = 2*(off + 1)
return len(groups[d+1]) if i < d else len(groups[i-d])
def maximal_reversal(s):
# example input: 'aabaababbaababbaabbbaa'
first_b = s.find('b')
start, rest = s[:first_b], s[first_b:]
# 'aa', 'baababbaababbaabbbaa'
groups = [''.join(g) for _, g in groupby(rest)]
# ['b', 'aa', 'b', 'a', 'bb', 'aa', 'b', 'a', 'bb', 'aa', 'bbb', 'aa']
try:
max_length = max(len(g) for g in groups if g[0] == 'a')
except ValueError:
return s # no a's after the start, no reversal needed
indices = [i for i, g in enumerate(groups) if g[0] == 'a' and len(g) == max_length]
# [1, 5, 9, 11]
off = 0
while len(indices) > 1:
min_bs = min(get_next_bs(i, groups, off) for i in indices)
indices = [i for i in indices if get_next_bs(i, groups, off) == min_bs]
# off 0: [1, 5, 9], off 1: [5, 9], off 2: [9]
if len(indices) == 1:
break
max_as = max(get_next_as(i, groups, off) for i in indices)
indices = [i for i in indices if get_next_as(i, groups, off) == max_as]
# off 0: [1, 5, 9], off 1: [5, 9]
off += 1
i = indices[0]
groups[:i+1] = groups[:i+1][::-1]
return start + ''.join(groups)
# 'aaaabbabaabbabaabbbbaa'
TL;DR: Here's an algorithm that only iterates over the string once (with O(|S|)-ish complexity for limited string lengths). The example with which I explain it below is a bit long-winded, but the algorithm is really quite simple:
Iterate over the string, and update its value interpreted as a reverse (lsb-to-msb) binary number.
If you find the last zero of a sequence of zeros that is longer than the current maximum, store the current position, and the current reverse value. From then on, also update this value, interpreting the rest of the string as a forward (msb-to-lsb) binary number.
If you find the last zero of a sequence of zeros that is as long as the current maximum, compare the current reverse value with the current value of the stored end-point; if it is smaller, replace the end-point with the current position.
So you're basically comparing the value of the string if it were reversed up to the current point, with the value of the string if it were only reversed up to a (so-far) optimal point, and updating this optimal point on-the-fly.
Here's a quick code example; it could undoubtedly be coded more elegantly:
function reverseSubsequence(str) {
var reverse = 0, max = 0, first, last, value, len = 0, unit = 1;
for (var pos = 0; pos < str.length; pos++) {
var digit = str.charCodeAt(pos) - 97; // read next digit
if (digit == 0) {
if (first == undefined) continue; // skip leading zeros
if (++len > max || len == max && reverse < value) { // better endpoint found
max = len;
last = pos;
value = reverse;
}
} else {
if (first == undefined) first = pos; // end of leading zeros
len = 0;
}
reverse += unit * digit; // update reverse value
unit <<= 1;
value = value * 2 + digit; // update endpoint value
}
return {from: first || 0, to: last || 0};
}
var result = reverseSubsequence("aaabbaabaaabbabaaabaaab");
document.write(result.from + "→" + result.to);
(The code could be simplified by comparing reverse and value whenever a zero is found, and not just when the end of a maximally long sequence of zeros is encountered.)
You can create an algorithm that only iterates over the input once, and can process an incoming stream of unknown length, by keeping track of two values: the value of the whole string interpreted as a reverse (lsb-to-msb) binary number, and the value of the string with one part reversed. Whenever the reverse value goes below the value of the stored best end-point, a better end-point has been found.
Consider this string as an example:
aaabbaabaaabbabaaabaaab
or, written with zeros and ones for simplicity:
00011001000110100010001
We iterate over the leading zeros until we find the first one:
0001
^
This is the start of the sequence we'll want to reverse. We will start interpreting the stream of zeros and ones as a reversed (lsb-to-msb) binary number and update this number after every step:
reverse = 1, unit = 1
Then at every step, we double the unit and update the reverse number:
0001 reverse = 1
00011 unit = 2; reverse = 1 + 1 * 2 = 3
000110 unit = 4; reverse = 3 + 0 * 4 = 3
0001100 unit = 8; reverse = 3 + 0 * 8 = 3
At this point we find a one, and the sequence of zeros comes to an end. It contains 2 zeros, which is currently the maximum, so we store the current position as a possible end-point, and also store the current reverse value:
endpoint = {position = 6, value = 3}
Then we go on iterating over the string, but at every step, we update the value of the possible endpoint, but now as a normal (msb-to-lsb) binary number:
00011001 unit = 16; reverse = 3 + 1 * 16 = 19
endpoint.value *= 2 + 1 = 7
000110010 unit = 32; reverse = 19 + 0 * 32 = 19
endpoint.value *= 2 + 0 = 14
0001100100 unit = 64; reverse = 19 + 0 * 64 = 19
endpoint.value *= 2 + 0 = 28
00011001000 unit = 128; reverse = 19 + 0 * 128 = 19
endpoint.value *= 2 + 0 = 56
At this point we find that we have a sequence of 3 zeros, which is longer that the current maximum of 2, so we throw away the end-point we had so far and replace it with the current position and reverse value:
endpoint = {position = 10, value = 19}
And then we go on iterating over the string:
000110010001 unit = 256; reverse = 19 + 1 * 256 = 275
endpoint.value *= 2 + 1 = 39
0001100100011 unit = 512; reverse = 275 + 1 * 512 = 778
endpoint.value *= 2 + 1 = 79
00011001000110 unit = 1024; reverse = 778 + 0 * 1024 = 778
endpoint.value *= 2 + 0 = 158
000110010001101 unit = 2048; reverse = 778 + 1 * 2048 = 2826
endpoint.value *= 2 + 1 = 317
0001100100011010 unit = 4096; reverse = 2826 + 0 * 4096 = 2826
endpoint.value *= 2 + 0 = 634
00011001000110100 unit = 8192; reverse = 2826 + 0 * 8192 = 2826
endpoint.value *= 2 + 0 = 1268
000110010001101000 unit = 16384; reverse = 2826 + 0 * 16384 = 2826
endpoint.value *= 2 + 0 = 2536
Here we find that we have another sequence with 3 zeros, so we compare the current reverse value with the end-point's value, and find that the stored endpoint has a lower value:
endpoint.value = 2536 < reverse = 2826
so we keep the end-point set to position 10 and we go on iterating over the string:
0001100100011010001 unit = 32768; reverse = 2826 + 1 * 32768 = 35594
endpoint.value *= 2 + 1 = 5073
00011001000110100010 unit = 65536; reverse = 35594 + 0 * 65536 = 35594
endpoint.value *= 2 + 0 = 10146
000110010001101000100 unit = 131072; reverse = 35594 + 0 * 131072 = 35594
endpoint.value *= 2 + 0 = 20292
0001100100011010001000 unit = 262144; reverse = 35594 + 0 * 262144 = 35594
endpoint.value *= 2 + 0 = 40584
And we find another sequence of 3 zeros, so we compare this position to the stored end-point:
endpoint.value = 40584 > reverse = 35594
and we find it has a smaller value, so we replace the possible end-point with the current position:
endpoint = {position = 21, value = 35594}
And then we iterate over the final digit:
00011001000110100010001 unit = 524288; reverse = 35594 + 1 * 524288 = 559882
endpoint.value *= 2 + 1 = 71189
So at the end we find that position 21 gives us the lowest value, so it is the optimal solution:
00011001000110100010001 -> 00000010001011000100111
^ ^
start = 3 end = 21
Here's a C++ version that uses a vector of bool instead of integers. It can parse strings longer than 64 characters, but the complexity is probably quadratic.
#include <vector>
struct range {unsigned int first; unsigned int last;};
range lexiLeastRev(std::string const &str) {
unsigned int len = str.length(), first = 0, last = 0, run = 0, max_run = 0;
std::vector<bool> forward(0), reverse(0);
bool leading_zeros = true;
for (unsigned int pos = 0; pos < len; pos++) {
bool digit = str[pos] - 'a';
if (!digit) {
if (leading_zeros) continue;
if (++run > max_run || run == max_run && reverse < forward) {
max_run = run;
last = pos;
forward = reverse;
}
}
else {
if (leading_zeros) {
leading_zeros = false;
first = pos;
}
run = 0;
}
forward.push_back(digit);
reverse.insert(reverse.begin(), digit);
}
return range {first, last};
}
I have a categorized view, the first column is categorized and ascending.The last column calculates the total. Imagine the view name is called view1 and it looks like the following:
1st column(categorized) | 2nd column | 3rd column (total)
Australia | | 2
| Alex | 1
| Bill | 1
Brazil | | 3
| Alan | 1
| Alice | 1
| Tom | 1
Chile | | 1
| John | 1
Denmark | | 2
| Susan | 1
| Tim | 1
| | 8
I would like to retrieve the total of each categorized column, for example, the total for Australia is 2, the total for Brazil is 3. I browse on the internet and write some code but the result is not success.
Here is my first attempt: I intend to use #DbLookup, for loop and #Count.
var value = sessionScope.Value;
var lookupValue = (#DbLookup(#DbName(),"view1",value,1));
for(var a = 0; a < lookupValue.length; a++)
{
//try to display the lookupValue and the relevant total
writer.write("<tr><td>"+lookupValue[a]+"<td> "+ #Count(lookupValue[a]) + "</td></tr>");
}
When I run the code, I get the following result:
Australia 1
Australia 1
Brazil 1
Brazil 1
Brazil 1
Chile 1
Denmark 1
Denmark 1
After the first attempt, I start the second attempt: (please note, most of the code in this part I get from the internet and I just modify a little bit to suit my case)
var myView:NotesView = database.getView('view1');
var vec:NotesViewEntryCollection = myView.getAllEntries();
var viewEnt:NotesViewEntry = vec.getFirstEntry();
writer.write("<table>");
while (viewEnt != null)
{
// try to retrieve the total in the 3rd column
writer.write("<tr><td> "+ 'viewEnt.getColumnValues()[2]' +"</td></tr>");
var tmp = vec.getNextEntry(viewEnt);
viewEnt.recycle();
viewEnt = tmp;
}
writer.write("</table>");
writer.endDocument();
I run the code and the result is empty. I do not get any total of the categorized column.
I review the code from attempt 1 and attempt 2, I don't understand why I cannot get the total.
I search on the internet again and I found this similar post: Getting a summary count in a view export?
I try to apply the solution in the application:(please note, most of the code in this part I get from that post and I just add the view name in the function)
function getCategoryData(viewName, dataColumn, getChildren) {
var v = database.getView(view1); // change viewName to view1
var nav = v.createViewNav();
var ve = nav.getFirst();
var isFirst = true;
var result = "[";
while (ve) {
if (!ve.isTotal()) {
var curData = ve.getColumnValues();
if (!isFirst) {
result += ",";
}
result += "{label : \"";
result += curData[0];
result += "\", value : ";
result += curData[dataColumn];
/* for 2 level categories we fetch additional data */
if (getChildren) {
var childve = nav.getChild();
var firstChild = true;
result += ", children : [";
while (childve) {
var childData = childve.getColumnValues();
if (!firstChild) {
result += ",";
}
result += "{label : \"";
result += childData[1];
result += "\", value : ";
result += childData[dataColumn];
result += "}";
firstChild = false;
childve = nav.getNextSibling(childve);
}
result += "]"
}
result += "}";
isFirst = false;
}
ve = nav.getNextSibling(ve);
}
result += "]";
return result;
}
I run the code and I get the result like this:
[{label : "Australia", value : undefined},{label : "Brazil", value : undefined},{label : "Chile", value : undefined},{label : "Denmark", value : undefined}]
After the third attempt, I think the reason that I cannot get the total is I misunderstand the concept about how to get the total in the view.
In fact, what I would like to do is get the total of each categorized column in view1. Therefore, imagine the result will look like this:
Australia 2
Brazil 3
Chile 1
Denmark 2
Actually, I think the first attempt is very close to the result. I feel I missed something in the code (perhaps I guess in the for loop part).
How can I get the total of each categorized column in view?
Grateful for your advice on this issue please. Thank you very much.
You are on the right track with your third attempt. You definitely want to use NotesViewNavigator.
Here is a working example based on your requirements:
/*
* #param viewName Name of the view to navigate
* #param labelColumn Number of the column in the view containing the label
* Note: column count start at 0
* #param dataColumn Number of the column in the view containing
* totals. Note: column count starts at 0
* #return String representing a javascript array of objects containing label and value
*/
function getCategoryData(viewName, labelColumn, dataColumn) {
var v:NotesView = database.getView(viewName);
var nav:NotesViewNavigator = v.createViewNav();
var ve:NotesViewEntry = nav.getFirst(); // first category
var isFirst = true;
var result = "[";
while (ve) {
var curData = ve.getColumnValues();
if (curData[labelColumn] != "") { // skip the total at the bottom of the totals column
if (!isFirst) {
result += ",";
}
result += "{label : \"";
result += curData[labelColumn];
result += "\", value : ";
result += curData[dataColumn];
result += "}";
}
isFirst = false;
var tmpentry:NotesViewEntry = nav.getNextCategory();
ve.recycle();
ve = tmpentry;
}
result += "]";
nav.recycle();
v.recycle();
return result;
}
Based on the description of your view and assuming the name of your view is "countries" you would call this function like this:
getCategoryData("countries", 0, 2);
Note: I did not test this function with multiple categorized columns.