I'm trying to query contacts in Outlook that don't have a first name. I know they exist as can be seen by:
return (first name) of every contact whose (company = "SomeCompany")
returns: {missing value, "Francesca", missing value, missing value, missing value}
But I want only those where the first name is missing value. So I tried:
return (first name) of every contact whose (company = p_companyName and (first name is missing value))
But Applescript doesn't like that:missing value can't be converted to text" number -1700 from missing value to text
so my next attempt is:
return (first name) of every contact whose (company = p_companyName and (first name is missing value as text))
which runs fine, but doesn't return any records, which is incorrect. So what might be the correct way here?
Related
I have a query that must set a value if the number is between 2 values, but the output is not ok, I think because that column is a string. Any way to do it even it's string?
(In output I have value as , 5 witch is not ok).
All the values that are incorrect are Integer.
SET lkp_age_category_id = 7
WHERE
age BETWEEN '26' and '35.99';
I guess, only working on the SQL side, you could cast the values right into the query, e.g.:
SET lkp_age_category_id = 7
WHERE age BETWEEN '26'::float AND '35.99'::float;
Also check this answer https://stackoverflow.com/a/13809603/917617.
I have a TextBox in VB and a Command Button. I want to print the string upon clicking on command button.
I am using the following code, please tell what I am doing wrong:-
Dim name As String
name = Val(Text1.Text)
MsgBox ("Welcome" & Str(name))
When I input a string in Textbox and click on command button, result is:
Welcome 0
Leave out the val() around your Text1.Text, val() returns the numbers up to the first symbol it can't recognize as a number used in a String. See the documentation. I guess you used a 0 in your string in the TextField or no number at all, both would return 0.
Additionally, it is unnecessary to cast your String name to a String since it is already a String so you can also leave out the Str().
The val function returns the numeric representation of its argument, otherwise it returns "0". It's a bit hard these days to find the official VB6 documentation, but you may want to check: https://en.wikibooks.org/wiki/Visual_Basic/VB6_Command_Reference#Val
So, in your example, if you enter any number in the Text1 textbox control, you will see it in your message box. If you enter any text, you will get "Welcome 0", as you do now. Therefore, you have to remove the val function from your code, like:
Dim name As String
name = Text1.Text
MsgBox ("Welcome " & name)
maybe even simplifying it to:
MsgBox("Welcome " & Text1.Text)
So you declared a string varaible namewhich you want to fill with the text from the Text1box. So you need to spare the val(...)part.
Second, as namealready represents a string, leave out the strin the message box:
name = Text1.Text
MsgBox ("Welcome " & name)
I'm creating a function that displays a lot of variables with the format Variable + Variable Name.
Define LibPub out(list)=
Func
Local X
for x,1,dim(list)
list[x]->name // How can I get the variable name here?
Disp name+list[x]
EndFor
Return 1
EndFunc
Given a list value, there is no way to find its name.
Consider this example:
a:={1,2,3,4}
b:=a ; this stores {1,2,3,4} in b
out(b)
Line 1: First the value {1,2,3,4} is created. Then an variable with name a is created and its value is set to {1,2,3,4}.
Line 2: The expression a is evaluated; the result is {1,2,3,4}. A new variable with the name b is created and its value is set to `{1,2,3,4}.
Line 3: The expression b is evaluated. The variable reference looks up what value is stored in b. The result is {1,2,3,4}. This value is then passed to the function out.
The function out receives the value {1,2,3,4}. Given the value, there is no way of knowing whether the value happened to be stored in a variable. Here the value is stored in both a and b.
However we can also look at out({1,1,1,1}+{0,2,3,4}).
The system will evaluate {1,1,1,1}+{0,2,3,4} and get {1,2,3,4}. Then out is called. The value out received the result of an expression, but an equivalent value happens to be stored in a and b. This means that the values doesn't have a name.
In general: Variables have a name and a value. Values don't have names.
If you need to print a name, then look into strings.
This will be memory intensive, but you could keep a string of variable names, and separate each name by some number of characters and get a substring based on the index of the variable in the list that you want to get. For instance, say you want to access index zero, then you take a substring starting at (index of variable * length of variable name, indexofvariable *length + length+1).
The string will be like this: say you had the variables foo, bas, random, impetus
the string will be stored like so: "foo bas random impetus "
I am trying to hide a row in BIRT when a specific value equals a string. It works when the value equals something like 3 or (this_is_the_value) but it does not work when the value equals something like (this is the value) -> a value in the database with multiple words with spaces between them. How do I build the expression that hides the row when a value equals (this is the value) instead of (this_is_the_value)
Any help would be appreciated..
Michel
You have to enclose your value with quotes "this is the value" and use the double equals sign == to compare it to your row.
By the way, you can hide/show a visual text parameter in an expression like this:
!params[NAME_ID].value
Please consider the following XML--
<table class="rel_patent"><tbody>
<tr><td>Name</td><td>Description</td></tr>
<tr><td>A</td><td>Type-A</td></tr>
<tr><td>B</td><td>Type-B</td></tr>
<tr><td>C</td><td>Type-C</td></tr>
<tr><td>AC</td><td>Type-C
Type-A</td></tr>
<tr><td>D</td><td></td></tr>
</tbody></table>
Now I want to select and display all values of "Name" with corresp. values of "Description" element...even when Description element has null values viz element with name=D, and also, when description element has values separated by enter then I want those values (of Description) in separate rows- viz Type-C and Type-A for element with name=AC
This is the type of query I have written--
let $rows_data:= $doc//table[#class="rel_patent"]/tbody/tr[1]/following-sibling::tr
for $data_single_row in $rows_data
return
let $cited_name:= $data_single_row/td[1]
let $original_types_w_return:= $data_single_row/td[4]
let $original_types_list:= tokenize($original_types_w_return, '(\r?\n|\r)$')
for $cited_type_each at $pos2 in $original_types_list
return concat( $cited_name, '^', $original_type_each, '^', $pos2)
However, I am getting the following type of response--
A^Type-A^1
B^Type-B^1
C^Type-C^1
AC^Type-C
Type-A^1
Now, I need to get the following correct in the above code+response---
(1) The data for "AC" should be 2 separate rows with "Type-C" and "Type-A" being in each of the 2 rows along with corresp. value for last field in each row as 1 and 2 (because these are 2 values)
(2) The data for "D" is not being shown at all.
How do I correct the above code to conform with these 2 requirements?
This works:
for $data_single_row in $rows_data
return
let $cited_name:= $data_single_row/td[1]
let $original_types_w_return:= $data_single_row/td[2]
let $original_types_list:= tokenize(concat($original_types_w_return, " "), '(\r?\n|\r)')
for $cited_type_each at $pos2 in $original_types_list
return concat( $cited_name, '^', normalize-space($cited_type_each), '^', $pos2)
(The first change was to replace $original_type_each with $cited_type_each and [4] with [2] which may ).
The first problem can be solved by removing the $ at the end of the tokenize parameter, since in the default mode $ only match the end of the string.
The second one is solved by adding an space $original_types_w_return, so it is not empty and tokenize returns something, and then removing it again with normalize-space (in XQuery 3.0 it could probably be solved by using 'allowing empty' in the for expression)