I have a suggestion for a new block.

name: contains ?
category: sensing
type: Boolean
function: it would scan the first string to see if it contained the second strings input. if it did, it would come out true
block:

< [ ] contains [ ] ? :: sensing Boolean >

people who support (14):
Lythium
DownsGameClub
DaSpudLord
SpongyGames
pvz_pro
theonlygusti
Zekrom01
BurnedCrystal
sheep_maker
alexphan
Bookowl
Prscr
JonathanSchaffer
A-no-meep
Gooballs

people who don't support (0):

I actually support now, because I can't seem to find a valid workaround.

Lythium wrote:

<(answer) = [ whatever]>

Easy workaround.

what is the easy workaround?

Lythium wrote:

<(answer) = [ whatever]>

Easy workaround.

Uhh…. That doesnt work.

Support. But a quick revision

<[] contains []?>::sensing  // The entire block is supposed to be blue...  =/

Since this version of programming is somewhat like a sentence, and saying “if does (answer) contains (blank)”

DownsGameClub wrote:

Lythium wrote:

<(answer) = [ whatever]>

Easy workaround.

Uhh…. That doesnt work.

Support. But a quick revision

<[] contains []?>::sensing  // The entire block is supposed to be blue...  =/

Since this version of programming is somewhat like a sentence, and saying “if does (answer) contains (blank)”

I realised it after posting.

<[] contains [] ?::sensing>

Support.

Support

DaSpudLord wrote:

<[] contains [] ?::sensing>

Support.

SpongyGames wrote:

Support

reasons?
semi-support, this has a workaround with variables and lists

pvz_pro wrote:

semi-support, this has a workaround with variables and lists

Okay, for the record, here is the work-around:

set [substring found? v] to [false]
set [letters found v] to [0]
set [i v] to [1]
repeat until <<(length of (string)) = ((i)+(1))> or <(substring found?) = [true]>>
set [j v] to [1]
repeat until <<(length of (substring)) = ((j)+(1))> or <(substring found?) = [true]>>
if <(letter (j) of (substring)) = (letter ((i)+((j)-(1))) of (string))> then
change [letters found v] by (1)
end
change [j v] by (1)
end
if <(letters found) = (length of (substring))> then
set [substring found? v] to [true]
end
change [i v] by (1)
end
if <(substring found?) = [true]> then
else
end

As you can see, it is very simple and intuitive, especially for new scratchers.</sarcasm>

I don't see any reason not to support this idea, so, yeah, support!

<[] contains []? ::operators>

support

Support, as the workaround can have a considerable effect on performance.

That and I'm sure no one wants to redefine string manipulation functions that already exist in other languages

Support if it's case sensitive.

DownsGameClub wrote:

Lythium wrote:

<(answer) = [ whatever]>

Easy workaround.

Uhh…. That doesnt work.

Support. But a quick revision

<[] contains []? :: sensing>  // The entire block is supposed to be blue...  =/

Since this version of programming is somewhat like a sentence, and saying “if does (answer) contains (blank)”

For the block to be blue you have to put it inside the brackets.

Although you can simply add all of the letters into a list, and then using this:

<[ v] contains [ ]>

It would be very useful for finding substrings.
Support.

alexphan wrote:

Although you can simply add all of the letters into a list, and then using this:

<[ v] contains [ ]>

It would be very useful for finding substrings.
Support.

That only works if your looking for a single letter…
Support

alexphan wrote:

Although you can simply add all of the letters into a list, and then using this:

<[ v] contains [ ]>

It would be very useful for finding substrings.
Support.

That doesn't work.

Only this works:

theonlygusti wrote:

pvz_pro wrote:

semi-support, this has a workaround with variables and lists

Okay, for the record, here is the work-around:

set [substring found? v] to [false]
set [letters found v] to [0]
set [i v] to [1]
repeat until <<(length of (string)) = ((i)+(1))> or <(substring found?) = [true]>>
set [j v] to [1]
repeat until <<(length of (substring)) = ((j)+(1))> or <(substring found?) = [true]>>
if <(letter (j) of (substring)) = (letter ((i)+((j)-(1))) of (string))> then
change [letters found v] by (1)
end
change [j v] by (1)
end
if <(letters found) = (length of (substring))> then
set [substring found? v] to [true]
end
change [i v] by (1)
end
if <(substring found?) = [true]> then
else
end

As you can see, it is very simple and intuitive, especially for new scratchers.</sarcasm>

I don't see any reason not to support this idea, so, yeah, support!

<[] contains [] ?::sensing>

I support. I can't find an easy way to workaround this (above seems to have a workaround)
So, final answer,

support!::sensing

Sheep_maker wrote:

Support if it's case sensitive.

the scratch block's strings are not case sensitive

pvz_pro wrote:

Sheep_maker wrote:

Support if it's case sensitive.

the scratch block's strings are not case sensitive

I also defenitely support for that. That would make case detection SO much easier.