I have such a query:
WHERE x LIKE $1
, where $1 is a bindvar string built in the backend:
$1 = "%" + PATTERN + "%"
Is it possible to build a LIKE PATTERN in that way that special characters (% and _) are escaped, so I have the same functionality, but it works for all possible PATTERN values.
You would want to escape the literal % and _ with backslash. For example, in PHP we might try:
$pattern = "something _10%_ else";
$pattern = preg_replace("/([%_])/", "\\\\$1", $pattern);
echo $pattern; // something \_10\%\_ else
Related
Below is a input.
!{ID=34, ID2=35}
>
!{ID=99, ID2=23}
>
!{ID=18, ID2=87}
<
I am trying to make a final result like as following. That is, wanted to remove space,'{' and '}' character and check if the next line is '>' or '<'.
In fact, the input above is repeated. I also need to parse '>' and '<' character so I will put the parsed string(YES or NO) into database.
ID=34,ID=35#YES#NO
ID=99,ID=23#YES#NO
ID=18,ID=87#NO#YES
So, with 'sub' function I thought I can replace the space with blank but the result shows:
1#YES#NO
Can you let me know what is wrong?
If possible, teach me how to remove '{' and '}' as well.
Appreciated if you could show me the awk file version instead of one-liner.
BEGIN {
VALUES = ""
L_EXIST = "NO"
R_EXIST = "NO"
}
/!/ { VALUES = gsub(" ", "", $0);
getline;
if ($1 == ">") L_EXIST = "YES";
else if ($1 == "<") R_EXIST = "YES";
print VALUES"#"L_EXIST"#"R_EXIST
}
END {
}
Given your sample input:
$ cat file
!{ID=34, ID2=35}
>
!{ID=99, ID2=23}
>
!{ID=18, ID2=87}
<
This script produces the desired output:
BEGIN { FS="[}{=, ]+"; RS="!" }
NR > 1 { printf "ID=%d,ID=%d#%s\n", $3, $5, ($6==">"?"YES#NO":"NO#YES") }
The Field Separator is set to consume the spaces and other characters between the parts of the line that you're interested in. The Record Separator is set to !, so that each pair of lines is treated as a single record.
The first record is empty (the start of the first line, up to the first !), so we only process the ones after that. The output is constructed using printf, with a ternary to determine the last part (I assume that there are only two options, > or <).
Let's say you have this input:
input.txt
!{ID=34, ID2=35}
!{ID=36, ID2=37}
>
You can use the following awk command
awk -F'[!{}, ]' 'NR>1{yn="NO";if($1==">")yn="YES";print l"#"yn}{l=$3","$5}' input.txt
to produce this output:
ID=34,ID2=35#NO
ID=36,ID2=37#YES
I have a string that I build from a couple sources to do matching with later, the short of my code so far is:
$temp = "some\good"
if("some text" -match $temp)
My representation of $temp is simple but actually it is built, this is an example of how it can get built, so no, in this case changing " for ' to pass a literal to $temp won't work. If I hard code the if to use a literal string version of $temp, it works so its a matter of converting the value in $temp to a literal string.
I get the following error when I run my code:
parsing "some\good" - Unrecognized escape sequence \g.
At [not important]
+ if($temp2 -match $temp)
+ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
+ CategoryInfo : OperationStopped: (:) [], ArgumentException
+ FullyQualifiedErrorId : System.ArgumentException
"Converting to literal string" won't help you here. String is a string, it only matters how it's being used, i.e. if the content is being interpreted in some way.
-match operates on regex, and that's it, you can't change that, you'd have to escape every all characters that have a meaning in regex.
But you can use select-string instead, which has a switch for simple matching:
$text = "some text"
$patt = "some\good"
if (($text | Select-String -SimpleMatch -Pattern $patt | Measure-Object).count -gt 0) {
Write-Host "match"
} else {
Write-Host "nomatch"
}
In my below script, the str_replace(rtrim(c_manager),'''','_') doesn't seem to work.
I want to replace single quotes with underscores for my arguments. For example:
Input: `S'achin`
Result: `S_achin`
$sql = 'select rtrim(f_admin_disabled),'."\n".
' convert(varchar,t_password,101),'."\n".
' rtrim(c_email),'."\n".
' str_replace(rtrim(c_manager),'''','_'),'."\n".
' rtrim(c_mgr_email)'."\n".
' from tuserprofile'."\n".
' where ic_user1 = '."'$user_id'"."\n";
If you want to have single quotes in a single quoted string to produce str_replace(rtrim(c_manager),'''','_'), you need to either escape them:
' str_replace(rtrim(c_manager),\'\'\'\',\'_\'),'
or use a different delimiter:
q! str_replace(rtrim(c_manager),'''','_'),!
to replace a character in string
$string=~s/'/_/g;
Syntax
$string=~s/<string>/<replace_string>/g;
Use this subroutine.
sub str_replace {
my ($s) = #_;
$s =~ s/'/_/g;
return $s;
}
I'm trying to replace a word in a string. The word is stored in a variable so naturally I do this:
$sentence = "hi this is me";
$foo=~ m/is (.*)/;
$foo = $1;
$sentence =~ s/$foo/you/;
print $newsentence;
But this doesn't work.
Any idea on how to solve this? Why this happens?
Perl lets you interpolate a string into a regular expression, as many of the answers have already shown. After that string interpolation, the result has to be a valid regex.
In your original try, you used the match operator, m//, that immediately tries to perform a match. You could have used the regular expression quoting operator in it's place:
$foo = qr/me/;
You can either bind to that directory or interpolate it:
$string =~ $foo;
$string =~ s/$foo/replacement/;
You can read more about qr// in Regexp Quote-Like Operators in perlop.
You have to replace the same variable, otherwise $newsentence is not set and Perl doesn't know what to replace:
$sentence = "hi this is me";
$foo = "me";
$sentence =~ s/$foo/you/;
print $sentence;
If you want to keep $sentence with its previous value, you can copy $sentence into $newsentence and perform the substitution, that will be saved into $newsentence:
$sentence = "hi this is me";
$foo = "me";
$newsentence = $sentence;
$newsentence =~ s/$foo/you/;
print $newsentence;
You first need to copy $sentence to $newsentence.
$sentence = "hi this is me";
$foo = "me";
$newsentence = $sentence;
$newsentence =~ s/$foo/you/;
print $newsentence;
Even for small scripts, please 'use strict' and 'use warnings'. Your code snippet uses $foo and $newsentence without initialising them, and 'strict' would have caught this. Remember that '=~' is for matching and substitution, not assignment. Also be aware that regexes in Perl aren't word-bounded by default, so the example expression you've got will set $1 to 'is me', the 'is' having matched the tail of 'this'.
Assuming you're trying to turn the string from 'hi this is me' to 'hi this is you', you'll need something like this:
my $sentence = "hi this is me";
$sentence =~ s/\bme$/\byou$/;
print $sentence, "\n";
In the regex, '\b' is a word boundary, and '$' is end-of-line. Just doing 's/me/you/' will also work in your example, but would probably have unintended effects if you had a string like 'this is merry old me', which would become 'this is yourry old me'.
I have sentences like :
" a"
"a "
" a "
I would like to catch all this examples (with lex), but I don't how to say the beginning of the line
I'm not totally sure what exactly you're looking for, but the regex symbol to specify matching the beginning of a line in a lex definition is the caret:
^
If I understand correctly, you're trying to pull the "a" out as the token, but you don't want to grab any of the whitespace? If this is the case, then you just need something like the following:
[\n\t\r ]+ {
// do nothing
}
"a" {
assignYYText( yylval );
return aToken;
}