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I've seen a couple posts for this, like this one, but none are helping me in my particular situation.
scriptsPath="/var/db/gbi/scripts/"
echo "$scriptsPathawesome.csv";
I would expect this to echo /var/db/gbi/scripts/awesome.csv
Instead I get .csv
Seems like it thinks I'm trying to reference a variable named $scriptsPathawesome. How can I concatenate the $scriptsPath variable to the "awesome.csv" string literal?
358
String literalsThe code concatenates the smaller strings to create the long string literal. The parts are concatenated into a single string at compile time.
A string literal is a sequence of characters, enclosed in double quotation marks (" ") , which is used to represent a null-terminated string in C/C++. If we try to concatenate two string literals using the + operator, it will fail. For instance, consider the following code, which results in a compilation error: C.
A string literal is a sequence of zero or more characters enclosed within single quotation marks. The following are examples of string literals: 'Hello, world!' 'He said, "Take it or leave it."'
char *amessage = "This is a string literal."; All escape codes listed in the Escape Sequences table are valid in string literals. To represent a double quotation mark in a string literal, use the escape sequence \".
You need to surround your variable with curly braces like so:
scriptsPath="/var/db/gbi/scripts/"
echo "${scriptsPath}awesome.csv";
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