Command not found error when passing filename to a variable

Question

#!/bin/bash

file_name = $1
ext = (file -b --mime-type $file_name | sed 's/application\///g')
echo $file_name $ext

Following error messages are seen while executing the code:

file_extractor.sh: line 3: file_name: command not found

file_extractor.sh: line 4: ext: command not found

I have tried adding quotes to the variable, file_name = "$1" didn't work, then substituting $(file -b --mime-type $file_name | sed 's/application\///g'), using double quotes.

Nothing worked.

Possible duplicate of [Command not found error in Bash variable assignment](https://stackoverflow.com/q/2268104/608639). — jww, Jul 01 '18 at 13:33
Have you ever looked at a working shell script? Please don't try to learn a language by simply *guessing* at what you believe to be the correct syntax. — chepner, Jul 01 '18 at 14:16

score 0 · Accepted Answer · answered Jul 01 '18 at 12:15

0

You have two problems in that script:

Assignment to a variable does not allow for spaces around =
```
var="val"  # OK
var = "val"  # Not OK
```
You are almost cetainly lookging for command substitution:
```
var=$(something that produces output)
```
Even though that is not the source of your errors. The former one is.

answered Jul 01 '18 at 12:15

Ondrej K.

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Thank brother !! – Bharath B Jul 01 '18 at 15:05

Command not found error when passing filename to a variable

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