BASH: supporta variabili condizionali come var = "test"? "1": "2"

Question

Sono nuovo di bash ma ho fatto molto PHP e Javascript.

Esiste una sorta di equivoco a questo PHP?

$default = 10; 
$var = (!$var) ? $default : $var; 

Grazie

Answer 1

Sì, lo fa:

var=${var:-10} 

anche con altre variabili:

unset var 
export def=99 
echo ${var:-${def}} # gives '99' 
export var=7 
echo ${var:-${def}} # gives '7' 

Answer 2

Sì!

Dalla pagina man:

${parameter:-word} 
     Use Default Values. If parameter is unset or null, the expansion of word 
     is substituted. Otherwise, the value of parameter is substituted. 
${parameter:=word} 
     Assign Default Values. If parameter is unset or null, the expansion of word 
     is assigned to parameter. The value of parameter is then substituted. 
     Positional parameters and special parameters may not be assigned to in this way. 
${parameter:?word} 
     Display Error if Null or Unset. If parameter is null or unset, the expansion of word (or a message to 
     that effect if word is not present) is written to the standard error and the shell, if it is not inter‐ 
     active, exits. Otherwise, the value of parameter is substituted. 
${parameter:+word} 
     Use Alternate Value. If parameter is null or unset, nothing is substituted, otherwise the expansion of 
     word is substituted. 

Answer 3

$ default=10 
$ var=${var:-$default} 
$ echo $var 
10 
$ var=9 
$ var=${var:-$default} 
$ echo $var 
9