condition's statement gets skipped

i am new to scripting.how do i correct this bash script. for all values it gives me the "else" condition - "wrong value"

 #!/bin/bash
 read -p "First Name: " $first
 read -p "Last Name: " $last
 read -p "profession 1 for software and 2 for hardware (1/2): " $pro
 echo $first
 if [ "$pro" == "1" ];then
 echo $first
 echo $last
 elif [ "$pro" == "2" ];then
 echo $first
 echo $last
 echo $pro
 else
 echo "wrong value!"
 fi

edited Nov 11 at 7:59

Red Cricket

4,24483382

asked Nov 11 at 7:43

vamsi

2

Please take a look: shellcheck.net
– Cyrus
Nov 11 at 7:47

you need to pass the variable name to read without a dollar sign.
– oguzismail
Nov 11 at 7:48

add a comment |

i am new to scripting.how do i correct this bash script. for all values it gives me the "else" condition - "wrong value"

 #!/bin/bash
 read -p "First Name: " $first
 read -p "Last Name: " $last
 read -p "profession 1 for software and 2 for hardware (1/2): " $pro
 echo $first
 if [ "$pro" == "1" ];then
 echo $first
 echo $last
 elif [ "$pro" == "2" ];then
 echo $first
 echo $last
 echo $pro
 else
 echo "wrong value!"
 fi

edited Nov 11 at 7:59

Red Cricket

4,24483382

asked Nov 11 at 7:43

vamsi

2

Please take a look: shellcheck.net
– Cyrus
Nov 11 at 7:47

you need to pass the variable name to read without a dollar sign.
– oguzismail
Nov 11 at 7:48

add a comment |

i am new to scripting.how do i correct this bash script. for all values it gives me the "else" condition - "wrong value"

 #!/bin/bash
 read -p "First Name: " $first
 read -p "Last Name: " $last
 read -p "profession 1 for software and 2 for hardware (1/2): " $pro
 echo $first
 if [ "$pro" == "1" ];then
 echo $first
 echo $last
 elif [ "$pro" == "2" ];then
 echo $first
 echo $last
 echo $pro
 else
 echo "wrong value!"
 fi

edited Nov 11 at 7:59

Red Cricket

4,24483382

asked Nov 11 at 7:43

vamsi

i am new to scripting.how do i correct this bash script. for all values it gives me the "else" condition - "wrong value"

 #!/bin/bash
 read -p "First Name: " $first
 read -p "Last Name: " $last
 read -p "profession 1 for software and 2 for hardware (1/2): " $pro
 echo $first
 if [ "$pro" == "1" ];then
 echo $first
 echo $last
 elif [ "$pro" == "2" ];then
 echo $first
 echo $last
 echo $pro
 else
 echo "wrong value!"
 fi

bash

edited Nov 11 at 7:59

Red Cricket

4,24483382

asked Nov 11 at 7:43

vamsi

edited Nov 11 at 7:59

Red Cricket

4,24483382

asked Nov 11 at 7:43

vamsi

edited Nov 11 at 7:59

Red Cricket

4,24483382

edited Nov 11 at 7:59

Red Cricket

4,24483382

edited Nov 11 at 7:59

Red Cricket

4,24483382

asked Nov 11 at 7:43

vamsi

asked Nov 11 at 7:43

vamsi

asked Nov 11 at 7:43

vamsi

2

Please take a look: shellcheck.net
– Cyrus
Nov 11 at 7:47

you need to pass the variable name to read without a dollar sign.
– oguzismail
Nov 11 at 7:48

add a comment |

2

Please take a look: shellcheck.net
– Cyrus
Nov 11 at 7:47

you need to pass the variable name to read without a dollar sign.
– oguzismail
Nov 11 at 7:48

Please take a look: shellcheck.net
– Cyrus
Nov 11 at 7:47

you need to pass the variable name to read without a dollar sign.
– oguzismail
Nov 11 at 7:48

add a comment |

1 Answer
1

active

oldest

votes

You need to change …

read -p "First Name: " $first
read -p "Last Name: " $last
read -p "profession 1 for software and 2 for hardware (1/2): " $pro

… to …

read -p "First Name: " first
read -p "Last Name: " last
read -p "profession 1 for software and 2 for hardware (1/2): " pro

Note the lack of $ on first, last and pro. In shell you do not use the $ on a variable when assigning it a value.

answered Nov 11 at 7:51

Red Cricket

4,24483382

Further explanation: you use $ to get the value of a variable, but not to set it. E.g. read varname, varname=value, etc. BTW, you should also use double-quotes around variable references, e.g. echo "$pro" instead of just echo $pro.
– Gordon Davisson
Nov 11 at 8:53

It worked, and thank you so much for the explanation...
– vamsi
Nov 11 at 9:26

1

More information (and probably more information than you want): if first isn't set, then read $first is equivalent to read, which implicitly uses the variable REPLY to store the input. If first is set, then the expansion produces one (or possibly more) words to use as variables. For instance, first=foo; read $first sets the variable foo. first="foo bar"; read $first would set the variables foo and bar after performing word-splitting on the input. All of this is to say, it's not wrong to use read $first; it just doesn't do what you expected. :)
– chepner
Nov 11 at 20:18

add a comment |

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1 Answer
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active

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1 Answer
1

active

oldest

votes

You need to change …

read -p "First Name: " $first
read -p "Last Name: " $last
read -p "profession 1 for software and 2 for hardware (1/2): " $pro

… to …

read -p "First Name: " first
read -p "Last Name: " last
read -p "profession 1 for software and 2 for hardware (1/2): " pro

Note the lack of $ on first, last and pro. In shell you do not use the $ on a variable when assigning it a value.

answered Nov 11 at 7:51

Red Cricket

4,24483382

Further explanation: you use $ to get the value of a variable, but not to set it. E.g. read varname, varname=value, etc. BTW, you should also use double-quotes around variable references, e.g. echo "$pro" instead of just echo $pro.
– Gordon Davisson
Nov 11 at 8:53

It worked, and thank you so much for the explanation...
– vamsi
Nov 11 at 9:26

1

More information (and probably more information than you want): if first isn't set, then read $first is equivalent to read, which implicitly uses the variable REPLY to store the input. If first is set, then the expansion produces one (or possibly more) words to use as variables. For instance, first=foo; read $first sets the variable foo. first="foo bar"; read $first would set the variables foo and bar after performing word-splitting on the input. All of this is to say, it's not wrong to use read $first; it just doesn't do what you expected. :)
– chepner
Nov 11 at 20:18

add a comment |

You need to change …

read -p "First Name: " $first
read -p "Last Name: " $last
read -p "profession 1 for software and 2 for hardware (1/2): " $pro

… to …

read -p "First Name: " first
read -p "Last Name: " last
read -p "profession 1 for software and 2 for hardware (1/2): " pro

Note the lack of $ on first, last and pro. In shell you do not use the $ on a variable when assigning it a value.

answered Nov 11 at 7:51

Red Cricket

4,24483382

Further explanation: you use $ to get the value of a variable, but not to set it. E.g. read varname, varname=value, etc. BTW, you should also use double-quotes around variable references, e.g. echo "$pro" instead of just echo $pro.
– Gordon Davisson
Nov 11 at 8:53

It worked, and thank you so much for the explanation...
– vamsi
Nov 11 at 9:26

1

More information (and probably more information than you want): if first isn't set, then read $first is equivalent to read, which implicitly uses the variable REPLY to store the input. If first is set, then the expansion produces one (or possibly more) words to use as variables. For instance, first=foo; read $first sets the variable foo. first="foo bar"; read $first would set the variables foo and bar after performing word-splitting on the input. All of this is to say, it's not wrong to use read $first; it just doesn't do what you expected. :)
– chepner
Nov 11 at 20:18

add a comment |

You need to change …

read -p "First Name: " $first
read -p "Last Name: " $last
read -p "profession 1 for software and 2 for hardware (1/2): " $pro

… to …

read -p "First Name: " first
read -p "Last Name: " last
read -p "profession 1 for software and 2 for hardware (1/2): " pro

Note the lack of $ on first, last and pro. In shell you do not use the $ on a variable when assigning it a value.

answered Nov 11 at 7:51

Red Cricket

4,24483382

You need to change …

read -p "First Name: " $first
read -p "Last Name: " $last
read -p "profession 1 for software and 2 for hardware (1/2): " $pro

… to …

read -p "First Name: " first
read -p "Last Name: " last
read -p "profession 1 for software and 2 for hardware (1/2): " pro

Note the lack of $ on first, last and pro. In shell you do not use the $ on a variable when assigning it a value.

answered Nov 11 at 7:51

Red Cricket

4,24483382

answered Nov 11 at 7:51

Red Cricket

4,24483382

answered Nov 11 at 7:51

Red Cricket

4,24483382

answered Nov 11 at 7:51

Red Cricket

4,24483382

Further explanation: you use $ to get the value of a variable, but not to set it. E.g. read varname, varname=value, etc. BTW, you should also use double-quotes around variable references, e.g. echo "$pro" instead of just echo $pro.
– Gordon Davisson
Nov 11 at 8:53

It worked, and thank you so much for the explanation...
– vamsi
Nov 11 at 9:26

1

More information (and probably more information than you want): if first isn't set, then read $first is equivalent to read, which implicitly uses the variable REPLY to store the input. If first is set, then the expansion produces one (or possibly more) words to use as variables. For instance, first=foo; read $first sets the variable foo. first="foo bar"; read $first would set the variables foo and bar after performing word-splitting on the input. All of this is to say, it's not wrong to use read $first; it just doesn't do what you expected. :)
– chepner
Nov 11 at 20:18

add a comment |

Further explanation: you use $ to get the value of a variable, but not to set it. E.g. read varname, varname=value, etc. BTW, you should also use double-quotes around variable references, e.g. echo "$pro" instead of just echo $pro.
– Gordon Davisson
Nov 11 at 8:53

It worked, and thank you so much for the explanation...
– vamsi
Nov 11 at 9:26

1

More information (and probably more information than you want): if first isn't set, then read $first is equivalent to read, which implicitly uses the variable REPLY to store the input. If first is set, then the expansion produces one (or possibly more) words to use as variables. For instance, first=foo; read $first sets the variable foo. first="foo bar"; read $first would set the variables foo and bar after performing word-splitting on the input. All of this is to say, it's not wrong to use read $first; it just doesn't do what you expected. :)
– chepner
Nov 11 at 20:18

Further explanation: you use $ to get the value of a variable, but not to set it. E.g. read varname, varname=value, etc. BTW, you should also use double-quotes around variable references, e.g. echo "$pro" instead of just echo $pro.
– Gordon Davisson
Nov 11 at 8:53

It worked, and thank you so much for the explanation...
– vamsi
Nov 11 at 9:26

More information (and probably more information than you want): if first isn't set, then read $first is equivalent to read, which implicitly uses the variable REPLY to store the input. If first is set, then the expansion produces one (or possibly more) words to use as variables. For instance, first=foo; read $first sets the variable foo. first="foo bar"; read $first would set the variables foo and bar after performing word-splitting on the input. All of this is to say, it's not wrong to use read $first; it just doesn't do what you expected. :)
– chepner
Nov 11 at 20:18

add a comment |

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