if statement results in 'command not found error` [duplicate]

Question

I have this code:

ColumnInformation=`sqlite3 $database "PRAGMA table_info($table);"`

and it works perfect, but I need add the if statement:

if [$table != "Order"]; then
    ColumnInformation=`sqlite3 $database "PRAGMA table_info($table);"`
else
    ColumnInformation=`sqlite3 $database "PRAGMA table_info('Order');"`
fi

it throws command not found error. Could you tell me why ?

best, Tomek

You need spaces around [ and ].

devnull
– devnull

2014-01-10 14:56:30 +00:00
Commented Jan 10, 2014 at 14:56 — devnull
– devnull, Commented Jan 10, 2014 at 14:56

trojanfoe · Accepted Answer · 2014-01-10 14:56:51Z

2

Could be lack of space either side of [ and ] and you need to use "backticks" to get the output from a sub-command:

if [ $table != "Order" ]; then
    ColumnInformation=`sqlite3 $database "PRAGMA table_info('$table');"`
else
    ColumnInformation=`sqlite3 $database "PRAGMA table_info('Order');"`
fi

However the test appears to be pointless anyway as $table is "Order" in the second case anyway. Also note that you were missing single-quotes around $table in the first case.

answered Jan 10, 2014 at 14:56

trojanfoe

123k23 gold badges219 silver badges249 bronze badges

Sign up to request clarification or add additional context in comments.

Collectives™ on Stack Overflow

if statement results in 'command not found error` [duplicate]

1 Answer 1

Comments

Linked

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

Comments

Linked

Related