11-07-2008
Changing directories using variables.
I can't seem to solve this problem.
Please assist. Thanks!
#!/bin/bash
UserDir="$(echo ~$1)"
echo "Changing directory with variables"
cd "$UserDir"
echo "Changing directory without variables"
cd ~pearsn
pwd
Output:
pearsn$ sh -x ./test.bash pearsn
++ echo '~pearsn'
+ UserDir='~pearsn'
+ echo 'Changing directory with variables'
Changing directory with variables
+ cd '~pearsn'
./test.bash: line 6: cd: ~pearsn: No such file or directory
+ echo 'Changing directory without variables'
Changing directory without variables
+ cd /Users/pearsn
+ pwd
/Users/pearsn
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LEARN ABOUT PHP
parse_str
PARSE_STR(3) 1 PARSE_STR(3)
parse_str - Parses the string into variables
SYNOPSIS
void parse_str (string $str, [array &$arr])
DESCRIPTION
Parses $str as if it were the query string passed via a URL and sets variables in the current scope.
Note
To get the current QUERY_STRING, you may use the variable $_SERVER['QUERY_STRING']. Also, you may want to read the section on vari-
ables from external sources.
Note
The magic_quotes_gpc setting affects the output of this function, as parse_str(3) uses the same mechanism that PHP uses to populate
the $_GET, $_POST, etc. variables.
PARAMETERS
o $str
- The input string.
o $arr
- If the second parameter $arr is present, variables are stored in this variable as array elements instead.
RETURN VALUES
No value is returned.
EXAMPLES
Example #1
Using parse_str(3)
<?php
$str = "first=value&arr[]=foo+bar&arr[]=baz";
parse_str($str);
echo $first; // value
echo $arr[0]; // foo bar
echo $arr[1]; // baz
parse_str($str, $output);
echo $output['first']; // value
echo $output['arr'][0]; // foo bar
echo $output['arr'][1]; // baz
?>
SEE ALSO
parse_url(3), pathinfo(3), http_build_query(3), get_magic_quotes_gpc(3), urldecode(3).
PHP Documentation Group PARSE_STR(3)