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Top Forums Shell Programming and Scripting problem with sudo su and .sh script Post 302158153 by cleansing_flame on Monday 14th of January 2008 02:03:43 PM
Old 01-14-2008
problem with sudo su and .sh script

here is my script:

#!/bin/sh


cd /Users/a

echo "what is the name of the file u want?"
read var1
var1=$var1
var3=$var1

echo "is this $var1 okay?"
echo "enter the type of file u want"
echo "your choices are .java, .c, .sh, .ksh, .csh"
read var2
var2=$var2

var4=`find . -type f | grep "$var1$var2" | wc -l`
var5=0

if test $var4 -ne $var5
then
echo "next "$var2" prog"
exit 0
fi

if test "$var1$var2" = "$var3.java"
then
cd /Users/a/Javaprogs
mkdir ./$var1.java
emacs /Users/a/Javaprogs/$var1.java/$var1.java
cat ../defaults/javadefaultfile.txt >> ./Javaprogs/$var1.java/$var1.java
exit 0
fi

if test "$var1$var2" = "$var3.c"
then
cd /Users/a/cprogs
mkdir ./$var1.c
emacs /Users/a/cprogs/$var1.c/$var1.c
cat ../default/cprogdefaultfile.txt >> /Users/la/cprogs/$var1.c/$var1.c
exit 0
fi


if test "$var1$var2" = "$var3.sh"
then
`sudo su`
cd /bin
mkdir ./$var1
emacs ./$var1/$var1.sh
cat /Users/a/defaults/shdefaultfile.txt >> ./$var1/$var1.sh
exit 0
fi


the part in red contains the problem, the `sudo su` command does prompt me, but once I enter my passwd i change prompt as normal. I can type, but no commands are recognized and ctrl-c does nothing but enter a new blank line. Any help would be appreciated.
 

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DEBUG_ZVAL_DUMP(3)							 1							DEBUG_ZVAL_DUMP(3)

debug_zval_dump - Dumps a string representation of an internal zend value to output

SYNOPSIS
void debug_zval_dump (mixed $variable, [mixed $...]) DESCRIPTION
Dumps a string representation of an internal zend value to output. PARAMETERS
o $variable - The variable being evaluated. RETURN VALUES
No value is returned. EXAMPLES
Example #1 debug_zval_dump(3) example <?php $var1 = 'Hello World'; $var2 = ''; $var2 =& $var1; debug_zval_dump(&$var1); ?> The above example will output: &string(11) "Hello World" refcount(3) Note Beware the refcount The refcount value returned by this function is non-obvious in certain circumstances. For example, a developer might expect the above example to indicate a refcount of 2. The third reference is created when actually calling debug_zval_dump(3). This behavior is further compounded when a variable is not passed to debug_zval_dump(3) by reference. To illustrate, consider a slightly modified version of the above example: Example #2 <?php $var1 = 'Hello World'; $var2 = ''; $var2 =& $var1; debug_zval_dump($var1); // not passed by reference, this time ?> The above example will output: string(11) "Hello World" refcount(1) Why refcount(1)? Because a copy of $var1 is being made, when the function is called. This function becomes even more confusing when a variable with a refcount of 1 is passed (by copy/value): Example #3 <?php $var1 = 'Hello World'; debug_zval_dump($var1); ?> The above example will output: string(11) "Hello World" refcount(2) A refcount of 2, here, is extremely non-obvious. Especially considering the above examples. So what's happening? When a variable has a single reference (as did $var1 before it was used as an argument to debug_zval_dump(3)), PHP's engine opti- mizes the manner in which it is passed to a function. Internally, PHP treats $var1 like a reference (in that the refcount is increased for the scope of this function), with the caveat that if the passed reference happens to be written to, a copy is made, but only at the moment of writing. This is known as "copy on write." So, if debug_zval_dump(3) happened to write to its sole parameter (and it doesn't), then a copy would be made. Until then, the parameter remains a reference, causing the refcount to be incremented to 2 for the scope of the function call. SEE ALSO
var_dump(3), debug_backtrace(3), References Explained, References Explained (by Derick Rethans). PHP Documentation Group DEBUG_ZVAL_DUMP(3)
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