That would be ugly, glitch-prone, and a gaping security hole(never use eval), which is why I suggested the alternative.
This could work, if your variables don't have more than one line in them.
But again, the best way would be to avoid the problem completely by using a subshell -- that's one reason they exist. Or just don't overwrite those variables in the first place.
Hi all,
I've been trying to get this to work for ages to no avail. I've searched this site and googled but cannot find a satisfactory answer.
I've got a while loop, like this
while read line
do
echo "$line"
done < file_name
Now, my problem is that most of the lines in the file... (3 Replies)
I have a set of variables:
f1="./someFolder"
.
.
f10="./someOtherFolder"
And I'm trying to use the following loop
for (( i = 0; i <= 10; i++ ))
do
temp=f$i
done
I'm trying the get the values from my set of variable to make directories, but I can't seem the get those value... (3 Replies)
- I m retreving values from database and wish to use those values later in my shell script. I m placing these values in an array da_data but outside loop array is empty.Problem is its treating array as local inside loop hence array is empty outside loop.
Plz go through the script and suggest how... (1 Reply)
Hi there, I am checking disk spaced used on a box
# df -k | grep dsk | awk {'print $3'}
2055463
20165785
18310202
32274406
I want to somehow add them up but am no quite sure how to do this in a loop. ...i.e
for n in 'df -k | grep dsk | awk {'print $3}'
do
<some adding... (1 Reply)
I obviously haven't learned my lesson with shell and whitespace.
find /path/to/some/where/ -name "*.pdf" | awk '{print $5}'| uniq -d
results:
some Corporation
other Corporate junk
firmx
Works fine from cmdline but the whitespace turns into another FS in a for loop.
for... (7 Replies)
Hi All,
Thanks all of you for the help you provide to me. Well, I have one more problem, where I am trying to pull file system information in the loop and display the filesystem percentege. I am using following code to achive this, nut it's giving the weired output.
My file system is
... (1 Reply)
Hello All,
Maybe I'm Missing something here but I have NOOO idea what the heck is going on with this....?
I have a Variable that contains a PATTERN of what I'm considering "Illegal Characters". So what I'm doing is looping
through a string containing some of these "Illegal Characters". Now... (5 Replies)
Hi Everyone,
I am currently tasked with some reporting on various Unix based OSes. I have a script deployed that runs and grabs the information I am looking for, and has a bit of logic to output the desired result into a text file.
Example of my text file:
multiUsers=yes... (1 Reply)
Hi All
I am trying to fetch the size of three files into three separate variables within a for loop and am doing something like this:
for i in ATT1 ATT2 ATT3
do
size_$i=`ls -ltr $i | awk '{print $5}'`
echo ${size_$i}
done
but am getting the below error:
ksh: size_ATT1=522: not... (3 Replies)
Discussion started by: swasid
3 Replies
LEARN ABOUT PHP
debug_zval_dump
DEBUG_ZVAL_DUMP(3) 1 DEBUG_ZVAL_DUMP(3)debug_zval_dump - Dumps a string representation of an internal zend value to outputSYNOPSIS
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)