Sorry rbatte1, I am new to both UNIX and this forum.
Using a bash script and Ubuntu. I just figured out how to do what I needed 1 min ago.
If you have any suggestions on how to improve it, you are welcome :-)
Moderator's Comments:
Please use next time code tags for your code and data
I have to replace date in Control file every day i run the script
control file looks like this
$cat pharma.ctl
LOAD DATA
INFILE '/usr/bin/sqlscripts/SQL.PharmID.20071206.txt'
INTO TABLE uname.TEMP_TABLE
FIELDS TERMINATED BY '|'
TRAILING NULLCOLS
(col_names)
i tried
$sed -e... (2 Replies)
Hi
I have an XML file with strings XABCD, XEFGHX and XIJKLX. I would like to replace XABCDX with "This is the first string", XEFGHX with "This is the second string" and XIJKLX with "This is the third string".
What is the best way to implement this? Should I have a file with the data that is... (4 Replies)
Hi,
i call my shell like:
my_shell "my project name"
my script:
#!/bin/bash -vx
projectname=$1
sed s/'PROJECT_NAME ='/'PROJECT_NAME = '$projectname/ <test_config_doxy >temp
cp temp test_config_doxy
the following error occurres:
sed s/'PROJECT_NAME ... (2 Replies)
Can someone tell me how I can do this?
e.g:
a=$(echo -e wert trewt ertert ertert ertert erttert
erterte
rterter
tertertert
ert)
How do i replace the STRING with $a?
I try this:
sed -i 's/STRING/'"$a"'/g' filename.ext
but this don' t work (2 Replies)
Hi,
I have a text file
6 00 01 ww code 40 80 30 NIK MOT 001
7 02 01 ww perl 40 80 30 NIK MOT 002
8 06 01 ww shell (unix) 40 80 30 LG MOT 007
4 03 04 xx Java (McGraw) 99 77 33 hhh uuu X
I have to replace... (11 Replies)
Hi all,
Hereby wish to have your advise for below:
Main concept is
I intend to get current directory of my script file.
This script file will be copied to /etc/init.d.
A string in this copy will be replaced with current directory value.
Below is original script file:
... (6 Replies)
here is what i want to achieve... consider a file contains below contents. the file size is large about 60mb
cat dump.sql
INSERT INTO `table1` (`id`, `action`, `date`, `descrip`, `lastModified`) VALUES (1,'Change','2011-05-05 00:00:00','Account Updated','2012-02-10... (10 Replies)
Sorry for the long/weird title but I'm stuck on a problem I have. I have this XML file:
</member>
<member>
<name>TransactionID</name>
<value><string>123456789123456</string></value>
</member>
<member>
<name>Number</name>
... (9 Replies)
Hi Team,
I have a file and need to replace string. Out of 20 rows, there is one row given below
$Paramsoqlfilter=Systemmodstamp > 1900-01-01T00:00:00.000Z
in a Shell Script, I have a variable
HIST_DATE="1900-01-01T00:00:00.000Z"
INC_DATE="2018-10-04T09:18:43.000Z"
Now I need to... (14 Replies)
Discussion started by: ace_friends22
14 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)