I'm trying to pass a JSON by reference and then add something to it. For example, I want to get the same thing as:
$json.commontokens.search.tokens.token += $newItem
I keep the keys following $json in a list of strings in $jsonList (since the keys are unknown and can change) to build $json.commontokens.search.tokens.token:
$jPath = $json
foreach($item in $jsonList){
$jPath = $jPath.$item
}
As a result, $jpath += $newItem doesn't keep the change made to $json (but works for $jpath). So I thought I needed to pass by reference to keep the change. Something like
function adder{
param(
[ref]$j
$stringList
)
if ($stringList -ne $null) {
$first = $stringList.first.value
$stringList.removefirst()
adder ([ref]$j.value.$first) $stringList
} else {
return $j
}
}
$jpath = adder ([ref]$json) $jsonList
$jpath.value += $newItem
Again, this works for $jpath, but nothing happens to $json. When I try $jpath.value[70], I get the expected answer, but $json[70] doesn't exist. In fact, I tried it without the function:
$jpath = [ref]$json.commontokens.search.tokens.token
$jpath.value += $newItem
But it still doesn't work. Am I doing the referencing wrong? Or does it have something to do with the keys?
So, I don't really understand it, but I managed to get it to work.
In adder, I changed the recursive line from
([ref]$j.value.$first)
to
([ref]$($jvalue.value.$first))
I also added a return to make it more obvious I was doing recursion.
For some reason, I still couldn't do $jpath.value += $newItem; However by removing $token from $jpath, $jpath.value.token += $newItem works.
So instead of adding all the strings from stringlist, I skipped the last one.
Adder was changed to skip the last string:
($stringList -ne $null)
was changed to
($stringList-ne $stringList.list.last)
where $stringlist is actually just a node now (I should probably change the name). So the final code is $jpath.value.($stringList.Last.value) += $newItem.
I guess I just can't straight up modify $jpath.value? Or something?
Related
Not sure I am doing this correctly as it doesn't appear to be working.
Is this the correct way to declare my users, and is the If statement correctly formated?
At the top I have declared:
my $las = 'jpietrza hpietrza oszones';
These are employees we are checking against current users.
Further down in the code, I want to change the text color that is printed if the user is in the list vs. someone else.
while ( $sth->fetch() ) {
next unless defined $currentuser;
$lastlogin =~ s/ .*$//;
$host_name =~ s/1408//;
foreach ( #las ) {
if ( $currentuser eq "$_" ) {
$lacolor = "black";
last;
}
else {
$lacolor = "red";
}
}
print "<tr>";
print "<td>$host_name</td>";
print "<td><font color=\"$lacolor\">System In-Use (User Undisclosed)</font></td><td> </td>";
}
maybe there was nothing wrong with the if statement, but the whole 9 lines of code can be condensed to 1 very readable line:
$lacolor = any { /^$currentuser$/ } #las ? "black" : "red";
please
use List::Util qw/any/;
while($sth->fetch()) {
# $currentuser is assigned between the 'while' and this 'next' statement ?
# if not, then outside the loop and do not loop at all unless defined
next unless defined $currentuser;
$lastlogin=~s/ .*$//;
$host_name=~s/1408//;
$lacolor = any { /^$currentuser$/ } #las ? "black" : "red";
print "<tr>";
print "<td>$host_name</td>";
print "<td><font color=\"$lacolor\">System In-Use (User Undisclosed)</font></td><td> </td>";
}
please, also use strict; and use warnings;
I figured out how to put the array together correctly:
my #las = qw(
jpietrza
hpietrza
oszones
);
instead of:
my $las='
jpietrza
hpietrza
oszones
';
Firstly, as I think you have worked out now, the scalar variable $las and the array #las are completely different. As you've seen, you should declare and initialise your array like this:
my #las = qw(
jpietrza
hpietrza
oszones
);
Actually, I suspect this all gets easier if you store this in a hash, not an array;
my %las = map { $_ => 1 } qw(jpietrza hpietrza oszones);
Then your check just becomes:
my $lacolour = $las{$currentuser} ? 'black' : 'red';
A few more points:
Please add use strict and use warnings. And understand and fix the problems they reveal.
The quotes are unnecessary in if ($currentuser eq "$_").
Using a templating system to create the output will make your life a lot easier.
Update: Oh, and one I forgot earlier. It's 2017. No-one has used the font element in HTML for fifteen years. Take a look at CSS.
The code below works for a string value but not when I try to access the variable directly.
The data being accessed is a table at http://webrates.truefx.com/rates/connect.html?f=html
My code strips it of tags and put it in an array $row0
And puts it in a function. But I can't get it out. The function is simplified for this question. I intend to concatenate some of the variables inside the function once I find out what I'm doing wrong.
$row0 = array();
include "scrape/simple_html_dom.php";
$url = "http://webrates.truefx.com/rates/connect.html?f=html";
$html = new simple_html_dom();
$html->load_file($url);
foreach ($html->find('tr') as $i => $row) {
foreach ($row->find('td') as $j => $col) {
$row0[$i][$j]= strip_tags($col);
}
}
myArray($row0); //table stripped of tags
function myArray($arr) {
$a = 'hello'; //$arr[0][0]; HELLO will come out but not the variable
$b = $arr[1][0];
$r[0] = $a;
$r[1] = $b;
//echo $r[1]; If the //'s are removed one can see the proper value here but not outside the function.
return $r;
}
$arrayToEcho = myArray($arr);
echo $arrayToEcho[0]; // will echo "first"
I have tried all the suggestions from here:
http://stackoverflow.com/questions/3451906/multiple-returns-from-function
http://stackoverflow.com/questions/5692568/php-function-return-array
Suggestion appreciated please and more info available if required. Thank you very much for viewing.
You need to get the innertext of $col in your loop. Like this:
$row0[$i][$j]= $col->innertext;
The next thing is:
myArray($row0);
This call will correctly return the parsed array; try echoing it and you'll see. But when you do this:
$arrayToEcho = myArray($arr);
...you're referencing to $arr which is a local variable (a parameter, actually) inside your function myArr. So what you probably meant was this:
$arrayToEcho = myArray($row0);
Hope this helps!
UPDATE
Look, I show you what happens when you call a function:
In order to condense my code I am trying to make one of my while loops into a function. I have tried numerous times and have yet to receive the same result upon compiling as I would just leaving the while loop.
Here's the while loop:
while True:
i = find_lowest_i(logs)
if i == -1:
break
print "i=", i
tpl = logs[i].pop(0)
print tpl
out.append(tpl)
print out
And here's what I have so far for my function:
def mergesort(list_of_logs):
i = find_lowest_i(logs)
out = []
while True:
if i == -1:
break
print "i=", i
tpl = logs[i].pop(0)
print tpl
out.append(tpl)
print out
return out
Thanks in advance. This place is a safe-haven for a beginner programmer.
It looks like the parameter to your function is list_of_logs and you're still using logs inside the function's body. The simplest fix is probably to rename your parameter to mergesort from list_of_logs to logs. Otherwise, looks completely correct to me.
I have an Array (twodimensional) and i insert it into my database.
My Code:
$yourArr = $_POST;
$action = $yourArr['action'];
$mysql = $yourArr['mysql'];
$total = $yourArr['total'];
unset( $yourArr['action'] , $yourArr['mysql'] , $yourArr['total'] );
foreach ($yourArr as $k => $v) {
list($type,$num) = explode('_item_',$k);
$items[$num][$type] = $v;
$pnr= $items[$num][pnr];
$pkt= $items[$num][pkt];
$desc= $items[$num][desc];
$qty= $items[$num][qty];
$price= $items[$num][price];
$eintragen = mysql_query("INSERT INTO rechnungspositionen (artikelnummer, menge, artikel, beschreibung,preis) VALUES ('$pnr', '$qty', '$pkt', '$desc', '$price')");
}
I get 5 inserts in the Database but only the 5th have the informations i want. The firsts are incomplete.
Can someone help me?
Sorry for my english.
check if You have sent vars from browser in array (like
input name="some_name[]" ...
also You can check, what You get at any time by putting var_dump($your_var) in any place in script.
good luck:)
You probably want to have your query and the 5 assignments above that outside of the foreach. Instead in a new loop which only executes once for every item instead of 5 times. Your indentation even suggests the same however your brackets do not.
Currently it is only assigning one value each time and executing a new query. After 5 times all the variables are assigned and the last inserted row finally has everything proper.
error_reporting(E_ALL);
$items = array();
foreach($yourArr as $k => $v) {
// check here if the variable is one you need
list($type, $num) = explode('_item_', $k);
$items[$num][$type] = $v;
}
foreach($items as $item) {
$pnr = mysql_real_escape_string($item['pnr']);
$pkt = mysql_real_escape_string($item['pkt']);
$desc = mysql_real_escape_string($item['desc']);
$qty = mysql_real_escape_string($item['qty']);
$price = mysql_real_escape_string($item['price']);
$eintragen = mysql_query("INSERT INTO rechnungspositionen (artikelnummer, menge, artikel, beschreibung,preis) VALUES ('$pnr', '$qty', '$pkt', '$desc', '$price')");
}
Switching on your error level to E_ALL would have hinted in such a direction, among else:
unquoted array-keys: if a constant of
the same name exists your script will
be unpredictable.
unescaped variables: malformed values
or even just containing a quote which
needs to be there will fail your
query or worse.
naïve exploding: not each $_POST-key
variable will contain the string
item and your list will fail, including subsequent use of $num
What I am trying to do is: (programmatically)
Update status where id is something, if no rows where updated, give error: we cannot find the record with id something, otherwise give message success.
Here I am using mysql_affected_rows() to know if a row was updated or not, but it always return 1, so the user gets a success message, even though there was no row updated.
Can anyone tell me what could it be?
Here's the code:
function update_sql($sql) {
$this->last_query = $sql;
$r = mysql_query($sql);
if (!$r) {
$this->last_error = mysql_error();
return false;
}
$rows = mysql_affected_rows();
if ($rows == 0) return true; // no rows were updated
else return $rows; }
This code returns 1.
That is because true will print out as "1" if you use echo. For debugging try using var_dump(), or let your function return 0 (which seems to me, in this case, the better option).
One little note; I think you should try to make your code a bit more readable (if the code in your question has the same layout as the code in your file). Try to indent code blocks, use separate lines for closing curly brackets, etc...
This is just a guess...
Maybe your function works as excepted? Maybe this piece of code if ($rows == 0) return true; works fine, and returns true but you treat that value as integer (boolean true can be displayed as 1)? Do: var_dump(uddated_sql('YOUR QUERY')) and check whether it returns boolean true or integer 1 value.