Invalid argument supplied for foreach() issue in laravel - mysql

My query in controller page
$d= "DB::table('users')->where('gender',$gender)";
if(!empty($age)){
$d.= "->where('age','>',$age)";
}
$d.="->paginate(20)";
if age is not empty that time only want add age to the query. but query showing Invalid argument supplied for foreach() error. Without double quotes query running but unable to avoid empty age variable.

Shorter version
$d = DB::table('users')->where('gender',$gender);
$d = empty($age) ? $d->paginate(20) : $d->where('age','>',$age)->paginate(20);

Please try it:
$d = DB::table('users')->where('gender',$gender);
if(! empty($age)) {
$d = $d->where('age','>',$age);
}
$d = $d->paginate(20);

Related

Return variable from one function to another in powershell

I am inside function #1 and triggering function #2. Function #2 has a variable that I want to get back into function #1 and use it.
My output ends up being:
hey there
var is:
What I want is the output to be:
var is: hey there
Why is it that I can feed a function a variable, and it uses it, but when I change that variable in the #2 function, it does not change the variable after it returns?
$var = $null
function one() {
two($var)
write-host "var is:" $var
}
function two($var){
$var = "hey there"
return($var)
}
clear
one
First, change your two function to actually return a value:
function two {
$var = "hey there"
return $var
}
and then update the one function to actually "capture" the output value by assigning it to a variable:
function one {
# PowerShell doesn't use parentheses when calling functions
$var = two
Write-Host "var is:" $var
}
If you really do want to reach up and change a variable in the scope of your caller, you can...
Set-Variable -Name $Name -Value $Value -Scope 1
It's a little meta/quirky but I've run into cases where it seemed proper. For example, I once wrote a function called RestoreState that I called at the beginning of several functions exported from a module to handle some messy reentrant cases; one of the things it did is reach up and set or clear $Verbose in the calling function's scope.

Trying to pass a string back from a function in PowerShell

function ExtractLocations([ref]$lp_Locations) {
$lp_Locations.Value = "A STRING VALUE"
return 0
}
...
$Locations = ""
if (#(ExtractLocations([ref]$Locations)) -ne 0) {
RecordErrorThenExit
}
$Locations always ends up as a blank string.
Apart from what #mklemen0 said, to set the value, you need not to do it like $Variable.Value = 'Something' , its just $Variable = 'Something'
With the #() expression, you are converting the output to an array which is not what you need here. Declaring functions similar to methods in c# is not suggested in PowerShell. You could do it like below.
function ExtractLocations{
Param([ref]$lp_Locations)
$lp_Locations = "A STRING VALUE"
return 0
}
ExtractLocations -lp_Locations ([ref]$Locations)
Could not get it working with parameters no matter what I tried so returned the value as a string and that works:
function ExtractLocations
{
....
return $lp_Locations
}
$Locations = ExtractLocations
write "MODIFY $TABLE TO REORGANIZE WITH LOCATION = ($($Locations.value)) \p\g" | sql $SQLFLAGS $TARGETDB

Strange behaviour from MySQL in Powershell

I'm new to PowerShell and have a specific question about working with MySQL in PowerShell.
I got this function:
Function run-mySQLInsertQuery{
param(
$connection,
[string[]]$insertQuery
)
foreach ($command in $insertQuery){
$MySQLCommand = $connection.CreateCommand()
$MySQLCommand.CommandText = $command
$rowsInserted = $MySQLCommand.ExecuteNonQuery()
if ($rowsInserted) {
return $rowsInserted
} else {
return $false
}
}
}
With this version of the function i get the following Error:
Cause:
"The CommandText property has not been properly initialized."
Errorline:
$rowsInserted = $MySQLCommand.ExecuteNonQuery()
I searched for a solution and edited my function a bit to the following (for testing purpose):
Function run-mySQLInsertQuery{
param(
$connection,
[string[]]$insertQuery
)
$abcd = $insertQuery[1]
foreach ($command in $insertQuery){
$MySQLCommand = $connection.CreateCommand()
$MySQLCommand.CommandText = $abcd
$rowsInserted = $MySQLCommand.ExecuteNonQuery()
}
}
With this code, the function executes the query without a problem. My question now is, why? i cant really see a difference, because in $command should be the exact same query like it is in $abcd. Or am I getting something wrong?
EDIT:
As its asked it the comments, here is how i call the function:
[String[]]$statements = ""
foreach($key in $arrayStatus.Keys){
$item = $arrayStatus[$key]
$insertStatus = "INSERT INTO tx_tphbusinessofferings_domain_model_status (status_id, status) VALUES ('$key', '$item')"
$statements += $insertStatus
}
$Rows = run-mySQLInsertQuery -connection $mySQLconnection -insertQuery $statements
The problem is that you are initializing your array (the one you are passing in) with an empty string:
[String[]]$statements = ""
And then adding elements to it... so your first iteration of the passed array is an empty string, which won't work (it'll set the command text as empty, that's the error you are getting). It works on the second code because you are grabbing the second object of the array (which is your insert statement).
Initialize your array to empty and it should work:
[String[]]$statements = #()
Apart from that, your first script always returns on the first iteration, so it'll only work once (not for every insert you pass). Not sure what do you want to return if you are passing in more than one query, but that's up to your design decisions

'Not an ARRAY reference' error thrown

I'm writing a Perl script that is meant to deal with an API which returns metrics about a set of URLs that I pull from MySQL then post these metrics back into a different table. Currently this piece of code:
my $content = $response->content;
my $jsontext = json_to_perl($content);
my $adsql = 'INSERT INTO moz (url_id,page_authority,domain_authority,links,MozRank_URL,MozRank_Subdomain,external_equity_links) VALUES (?,?,?,?,?,?,?)';
my $adrs = $db->prepare( $adsql );
my $adsql2 = 'UPDATE url
SET moz_crawl_date = NOW()
where url_id = ?;';
my $adrs2 = $db->prepare( $adsql2 );
my $currentUrlId = 0;
foreach my $row (#$jsontext){
$adrs->execute($url_ids[$currentUrlId], $row->{'fmrp'}, $row->{'upa'}, $row->{'pda'}, $row->{'uid'}, $row->{'umrp'}, $row->{'ueid'});# || &die_clean("Couldn't execute\n$adsql\n".$db->errstr."\n" );
$adrs2->execute($url_ids[$currentUrlId]);
$currentUrlId++;
}
is throwing this error:
Not an ARRAY reference at ./moz2.pl line 124.
this is line 124:
foreach my $row (#$jsontext){
this whole chunk of code is in a while loop. I am actually able to iterate a couple times and fill my MySQL table before the script fails (technically the program works, but I don't want to just leave an error in it).
Anybody have any suggestions?
Perl gave you the correct answer
Not an ARRAY reference: #$jsontext
You are dereferencing $jsontext, which is the result of json_to_perl(string), to an array.
But json_to_perl() didn't return an arrayref.
json_to_perl seems to be from this API: http://search.cpan.org/~bkb/JSON-Parse-0.31/lib/JSON/Parse.pod#json_to_perl
which returns according to the doc either an arrayref or a hashref.
Apparently it did return a hashref in your case, so you have to add the logic to deal with the HASH case. Which seems to be a single row.
if (ref $jsontext eq 'HASH') {
# seems to be a single row
$adrs->execute($url_ids[$currentUrlId], $jsontext->{'fmrp'}, $jsontext->'upa'}, $jsontext->'pda'}, $jsontext->'uid'}, $jsontext->'umrp'}, $jsontext->'ueid'});# || &die_clean("Couldn't execute\n$adsql\n".$db->errstr."\n" );
$adrs2->execute($url_ids[$currentUrlId]);
$currentUrlId++;
} elsif (ref $jsontext eq 'ARRAY') {
foreach my $row (#$jsontext){
$adrs->execute($url_ids[$currentUrlId], $row->{'fmrp'}, $row->{'upa'}, $row->{'pda'}, $row->{'uid'}, $row->{'umrp'}, $row->{'ueid'});# || &die_clean("Couldn't execute\n$adsql\n".$db->errstr."\n" );
$adrs2->execute($url_ids[$currentUrlId]);
$currentUrlId++;
}
}

PHP force a var to have a certain type

This may be a stupid question but I might aswell as it :)
is there away to force
$tel1 = '05';// string
settype($tel1,'string');
$tel1 = 06;//either throw error, or convert it to string automatically.
var_dump($tel1);//(string [2]) 05
The above code is of the top of my head so might not be accurate but I need to keep a variable as a string not numeric, because of some silly thing I have done, now my phone numbers lose the leading 0s :-(
n I cn't rewrite it because it will mess up with other numeric types,b4 u ask it was an automated service for db to check if it was a numeric value or not,
UPDATE
This is the problem
function escape($str){
if(is_numeric($str)){
return $str;
}else{
return "'".mysql_real_escape_string($str).'\'';
}
}
$tel1 = "06";
$sql = 'SELECT * FROM blabla WHERE id = '.escape($tel1).'';
//above is same as below
$sql = 'SELECT * FROM blabla WHERE id = 06 ';
I can't change anything inside the scape function because other inputes thruout the website are using this function, I dont wanna mess their validations.
Your use of is_numeric tests for numeric content, not an integer type. But then you take a variable called $str which implies you want it to be a string.
Perhaps use:
function escape($val) {
if (is_numeric($val) && !is_string($val)) {
return $val;
}
else{
return "'" . mysql_real_escape_string($val) . '\'';
}
}
Now strings will be escaped and quoted, but not if they contain only numeric content.
you can do something like:
$string = (string) $int;
or use a function
$string = strval($int);
You can't force a variable to a specific type in the global scope.
You can force Arrays and Objects in a function.
function getElementsByClassName(DOMNode $parentElement, Array $classNames) {
...
}
If you pass an object that is not an instantiation of DOMNode (or a subclass), or if you don't pass an Array as the second argument, you'll get an error.
You can of course cast any variable, e.g. (string) $tel1.
You shouldn't be treating phone numbers as Ints anyway, because of leading zeroes and possible parenthesis and dashes. Also, once your telephone number is an Int, it won't know its 0 padding anymore because it will be discarded, so casting it back won't give you the original String.
To cast a variable you can use something like:
$i = 1;
$s = (string) $i;
Depending on the db adaptor you might not be able to detect the type being returned from the database. I believe it's PDO that returns everything (even INT values) as strings.
The number_format() function may be of use to you too.
If you declare a variable as:
$var = 06;
it immediately becomes 6 without leading zero because leading zero when it comes to integers is meaningless and therefore it's cut out.
In other words, your variable has to be created as string, which is what you probably deduced yourself.
Quick fix would be the following: you can add another parameter to your escape() function.
For example:
function escape($str, $force_str = false)
{
if($force_str)
{
// do your conversion, the rest of the site will by default pass false so nothing will be broken
}
}
As alex said, start by making sure the phone number is never converted from string to int in your own code. Then, you need to make sure it will not be converted when sent to your SQL DB.
It ought to work if you do it this way:
$sql = "SELECT * FROM blabla WHERE id = '" . mysql_real_escape_string($tel1) . "'";
This is the same as
$sql = "SELECT * FROM blabla WHERE id = '06'";