I’m trying to store an array in a session variable then use it latter like this:
Session["sessionVariable"] = searchString;
Now here I’m trying to store the session variable into a string variable.
String[] sv = Session["sessionVariable"];
When I do I get his error.
Cannot explicitly convert type ‘Object’ to ‘String[]’, An explicit conversion exists, (are you mission a cast?)
I’ve tried various conversions but can’t find the correct one. Can you please help me to find the correct conversion? Thanks.
Try this:
Set:
Session["test"] = new string[] { "1", "2", "3" };
Get:
string[] array = Session["test"] as string[];
Related
I am trying to get the first row and the last row of a query result. I see from Ci4's docs, there are two methods to help, namley, getFirstRow([$type = 'object']) and getLastRow([$type = 'object']) but I am having difficulty using them. Here is my method so far:
function getLoginFailCount($login_fail_ip, $max_login_attempts = 3, $within_seconds = 320){
$builder = $this->builder('login_fail');
$builder->where('login_fail_ip', $login_fail_ip);
$builder->orderBy('login_fail_created_at','DESC');
$query = $builder->get(3);
print_r($query->getFirstRow($query));
}
I get an error at getFirstRow as follows;
Argument 1 passed to CodeIgniter\Database\BaseResult::getFirstRow() must be
of the type string, object given
How can I get getFirstRow() to work? Doesn't this doc definition say I need to pass it an object? Why does the error say it my be of type string
Well in the documentation for getFirstRow() it states that you can use
$row = $query->getFirstRow() // which will give you an object
OR
$row = $query->getFirstRow(‘array’) // which will give you an Array
So your error message, which states...
Argument 1 passed to CodeIgniter\Database\BaseResult::getFirstRow()
must be of the type string, object given
Would make you look and say to yourself, I had better go and read the documentation. So you can either pass in nothing, or a String 'array'.
So now can you see why
$query->getFirstRow($query))
does not make any sense! Why would you pass in the $query object as parameter.
You may have misread the documentation. I see you stated getFirstRow([$type = 'object'])
You might have got a little confused by that...
[$type = 'object'] means that the $type is defaulted to be the string 'object' so the returned type is an object by default with No Parameter being passed in.
If you want it to return an array, then you would specify the string 'array'. So then the $type parameter would be set to the string 'array' and return an array instead of an object.
Does that help!
I have a String based Key-Value Array inside of a String, and I want to decoded it and assign the value to an existing array in Swift 4.2. For example:
let array: [String:String] = []
let stringToDecode = “[\“Hello\”:\”World\”, \"Key\":\"Value\"]”
// I want ‘array’ to be assigned
// to the value that is inside
// ‘stringToDecode’
I’ve tried the JSON decoder, but it couldn’t decode it. Is there a simple way to do this? Thank you.
Try using a library like SwiftyJson, it makes working with json much easier.
I need to get a list of all existing servers in the application.conf file, I take a look to EBean class, but i only found how to get an specific server Ebean.getServer("test"), also this returns an EbeanServer object, and i need a string value.
This is part of my application.conf:
db.default.driver=oracle.jdbc.OracleDriver
db.default.url="jdbc:oracle:thin:#//178.20.26.25:1521/orcl"
db.default.user="TEST1"
db.default.password="test1"
db.test.driver=oracle.jdbc.OracleDriver
db.test.url="jdbc:oracle:thin:#//178.20.26.26:1521/orcl"
db.test.user="TEST"
db.test.password="test"
ebean.default="models.*"
ebean.test="models.*"
My expected output is a list that contains (default,test). Does anybody know a way to get this without parsing hole file?
Thanks in advance.
Following code will give set instead of list:
Map<String, String> map = (Map<String, String>) play.Play.application().configuration().getObject("db");
Set<String> keys = map.keySet();
If you want to do it in type safe way and get rid of compiler warning:
Set<String> keys = play.Play.application().configuration().getConfig("db").subKeys();
Both examples will return subkeys of db key which is [default, test].
I'm trying to do the following update using XPages Extension Library.
#{javascript:var mydata = {
product: getComponent("inputProduct").getValue()
};
var params = [1, 2];,
#JdbcUpdate("mssql","table_name",mydata,"order_no=? AND order_line_no=?",params)};
I get the error:
Error while executing JavaScript action expression
Script interpreter error, line=6, col=1: Error while executing function '#JdbcUpdate'
Invalid column name 'PRODUCT'.
The problem is that XPages when it converts the JSON it puts product to PRODUCT.
Can you set the extension library to respect the case of the JSON and not convert to Uppercase? Or can anyone point to where this setting could be set if not the extension library?
Thanks
The problem is com.ibm.xsp.extlib.util.JdbcUtil.appendColumnName()
public static void appendColumnName(StringBuilder b, String colName) {
colName = colName.toUpperCase();
b.append(colName);
}
This just needs changing to not upper case the variable.
There may be other methods that need changing if other variables are getting upper cased.
When using the MongoDB shell, how do I use a guid datatype (which I have used as the _id in my collection).
The following format doesn't work:
>db.person.find({"_id","E3E45566-AFE4-A564-7876-AEFF6745FF"});
Thanks.
You can use easily:
.find({ "_id" : CSUUID("E3E45566-AFE4-A564-7876-AEFF6745FF")})
You have to compare the _id value against an instance of BinData (not against a string). Unfortunately the BinData constructor takes a Base64 string instead of a hex string.
Your GUID value is missing two hex digits at the end, so for the purposes of this example I will assume they are "00". The following values are equivalent:
hex: "E3E45566-AFE4-A564-7876-AEFF6745FF00" (ignoring dashes)
base64: "ZlXk4+SvZKV4dq7/Z0X/AA=="
So your query should be:
>db.person.find({_id : new BinData(3, "ZlXk4+SvZKV4dq7/Z0X/AA==")})
I am assuming that the binary subtype was correctly set to 3. If not, what driver was used to create the data?
You could use the following js function in front of your query like so:
function LUUID(uuid) {
var hex = uuid.replace(/[{}-]/g, ""); // removes extra characters
return new UUID(hex); //creates new UUID
}
db.person.find({"_id" : LUUID("E3E45566-AFE4-A564-7876-AEFF6745FF"});
You could save the function in .js file and load it or open it before you make your query and if you copy the value from your results you should rename the function with:
LUUID for Legacy UUID
JUUID for Java encoding
NUUID for .net encoding
CSUUID for c# encoding
PYUUID for python encoding
I know it's an old issue, but without any additional needs you can use this one:
find({_id:UUID('af64ab4f-1098-458a-a0a3-f0f6c93530b7')})
You can fix this issue by using split() and join() workaround:
for instance if I use "E3E45566-AFE4-A564-7876-AEFF6745FF" hex value with - inside UUID() function, it does not return BinData in mongo so please try removing all the - before passing to UUID function.
db.person.find({"_id":UUID("E3E45566-AFE4-A564-7876-AEFF6745FF".split("-").join(''))});
Or by defining a variable to do it in multiple line:
var uuid = UUID("E3E45566-AFE4-A564-7876-AEFF6745FF".split("-").join(''))
db.person.find({"_id":uuid});
or by creating a simple function:
function BUUID(uuid){
var str = uuid.split("-").join('');
return new UUID(str);
}
db.person.find({"_id": BUUID("E3E45566-AFE4-A564-7876-AEFF6745FF")}).pretty();