I/m facing this issue. Can anyone assist? Thanks
You have a null safety. Just add '!' after data. Example snapshot!.data!.[i].name
To complement KOrra's answer,
In line 44, you have AsyncSnaphot.data.lenght. It should be snapshot.data.length
I am trying to set up a PayPal checkout with this tutorial:
https://developer.paypal.com/docs/checkout/integrate/#
I am facing a problem with step 6. Verify the transaction
The php script throws:
Parse error: syntax error, unexpected 'class' (T_CLASS), expecting identifier (T_STRING) or variable (T_VARIABLE) or '{' or '$' in /vendor/braintree/braintreehttp/lib/BraintreeHttp/HttpClient.php on line 38
in the Browser console.
function __construct(Environment $environment)
{
$this->environment = $environment;
$this->encoder = new Encoder();
$this->curlCls = Curl::class;
}
The last line is 38. Anybody an idea what is going wrong here?
I had the same error and commented out line that sets curlCls
$this->curlCls = Curl::class;
It looks it is not used anywhere and the basic PayPal example started to work.
<?php
$con=mysqli_connect("localhost","root");
if (!$con){
die("X");
}
$select=mysqli_select_db("$con","template1");
if (!$select){
die("XX");
}
?>
Catchable fatal error: Object of class mysqli could not be converted to string in C:\xampp\htdocs\login\login.php on line 7
So basically it is just a try out however i get the above error on my laptop and i would like to know why it happen.
Thanks in advance
Here's the line that needs to be fixed.
$select=mysqli_select_db("$con","template1");
By doing "$con" you're trying to convert $con to string. Just remove the quotes.
$select=mysqli_select_db($con,"template1");
Whenever I run the code, it says “Consecutive Declarations On A Line Must Be Separated By ','”
var obstacleOffset = CGFloat(movementAmount) - self.frame.size.width / 4)
Can someone please help me? I don't understand what is wrong.
It looks to me like you have an extra parenthesis at the end of your statement.
Good Luck!
I have a program that is dependant on PostgreSQL, It is suddenly giving me the error
SQLSTATE[22007]: Invalid datetime format: 7 ERROR: invalid input syntax for type time: "00:00:01."
LINE 1: ...:55:02.027392', '00:00:00.042608', '00:55:02.07', '00:00:01....
It occurred after loading some data into a folder that is monitored for changes to update the database. I have since removed ALL the data from that folder. I have searched for several hours, I am unable to find anything giving me direction where to start.
postgresql-9.1-main.log
2015-04-05 12:26:15 EDT ERROR: invalid input syntax for type time: "00:00:01." at character 234
2015-04-05 12:26:15 EDT STATEMENT: INSERT INTO cc_schedule (starts, ends, cue_in, cue_out, fade_in, fade_out, clip_length, position, instance_id, file_id, stream_id) VALUES ('2015-04-06 17:00:00.000000', '2015-04-06 17:55:02.027392', '00:00:00.042608', '00:55:02.07', '00:00:01.', '00:00:01.', '00:55:02.027392', 0, 25, 1717, null) RETURNING id
I would appreciate any advice. I am thinking there is just a bungled entry somewhere that could be removed? or do I need to reinstall PostgreSQL ?
you have two time entries with a dot at the end of time, that cause the error:
00:00:01.
fix the data to
00:00:01
I found the issue, I had changed a configuration as related to Track Fade time. When doing so from "0.50" to "1." leaving . at the end bungled it. Thank You for your time and assistance.