[Solved] Auto Insert into MySql Data Base

This is quite simple: 1st what you want to do is to query the database for the previous registration no /ID. something like this: $query = mysqli_query($connect,’SELECT regCode From data ORDER BY regCode DESC LIMIT 1′); then you check if there is any result like this: $results = count(mysqli_num_rows($query)); if it $results returns 0 (it … Read more

[Solved] Select preferred address if exists otherwise select ‘work’ address

This should do: SELECT * FROM peopleaffiliation pa LEFT OUTER JOIN addresses addr ON pa.addressid = addr.addressid LEFT OUTER JOIN addresstypes addtyp ON addtyp.addresstypeid = addr.addresstypeid WHERE (addr.preferredaddress = 1 ) OR( COALESCE(addr.preferredaddress, 0) = 0 AND addtyp.addresstypeconst=”Work” ) 0 solved Select preferred address if exists otherwise select ‘work’ address

[Solved] DIV on Google map [closed]

Try <div class=”map-container”> <div class=”b-map-content” id=”map-content”></div> <div class=”temp”></div> </div> When Google Maps initializes, I am guessing that it removes all inner elements within id=”map-content”, which is why you are seeing the flash of red. 1 solved DIV on Google map [closed]

[Solved] How to read and parse this JSON using Jackson? [closed]

Jackson provides many ways of reading this JSON. One simple way would be doing something like this: Map<String, Object> result = new ObjectMapper().readValue(“JSON_Input_Here”, Map.class); Additionally, you could do something like: JsonNode input = new ObjectMapper().readTree(“JSON_Input_Here”); I’m not sure how a map would handle a Json array, but the JsonNode object lets you check the type … Read more

[Solved] Jquery getScript showing ReferenceError: $ is not defined [duplicate]

Make sure that jQuery is properly loaded. Use firebug as a Firefox debugger or the Chrome console. Are you accessing a different domain? If so you will need to use a json callback. $.ajax({ url: “http://ihound.com.au/livechat/php/app.php/widget-init.js”, jsonpCallback: “jsonpcallback”, jsonp: false, dataType: “jsonp” }).done(function(data){ console.log(data); // array of objects }); 4 solved Jquery getScript showing ReferenceError: … Read more

[Solved] i want to scrape this part

This data is taken from additional request to https://www.seloger.com/detail,json,caracteristique_bien.json?idannonce=142632059. There you will get json with whole information. UPD: url_id = re.search(r’/(\d+)\.htm’, response.url).group(1) details_url=”https://www.seloger.com/detail,json,caracteristique_bien.json?idannonce={}” # make request to url yield Request(details_url.format(url_id)) 5 solved i want to scrape this part

[Solved] Getting string between exact letters by regex in Swift

As per my comment, this is a task for DateFormatter rather than RegeX. I threw this together in a playground quickly to demonstrate what I mean. let inFormatter = DateFormatter() inFormatter.locale = Locale(identifier: “en_US_POSIX”) inFormatter.dateFormat = “yyyy-MM-dd’T’HH:mm:ssZZZZZ” let input = “2019-03-11T17:04:00+0100” let dateFromInput = inFormatter.date(from: input)! // This should be unwrapped properly in your code. … Read more

[Solved] How to swap rows in SQL Server

Put the original values in a temporary table with the IDs swapped then join to the temporary table whilst updating the table, example code below: –drop temp table if exists IF OBJECT_ID(‘tempdb..#Temp’, ‘U’) IS NOT NULL DROP TABLE #Temp –need to store the original values SELECT *,CASE WHEN Room_Number=9104 then 9103 ELSE 9104 END AS … Read more

[Solved] Wondering why my app crashes

Looking at your code it seems that you made a typo here: TextView DialogText=(TextView) findViewById(R.id.imageViewDialog); It should be TextView DialogText=(TextView) findViewById(R.id.textViewDialog); or whatever the name of the id for this TextView is. Apart from that, you need to learn to read the messages from the LogCat as they are often very informative. Cleary it says: … Read more

[Solved] Display table values in textboxes [closed]

try something like this…. using (SqlConnection con = new SqlConnection(strConnect)) { con.Open(); using (SqlCommand com = new SqlCommand(“SELECT FirstName, LastName, EmailId, MobileNUmber FROM myTable WHERE id=@ID”, con)) { com.Parameters.AddWithValue(“@ID”, iD); DataTable dt = new DataTable(); SqlDataAdapter da = new SqlDataAdapter(com); da.Fill(dt); txtFirstName.Text = dt.Rows[0][“FirstName”]; txtName.Text = dt.Rows[0][“LastName”]; } } 3 solved Display table values in … Read more