[ad_1] Try this : DEMO HERE HTML: <ul id=’menu’><li></li><li></li> </ul> jQuery: var array = [“one”, “two”]; $(‘ul#menu li’).append(function(index){ $(this).text(‘Number ‘+array[index]); }); 3 [ad_2] solved Populate li elements with values from array
[ad_1] There are a lot if issues with the code you have, so I’ve just written something new… $result = [“<mo>+</mo><mi>x</mi><mo>=</mo><mfrac><mrow><mo>-</mo><mi>b</mi><mo>±</mo><msqrt><msup><mi>b</mi><mn>2</mn></msup><mo>-</mo><mn>4</mn><mi>a</mi><mi>c</mi></msqrt></mrow><mrow><mn>2</mn><mi>a</mi></mrow></mfrac>”, “<mo>+</mo><mi>x</mi><mo>+</mo><mn>2</mn><mo>=</mo><mn>3</mn>”, “<mfrac><mrow><mn>2</mn><mi>x</mi><mo>+</mo><mn>2</mn></mrow><mn>3</mn></mfrac><mo>-</mo><mn>3</mn><mo>=</mo><mn>2</mn>”, “<mo>-</mo><mn>3</mn><mo>+</mo><mn>2</mn><mo>=</mo><mi>x</mi>” ]; $arr_result=[]; for ($i=0; $i < count($result) ; $i++) { if (substr($result[$i],0,4)!=”<mo>”) { $arr_result[]= “<mo>+</mo>”.$result[$i]; } else { $arr_result[]= $result[$i]; } } print_r($arr_result); This just goes through each line at a … Read more
[ad_1] Okay, reviewing your HTML-template I’d suggest the following solution: extend your nagelpaletten-model by adding the field bestellmenge or just menge. e.g. export class nagelpaletten { constructor( bestellmenge: number, PKArtikelID: string, laenge: number, Gewicht: number, ME: number, Preis: number ) {} } Maybe your model is structured a little differently but mine is just supposed … Read more
[ad_1] for (int i = 0; i < arr.length; i++) { for (int a = 0; a < book.length; a++) { if (book[a].equalsIgnoreCase(arr[i])) { //IF I TRY TO PRINT ARR[I] HERE IT HAS A VALUE bi = 1; } } if (bi != 1) { dif[l] = arr[i]; //HERE ARR[I] IS NULL System.out.println(arr[i]); l = … Read more
[ad_1] You can do it : Create table Jquery: $(“#my_id_container”).html(“<table><thead></thead><tbody></tbody></table>”); Html: <div id=”my_id_container”></div> Add value of this tableau : $(“#my_id_container”).find(‘table tbody’).append(‘<tr>My new ligne</tr>’); [ad_2] solved how to create dynamic array and adding dynamic values using jQuery.
[ad_1] You need to transpose X and ‘B’ to get the correct result. Try the following: C = A * (X.’) + B.’; 0 [ad_2] solved multiply two arrays with different size in matlab
[ad_1] I assume you want to search in your AirbnbListing list. You can use Java Stream. Use the filter method for that: List<AirbnbListing> matchingListings = listings.stream() .filter(l -> “Surrey”.equals(l.getCity())) .collect(Collectors.toList()); If you want a list of all cities, you can use the map method: List<String> matchingListings = listings.stream() .map(l -> l.getCity()) .collect(Collectors.toList()); Additionally here is … Read more
[ad_1] So it looks like you are trying to string manipulate JSON to try to make some sort of associative array. Replacing the : with => is not the right move here… <?php include_once(‘simple_html_dom.php’); ?> <!DOCTYPE html> <html> <head> <title> HTML DOM Parser</title> </head> <body> <?php header(‘Content-Type: text/html; charset=utf-8’); set_time_limit(0); $html=file_get_html(‘https://www.monreseauplus.com/villes/’); $array[]=array(); $array3[]=array(); foreach($html->find(‘.ul. li.cat-item … Read more
[ad_1] Why is the first element of a class returned by Jarray.ToObject always null, even if the JSON itself has data assigned there? [ad_2] solved Why is the first element of a class returned by Jarray.ToObject always null, even if the JSON itself has data assigned there?
[ad_1] Use Array reduce() var result = res.reduce((accu, obj) => { accu[obj.url] = (accu[obj.url] || 0) + obj.views; return accu; }, {} ); Output from node CLI: > var result = res.reduce((accu, obj) => { accu[obj.url] = (accu[obj.url] || 0) + obj.views; return accu; }, {}); undefined > result { ‘/page1’: 16, ‘/page2’: 5, ‘/page3’: … Read more
[ad_1] You may want to pass a single dimension array: bool latin(int const * const arr, unsigned int capacity) { int counter = 0; int s; // Please initialize this. for (int row = 0; row < MAXIMUM_ROWS; ++row) { for (int column = 0; column < MAXIMUM_COLUMNS; ++column) { const int value = *(arr … Read more
[ad_1] This is how you would do it: Scanner scanner = new Scanner(System.in); String input = scanner.nextLine(); int[] array = new int[input.length()]; for(int i = 0; i < input.length(); i++){ array[i] = Integer.parseInt(Character.toString(input.charAt(i))); } String.charAt(int position) gets the Character located at a certain position in the String. Character.toString(Characters ch) converts a Character to a String. … Read more
[ad_1] Basically I want to look at the position in where all my “X” are. You can use np.where to get the indices for each axis, then zip them to get index tuples for all the locations: >>> list(zip(*np.where(Arr == “X”))) [(3, 2), (5, 4), (6, 1)] If you want the (row, column) “locations”, you … Read more
[ad_1] If you want to find the maximum sub array length which contains only continious r and b, here is a solution. The basic idea is using two cursor and greedy search. public static int findMaximum(char[] input) { int result = 0; int first = 0; int second = 0; while (input[first] == input[second]) { … Read more
[ad_1] Try this @Override public View getView(int position, View convertView, ViewGroup parent) { LayoutInflater inflater = LayoutInflater.from(mCtx); // View view = inflater.inflate(R.layout.list_publikasi,null,true); if(converView == null) { convertView = inflater.inflate(R.layout.item_user, parent, false); } TextView textNama = (TextView) convertView.findViewById(R.id.textNama); TextView textDetail = (TextView) convertView.findViewById(R.id.textDetail); TextView textStatus = (TextView) convertView.findViewById(R.id.textStatus); TextView textPeriode = (TextView) convertView.findViewById(R.id.textPeriode); Publikasi publikasi = … Read more