Заполнить поле SELECT из массива
Вывод массива из БД в поле select
Здравствуйте. В проекте есть поле "Выбор фотосессий", в БД записано как `ps_types`. Выбирается из.
Как при выборе определённого значения поля select показать новое поле select
Такая задача, у меня есть список значений в поле типа select. Value1 Value2 Value3 Value4 При.
Как связать поле список-() поле text и checkbox
есть поле text — фамилия, потом поле с датой, поле список — наименование товара, поле text — кол-во.
foreach ($array as $key => $value) { echo $key; }
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$array = array ( "Ru" => array ( "1"=>"a", "2"=>"б", "3"=>"в", "4"=>"г", "5"=>"д" ), "En" => array ( "6"=>"a", "7"=>"b", "8"=>"c", "9"=>"d", "10"=>"e" ) ); ?> select> option value=""> /option> ($array as $k1=>$v1)< echo "optgroup label='$k1'>"; foreach($v1 as $k2=>$v2) echo "option value='$k2'>$v2/option>"; echo "/optgroup>"; >?> /select>
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$array = array ( "Ru" => array ( "1"=>"a", "2"=>"б", "3"=>"в", "4"=>"г", "5"=>"д" ), "En" => array ( "6"=>"a", "7"=>"b", "8"=>"c", "9"=>"d", "10"=>"e" )); echo '; echo ''; foreach($array as $a => $b ){ echo '.$a.'>'; foreach($b as $k => $v){ echo '.$v.''; } } echo '';
Накинулись все Про лейбл забыли — вроде как ТС спрашивал
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. echo ''; echo '; echo ' '; $i = 0; foreach ($array as $key => $value){ echo '.$key.'">'; foreach ($value as $k => $v){ echo '.$v.''; $i++; } } echo '';
Запрос sql UPDATE .SET поле=м.поле FROM (SELECT.) AS м
Вечер добрый, граждане, подскажите пожалуйста. можно ли в Access реализовать запрос вида .
Заполнить поле случайным числом, если другое соответствующее поле не пустое
est table. mne neobhodimo(v sql server) zapolnit pole ‘Number_of_Reviews’ sluchainimi chislami v.
Заполнить select значениями
На странице есть пустой объект select <select При нажатии на кнопку.
Заполнить тег select опциями
Добрый день. делаю гаджет на Win8 — 10 — отправка отзывов на сотрудников. В форме гаджета, в HTML.
Alec’s Web Log
Here’s a super obvious php snippet that generates a list of options for a “select” html form tag on the fly:
Combined with the javascript from an earlier post, it should be very easy to generate selectors for php based forms.
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7 Responses to “Php generate html select options from array”
I’d prefer this version which is simpler for web integrators: Select…
$v): ?>
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How To Fill A DropDown List From Array Using Javascript
Dear,
Thanks for your code, it’s great. But I want to select the Option where month[i][2] = «1», there can be more than one selected.
How can I do this? Here is my code.
var month = [[«1», «January», «»], [[«3», «Mars», «»], [«4», «April», «»], [«5», «May», «»], [«2», «February», «1»]]
var select = document.getElementById(‘select_table’);
for (var i=0; i < month.length;++i)var option = document.createElement("OPTION"),
txt = document.createTextNode(month[i][1]);
option.appendChild(txt);
option.setAttribute(«value»,month[i][0]);
select.insertBefore(option,select.lastchild);
>
Hello Sir,
How to pass the API response to select box of html form
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Show selected option value from Array & MySQL DB using PHP
In this tutorial, you will learn how to create an array of categories, display the values inside HTML select box and have selected options pre-defined with PHP and also I will show you how to get the selected options from a database using the ID of the record.
body < font-size: 2em; >.container < display: grid; grid-template-rows: repeat(2, 1fr); grid-template-columns: repeat(2, 1fr); grid-gap: 6rem; padding: 5rem; border: 1px solid #ccc; justify-content: space-evenly; >selectSelected From ArraySelected From DB Record"; foreach($options as $option)< if($selected == $option) < echo ""; > else < echo ""; > > echo ""; ?>if(isset($_GET['category'])) < $categoryName = $_GET['category']; $sql = "SELECT * FROM categories WHERE if($result = mysqli_query($link, $sql)) < if(mysqli_num_rows($result) >0) < while($row = mysqli_fetch_array($result))< $dbselected = $row['category']; >// Function frees the memory associated with the result mysqli_free_result($result); > else < echo "Something went wrong. "; >> else < echo "ERROR: Could not execute $sql." . mysql_error($link); >> $options = array('Comedy', 'Adventure', 'Drama', 'Crime', 'Adult', 'Horror'); echo ""; ?>