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나는 솔직히 이걸 어디서 시작해야할지 전혀 모른다. 기본 html로 처리하는 방법을 알고 있지만 데이터베이스에서 질문을 가져와야하지만이 방법이 훨씬 복잡하다고 느낍니다.라디오 선택에 따라 더 많은 라디오 버튼을 표시 하시겠습니까?
혼란스러운 점은 for 루프를 사용하여 모든 최상위 수준 질문을 표시해야한다는 것입니다. 그런 다음 최상위 레벨 질문에 예를 선택하면 하위 레벨 질문이 표시됩니다. 지금은 저 레벨 질문의 잠금을 해제하는 방법을 모르기 때문에 최상위 질문 만 표시합니다.
여기 내 HTML 코드입니다.
<!doctype html>
<html>
<head>
<script src="http://code.jquery.com/jquery-1.11.0.min.js"></script>
<script>
jQuery(function($){
// Prevents errors in old IE.
if(!window.console || !window.console.log) window.console = {log: function(){}};
// Enable debugging for complex logic.
var debug = true;
function checkConditions(inputs, views){
views.each(function(){
var view = $(this);
var conditions = view.attr('data-view-conditions').split(',');
var display = true;
// Loop over all required conditions to be met.
for(var i = 0; i < conditions.length; i++){
var name = conditions[i].split(':')[0];
var value = conditions[i].split(':')[1];
var input = inputs.filter('[name="' + name + '"]:checked');
if(debug) console.log('\nRecalculating view conditions.');
if(!input.length){
display = false;
if(debug) console.log('View not show as no :checked input with the name "' + name + '" was found.');
}else{
if(value != undefined && input.val() != value){
display = false;
if(debug) console.log('View not show as no :checked input with the name "' + name + '" has the value of "' + input.val() + '". Value needed was: "' + value + '".');
}
}
}
if(display){
view.css({display: 'block'});
}else{
view.css({display: 'none'});
}
});
}
$('form').each(function(){
var form = $(this);
var inputs = form.find('input[type="checkbox"], input[type="radio"]');
var views = form.find('[data-view-conditions]');
// Recalculate which views to be shown each time the inputs change.
inputs.on('change', function(e){
checkConditions(inputs, views);
});
// Check conditions to set up required view state on page load.
checkConditions(inputs, views);
});
});
</script>
<!--<link href="css/bootstrap.min.css" rel="stylesheet">-->
<meta charset="utf-8">
<title>Submit a Ticket</title>
<link rel="stylesheet" href="http://yui.yahooapis.com/pure/0.4.2/forms-min.css">
<style>
.myForm {
margin-left: auto;
margin-right: auto;
max-width: 768px;
}
</style>
</head>
<body>
<form name="ticket1" class="pure-form pure-form-stacked myForm" action="flashpoint_processing.php" method="post">
<h2>Submit A Ticket</h2>
<legend> Your Details </legend>
<label>
<span>First Name: </span>
<input id="FirstName" name="FirstName" type="text" placeholder="" class="input-xlarge" required>
</label>
<label>
<label>
<span>Last Name: </span>
<input id="LastName" name="LastName" type="text" placeholder="" class="input-xlarge" required>
</label>
<label>
<span> Business Name </span>
<input id="Busname" name="Busname" type="text" placeholder="" class="input-xlarge">
</label>
<label>Phone</label>
<input id="Phone" name="Phone" type="text" placeholder="" class="input-xlarge" required>
<label>Extension</label>
<input id="Ext" name="Ext" type="text" placeholder="" class="input-small">
<label> E-mail</label>
<input id="email" name="email" type="text" placeholder="" class="input-xlarge" required>
<label>Preferred Method of Contact</label>
<input type="radio" name="contact" id="contact-0" value="Phone" checked="checked">
Phone
<input type="radio" name="contact" id="contact-1" value="Email">
Email
<label>
<legend>Issue</legend>
<?
$topq = array();
$topid = array();
$results = mysql_query("select questions, ID from tbl_questions where top=1");
while($row = mysql_fetch_array($results)) {
$topq[] = $row['questions'];
$topid[] = $row['ID'];
};
for($i=0;$i<count($topq);$i++){
echo "<label>";
echo $topq[$i];
echo "</label>";
echo " <input type=\"radio\" name=\"question+$i\" id=\"$i\" value=\"1\" checked=\"checked\">No
<input type=\"radio\" name=\"question+$i\" id=\"$i+1\" value=\"2\">Yes";
echo $topid[$i];
echo "<br>";
};
?>
<br>
<input type="submit" name="ticket" value="Submit">
</form>
다음은 내가하려는 일의 예입니다. http://jcsites.juniata.edu/students/bookhjr10/flashpoint/test6.html
질문을 하드 코딩하면 쉽게 할 수 있지만 데이터베이스에서 가져올 수 있습니다.