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我有一個應用程序,我試圖讓單選按鈕和複選框將信息輸入到數據庫中。目前,只有文本字段能夠正確輸入數據庫中的信息,但複選框和單選按鈕僅將第一個單選按鈕的值輸入到數據庫中。因此,例如使用單選按鈕,即使選擇了「否」,「是」也會被輸入到數據庫中。我不確定我做錯了什麼。單選按鈕/複選框沒有使用AJAX輸入到MySQL數據庫中
下面是相關的HTML功能:
function saveUserInfo(userName, userEmail, optedIn, currentUser, catDog, otherBrand)
{
if(optedIn === "Yes")
{
finalAnswers = userName + "yes" + "~" + userEmail + "~" + optedIn + "~" + currentUser + "~" + catDog + "~" + otherBrand + "~";
}else{
finalAnswers = userName + "no" + "~" + userEmail + "~" + optedIn + "~" + currentUser + "~" + catDog + "~" + otherBrand + "~";
// finalAnswers = userName + "~" + userEmail + "~0" + optedIn + "~1" + currentUser + "~2" + catDog + "~3" + otherBrand + "~4";
}
try
{
$.ajax({
url: surveyServer + finalAnswers,
type: 'post',
success: function(evt){
console.log('success saving ' + finalAnswers);
}
});
}
catch(err)
{
alert(err.message);
}
}
function saveUser()
{
$('.wrapper').find("#userEmail").blur();
if (($('#userName').val().length > 0) && ($('#userEmail').val().length > 0))
{
var testEmail = /^[A-Z0-9._%+-][email protected]([A-Z0-9-]+\.)+[A-Z]{2,4}$/i
var valueToTest = $('#userEmail').val();
if (testEmail.test(valueToTest))
{
//pass
saveUserInfo($('#userName').val(), $('#userEmail').val(), $('#optedIn').val(), $('#currentUser').val(), $('#catDog').val(), $('#otherBrand').val());
$('.wrapper').slick('slickNext');
} else {
// Do whatever if it fails.
alert("Please enter a valid email address.");
$('.wrapper').find("#userEmail").focus();
}
}else{
alert("Please enter your name and email address then click Submit.");
$('.wrapper').find("#userName").focus();
}
}
這裏是surveySurver代碼:
if (isset($_GET['user']))
{
try
{
$dbh = new PDO('mysql:host=localhost; dbname=bp1234', 'bp1234','bp1234');
$userAnswerArray = explode("~", $_GET['user']);
$stmt = $dbh->prepare("INSERT INTO bpQuiz (userName, userEmail, optedIn, currentUser, catDog, otherBrand) VALUES (:userName, :userEmail, :optedIn, :currentUser, :catDog, :otherBrand)");
$stmt->bindParam(':userName', $userName);
$stmt->bindParam(':userEmail', $userEmail);
$stmt->bindParam(':optedIn', $userOption);
$stmt->bindParam(':currentUser', $currentUser);
$stmt->bindParam(':catDog', $catDog);
$stmt->bindParam(':otherBrand', $otherBrand);
// insert value
$userName = $userAnswerArray[0];
$userEmail = $userAnswerArray[1];
$userOption = $userAnswerArray[2];
$currentUser = $userAnswerArray[3];
$catDog = $userAnswerArray[4];
$otherBrand = $userAnswerArray[5];
$stmt->execute();
}
catch (PDOException $e)
{
echo 'ERROR: ' . $e->getMessage();
}
}
else
{
echo "no querystring...";
}
良好的措施這裏的表單字段:
<input type="text" class="contactFormInputBox" name="userName" id="userName"/>
<input type="text" class="contactFormInputBox" name="userEmail" id="userEmail"/>
<input type="checkbox" name="optedIn" id="optedIn" value="Yes">
<input type="radio" name="currentUser" value="1" id="currentUser">Yes
<input type="radio" name="currentUser" value="0" id="currentUser">No
<input type="radio" name="catDog" value="cat" id="catDog">Cat
<input type="radio" name="catDog" value="dog" id="catDog">Dog
<input type="radio" name="catDog" value="both" id="catDog">Both
</div>
<div class="surveyOptions" id="Q2-b">
<input type="text" class="contactFormInputBox" name="otherBrand" id="otherBrand"/>
從無線電卸下'ID ='鈕釦。 ID應該是唯一的。無線電的'name ='not id匹配,並且按名稱發佈,而不是id,但是id可能會令人困惑。 –
謝謝,我嘗試了這一點,它將這三個值設置爲undefined。但後來我用JQuery來拉取單選按鈕的值,它工作!在下面發佈答案。 – Hennessey
對不起,應該看起來更接近於你解析表單的JS,問題是用'$('#userName').val()' - 這總是會給你第一個匹配的元素(不管它是否是ID) ,而不是* selected *元素。你可能已經改成'$(「#userName:selected」).val()'。但關鍵是使用':selected'。 –