0
這是我第一週的PHP(所以不要欺負)。我試圖抓住選擇表單的值並插入到我的數據庫中。插入是好的,但我似乎無法通過的價值。如何在AJAX中獲取選擇表單的值?
腳本:
<script type="text/javascript">
//add to top 10
function addTop10()
{
var show_id = $('#show_id').val();
var user_id = $('#user_id').val();
var rank = $('select.rank').val();
//Storing the value of textbox into a variable
if(show_id == '') //Checking for NULL
{
$('#propspectDiv').html('Enter A Valid Name'); //Prints the progress text into our Progress DIV
$('#show_id').addClass('error'); //Adding the error class to the progress DIV
return;
}
else{
$('#show_id').removeClass('error');
$('#propspectDiv').removeClass('error'); //Removing the error class from the progress DIV
$('#propspectDiv').html('Submitting your Request.<img src="ajax.gif" />');//Prints the progress text into our Progress DIV
$.ajax({
url : 'top10ajax.php', //Declaration of file, in which we will send the data
data:{
"show_id" : show_id,
"user_id" : user_id,
"rank" : rank //we are passing the name value in URL
},
success : function(data){
window.setTimeout(function(){
$('#propspectDiv').html('Added'); //Prints the progress text into our Progress DIV
}, 2000);
}
});
}
}
</script>
和HTML
<div class="row">
<div class="twelve columns">
<h4>Add to your top 10 </h4>
<div class="two columns">
<form>
<select name="rank">
<option value="1">1</option>
<option value="2">2</option>
<option value="3">3</option>
<option value="4">4</option>
<option value="5">5</option>
<option value="6">6</option>
<option value="7">7</option>
<option value="8">8</option>
<option value="9">9</option>
<option value="10">10</option>
</select>
</div>
<div class="four columns">
<button class="large button" id="rank" type="button" value="Add to Top 10" onclick="addTop10()"></form>
</div>
我在做什麼錯,除了糟糕的格式?
我看不出有任何PHP .. – dbf 2013-05-05 13:17:33
我看到你提到的格式問題 - 在一般情況下,它的修復是一個好主意在發佈之前,爲了使您的代碼更易於閱讀。 – halfer 2013-05-05 13:24:30
我假設'$'是jQuery對象,並已相應地重新簽署了您的問題。 – halfer 2013-05-05 13:27:20