我构建的JS/Ajax函数没有按钮单击或页面刷新。该函数获取输入字段的值,并用php回显结果。但每次变量都被回显时,下一个变量会擦除前一个变量的值。如何避免这种情况? EXAMPLEPHP/JS:在不丢失值的情况下响应多个变量
JS
<script>
$(document).ready(function() {
var timer = null;
var dataString;
function submitForm(){
$.ajax({ type: "POST",
url: "index.php",
data: dataString,
success: function(result){
$('#special').html('<p>' + $('#resultval', result).html() + '</p>');
}
});
return false; }
$('#contact_name').on('keyup', function() {
clearTimeout(timer);
timer = setTimeout(submitForm, 050);
var name = $("#contact_name").val();
dataString = 'name='+ name;
});
$('#email').on('keyup', function() {
clearTimeout(timer);
timer = setTimeout(submitForm, 050);
var name = $("#email").val();
dataString = 'name='+ name;
});
$('#phone').on('keyup', function() {
clearTimeout(timer);
timer = setTimeout(submitForm, 050);
var name = $("#phone").val();
dataString = 'name='+ name;
});
$('#address').on('keyup', function() {
clearTimeout(timer);
timer = setTimeout(submitForm, 050);
var name = $("#address").val();
dataString = 'name='+ name;
});
$('#website').on('keyup', function() {
clearTimeout(timer);
timer = setTimeout(submitForm, 050);
var name = $("#website").val();
dataString = 'name='+ name;
});
});
</script>
HTML/PHP
<form action="" method="post" enctype="multipart/form-data" id="contact_form" name="form4">
<div class="row">
<div class="label">Contact Name *</div> <!-- end .label -->
<div class="input">
<input type="text" id="contact_name" class="detail" name="contact_name" value="<?php $contact_name ?>" />
<div id="special"><span id="resultval"></span></div>
</div><!-- end .input-->
</div><!-- end .row -->
<div class="row">
<div class="label">Email Address *</div> <!-- end .label -->
<div class="input">
<input type="text" id="email" class="detail" name="email" value="<?php $email ?>" />
<div id="special"><span id="resultval"></span></div>
</div><!-- end .input-->
</div><!-- end .row -->
</form>
谢谢!关于ID的一个问题:我如何使ID唯一? – CodingWonders90 2012-07-20 05:47:05
@cholomanCoding欢迎你,他们不应该是类似的,例如你有两个id为'#special'的元素,你可以把它们改成'#special1'和'#special2',或者简单地使用一个类如'.special '。 – undefined 2012-07-20 05:49:38
噢,好的。因此,如果我将div更改为'
'和'',我还需要添加:$('#special1')。append(''+ result +'
');' '$(' (''#special2')。append(''+ result +'
');' – CodingWonders90 2012-07-20 06:05:09