我希望您能够使用jQuery进行提交表单,而不是重新加载页面,但是数据并没有显示出来,并且表单正在重新加载,这是不必要的。
jQuery代码:
function submitFormData() {
var firstval = $("#first").val();
var second = $("#second").val();
//var operator = $("#myselect option:selected" ).text();
$.post("index.php",{first:first,second:second},
function(data){
$('#results').html(data);
$('#formcal')[0].reset();
});
}
以下是同一页面(index.php)的HTML代码:
<form action="" id="formcal" method="post">
<input type="number" id="first" name="first" placeholder="number"/>
<select name="operator" id="operator">
<option value="add">+</option>
<option value = "subtract">-</option>
<option value = "multiply">*</option>
<option value = "division">/</option>
</select>
<input type="number" id="second" name="second" placeholder="number 2"/>
<input type="button" id="submitFormData" onclick="SubmitFormData();" value="Calculate"/>
</form>
<br>
<?php //if(!empty($_POST['first']) && !empty($_POST['second'])){
$number = $_POST['first'];
$number2 = $_POST['second'];
echo "Answer: ";
if($_POST['operator'] == 'add'){
$complete = $number + $number2;
echo " $number + $number2 = $complete";
}
if($_POST['operator'] == 'subtract'){
$complete = $number - $number2;
echo "$number - $number2 = $complete";
}
if($_POST['operator'] == 'multiply'){
$complete = $number * $number2;
echo "$number X $number2 = $complete";
}
if($_POST['operator'] == 'division'){
$complete = $number / $number2;
echo "$number / $number2 = $complete";
}
//}
?>
</div>
<div id="results">
</div>