<form action="" method="post" id="ajax_form" >
<table id="myTab">
<tr>
<th>id</th>
<th>name</th>
<th>sex</th>
<th>age</th>
<th>password</th>
<th>operation</th>
</tr>
<tr>
<td><input type="text" name="adminId" class="data" value="" /></td>
<td><input type="text" name="adminName" class="data" value="" /></td>
<td><input type="text" name="adminSex" class="data" value="" /></td>
<td><input type="text" name="adminAge" class="data" value="" /></td>
<td><input type="text" name="adminPassword" class="data" value="" /></td>
<td><input type="button" value="保存" onclick="addAdmin(this)" />
<input type="button" value="更新" onclick="updateAdmin(this)" /></td>
</tr>
</table>
</form>function updateAdmin(input) {
debugger;
var formData = $('form').serialize();
var id = $('input[name="adminId"]').val();
var name = $('input[name="adminName"]').val();
var sex = $('input[name="adminSex"]').val();
var age = $('input[name="adminAge"]').val();
var password = $('input[name="adminPassword"]').val();
$.ajax({
url: "/demo/public/index.php/admin/index/postajax",
dataType: "json",
type: 'POST',
data: {
adminId:id,
adminName: name,
adminSex: sex,
adminAge: age,
adminPassword: password,
},
contentType: 'application/json;charset=utf-8',
success: function(result) {
console.log(result);
},
error:function () {
alert("error");
}
});
}下面是php控制器里的方法public function postAjax() {
$Admin = new Admin();
$data = [
'adminId' => input('param.adminId'),
'adminName' => input('param.adminName'),
'adminSex' => input('param.adminSex'),
'adminAge' => input('param.adminAge'),
'adminPassword' => input('param.adminPassword')
];
return $data;前台console出的数据{adminId: null, adminName: null, adminSex: null, adminAge: null, adminPassword: null} 最佳答案
