所以我有一个问题,
如果我有这样的php文件:
<?php
print
"<div class='col-lg-6 col-md-6 col-sm-6 col-xs-12 col-md-pull-6 col-sm-pull-6'>".
"<div class='form-group'>".
"<div class='col-md-12'><strong>Forename:</strong></div>".
"<div class='col-md-12'><input type='text' class='form-control' id='forename'></div>".
"</div>".
"<div class='form-group'>".
"<div class='col-md-12'><strong>Surname:</strong></div>".
"<div class='col-md-12'><input type='text' class='form-control' id='surname'></div>".
"</div>".
"<div class='form-group'>".
"<div class='col-xs-12 col-md-6'>".
"<button id='button' class='btn btn-success btn_add'>Add</button>".
"</div>".
"</div>".
"</div>".
"</div>";
}
?>
<script type="text/javascript">
$(document).on("click", 'button.btn_add', function(){
var forename = document.getElementById('forename').value;
var surname = document.getElementById('surname').value;
if((forename == null || forename == "") && (surname == null || surname == "")){
alert("Please fill in all fields");
}
else{
// DO SQL UPDATE VIA PHP
}
});
</script>
如何在jquery中的else case上进行sql更新?理想情况下,我想验证然后执行sql语句
答案 0 :(得分:1)
查看$.ajax
它可以让你做你想要的事。
.. code
else {
$.ajax({
url: some_url_to_execute_query.php,
method: 'post',
dataType: 'json',
data: 'forename=' + forename + '&surname=' + surname
})
.done(function(response) {
if (response.success) {
// notify user query worked
}
else {
// query didn't work; do something
}
}
}
PHP(some_url_to_execute_query.php)
<?php
$response = [];
// execute your MySQL query here
// make sure to connect to database, etc.
// access $_POST array values using $_POST['forename'] && $_POST['surname']
// then something like so (pseudo):
if ($query) // $query is pseudo variable holding (bool) response from query to determine whether is passed/failed
$response['success'] = true;
else
$response['success'] = false;
echo json_encode($response);
答案 1 :(得分:0)
你只能从服务器端进行SQL查询,你需要一些API来实现它。
您可以使用php打印JavaScript,但无法使用JavaScript从客户端执行php。
答案 2 :(得分:0)
你需要使用jquery进行ajax调用才能运行sql查询。
http://api.jquery.com/jquery.ajax/
所以你需要一个单独的php脚本来运行sql查询,然后你将使用ajax从你的表单中发布值。
答案 3 :(得分:0)
像这样:
请注意type =&#34;按钮&#34;
Imports Microsoft.Office.Interop
Dim Access_Object As New Access.Application
Access_Object = CreateObject("Access.Application")
Access_Object.Application.Visible = True