var table = $('#accessRequest').DataTable({
"dom": 'Blfrtip',
"ajax": "Editor-PHP-1.5.1/php/editor-serverside.php",
"columns": [
{ //special column for detail view
"className": 'details-control',
"orderable": false,
"data": null,
"defaultContent": ''
},
{ data: "DT_RowId",
render: function ( data ) {
return data.replace( 'row_', '' );
}
},
{ data: null,
render: function ( data, type, row ) {
// Combine the first and last names into a single table field
return data.first+' '+data.last;
}
},
{ "data": "phone" },
{ "data": "responsibleParty" },
{ "data": "email" },
{ "data": "building" },
{ "data": "typeOfWork" },
{ "data": "startTime" },
{ "data": "endTime" },
{ "data": "description" },
{ "data": "dockNeeded" },
{ "data": "numPeople" },
{ "data": "numTrucks" },
{ "data": "requestPlaced" },
{ "data": "updatedAt" },
{ "data": "approved" },
{ "data": "approvedBy" },
{ "data": "approvedAt" },
{ "data": null }
],
"aoColumnDefs": [
{
"aTargets": [-1],
"mData": null,
"mRender": function (data, type, full) {
return '<button id="ApprovalButton" onclick="$.post(\'extra.php\', \'approve_request\')" action="extra.php" method="post"> Process </button>';
//Send post request
}
}
],
"order": [[1, 'asc']],
select: true,
buttons: [
{ extend: "create", editor: editor },
{ extend: "edit", editor: editor },
{ extend: "remove", editor: editor }
]
});
我有一串代码(上面),我在其中创建一个包含信息的表。在“//发送帖子请求”行上,我试图使其上方的按钮将帖子请求发送到名为“extra.php”的单独文件,尽管我做了什么,但我还没有这样做没能做到这一点。
我尝试在输入表单中使用提交按钮,但这不起作用。我尝试了很多不同的事情,但我似乎无法让它发挥作用。任何帮助将不胜感激。
已经有一个文件(extra.php)能够接收发布请求并在收到请求后对其进行操作。我正在尝试在页面上的 HTML 中创建一个按钮,单击该按钮时会将“approve_request”发布到文件“extra.php”
抱歉,我对编码和这个网站都是新手,可能有一些简单的事情我没有做。
<!DOCTYPE html>
<html>
<head>
</head>
<body>
<?php
header('Access-Control-Allow-Origin: *');
$id = intval($_POST['id']);
$con = mysql_connect("RequestAccess.db.10160035.hostedresource.com","RequestAccess","br@6HeCher");
if (!$con){
die('Could not connect: ' . mysql_error());
}
mysql_select_db("RequestAccess", $con);
//if action sent is approve set request as apporved
if(isset($_POST['approve_request'])){
$sql = "UPDATE AccessRequests
SET approved='1', approvedAt=current_timestamp
WHERE AccessID='" . $_POST['approve_request']."'";
$results = mysql_query($sql);
}
//else action sent must be for child rows so populate child row
else {
}
这是负责处理发布请求的部分(如果有帮助的话)。
最佳答案
首先,某些浏览器会阻止使用 JS 发送请求,因此请将其添加到 PHP 代码之上
header('Access-Control-Allow-Origin: *');
现在您可以毫无问题地发送请求。
$.post(url ,{
postItem1 : postValue1,
postItem2: postValue2
//You have to define postItem1 and 2 in your PHP code -> $_POST["postItem1"] -> this will return postValue1
}, function(data, status){
//success
}).fail(function() {
//failed
});
关于javascript - 如何向 PHP 元素发送 post 请求,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/34189476/