我正在为我在学校的项目做一些添加、编辑和删除。添加模块中的代码运行良好,实际上我添加了一些记录。然后,这里是编辑模块,起初它非常好,从添加模块使用了类似的代码。但是当我尝试尝试时,编辑模块中的帖子是空的。
这是我的编辑代码:
$(".careersEdit").click(function () {
var careersTableSelect = encodeURIComponent($("input:radio[name=careersTableSelect]:checked").val());
if (careersTableSelect > 0) {
$(".careersEditForm_load").show();
$(".careersEditForm_error").hide();
$(".careersEditForm").hide();
var dataStringCareersEdit = 'careersTableSelect=' + careersTableSelect;
$.ajax({
type: "POST",
url: "admin/careers/process/careersEditGet.php",
data: dataStringCareersEdit,
beforeSend: function(){
alert(dataStringCareersEdit);
},
success: function () {
setTimeout("", 5000);
fetchResult();
},
error: function () {
alert("Post Error");
}
});
function fetchResult() {
$.ajax({
url: "admin/careers/process/careersEditGet.php",
type: "POST",
dataType: "json",
success: function (result) {
if (result) {
$("input#careersEditPosition").val(result['position']);
$("input#careersEditCompany").val(result['company']);
$("input#careersEditLocation").val(result['location']);
$(".careersEditForm_load").hide();
$(".careersEditForm").show();
}
},
error: function () {
alert("Fetch Error");
}
});
}
} else {
$(".careersEditForm").hide();
$(".careersEditForm_load").hide();
$(".careersEditForm_error").show();
}
});
以下是职业编辑获取.php:
<?php
include('connect.php');
error_reporting(0);
$careersTableSelect = $_POST['careersTableSelect'];
//$careersTableSelect = $careersTableSelect + 1;
//echo $careersTableSelect;
$query = "SELECT * FROM atsdatabase.admincareers WHERE refNum ='" . $careersTableSelect . "' LIMIT 0 , 30";
$runQuery = mysql_query($query);
if (!$runQuery) {
die('Could not enter data: ' . mysql_error());
}
$result = mysql_fetch_row($runQuery);
$array = array(
'position' => "" . $result[1] . "",
'company' => "" . $result[2] . "",
'location' => "" . $result[3] . "",
);
echo json_encode($array);
mysql_close($connection);
?>
是的,代码很丑/错误/废话,我对jquery的东西很陌生,大约3-4天。对于那些有帮助的人,请纠正我。我想学习这个jquery ajax的东西。Gracias
也许尝试以更常见的方式传递数据:
改变
data: dataStringCareersEdit,
自
data: { "careersTableSelect" : careersTableSelect },
调用你的ajax
函数一次,
$.ajax({
url: "admin/careers/process/careersEditGet.php",
type: "POST",
dataType: "json",
data: {careersTableSelect: careersTableSelect},
success: function (result) {
if (result) {
$("input#careersEditPosition").val(result.position);// json not array
$("input#careersEditCompany").val(result.company);// json not array
$("input#careersEditLocation").val(result.location);// json not array
$(".careersEditForm_load").hide();
$(".careersEditForm").show();
}
},
error: function () {
alert("Fetch Error");
}
});
感谢大家为回答这个问题所做的一切努力,我咨询了一位 Web 开发人员的朋友,教我如何在 jquery 中正确使用 ajax。 ;)
当你从jQuery.Ajax发布数据时,你做了一些根本错误的事情。数据应该是一个对象,键应该是服务器端 POST 变量的名称,稍后将在 PHP 中使用......例:
data : {"server_side_vriable" : "Your_data_to_Post" }
......
var dataStringCareersEdit = 'careersTableSelect=' + careersTableSelect + "&careersTableSelect=" + careersTableSelect;
$.ajax({
type: "POST",
url: "admin/careers/process/careersEditGet.php",
data: {"careersTableSelect" : dataStringCareersEdit},
beforeSend: alert(dataStringCareersEdit),
success: function () {
alert("Fetching Result");
setTimeout("", 3000);
$.ajax({
url: "admin/careers/process/careersEditGet.php",
type: "GET",
dataType: "json",
success: function (result) {
if (result) {
$("input#careersEditPosition").val(result['position']);
$("input#careersEditCompany").val(result['company']);
$("input#careersEditLocation").val(result['location']);
$(".careersEditForm_load").hide();
$(".careersEditForm").show();
}
},
error: function () {
alert("Fetch Error");
}
});
},
error: function () {
alert("Post Error");
}
});