[英]Empty $_POST when posting from jquery.ajax
我正在為學校的項目做一些添加,編輯和刪除操作。 add模塊中的代碼運行良好,實際上我已經添加了幾條記錄。 然后,出現了Edit模塊,起初它非常好,在add模塊中使用了類似的代碼。 但是,當我嘗試嘗試時,edit模塊中的帖子為空。
這是我的修改代碼:
$(".careersEdit").click(function () {
var careersTableSelect = encodeURIComponent($("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();
}
});
這是careersEditGet.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,代碼很丑/錯/胡扯,我對jquery東西還很陌生,大約3-4天。 對於那些有幫助的人,請糾正我。 我想學習這個jquery ajax的東西。 格拉西亞斯
也許嘗試以更常見的方式傳遞數據:
更改
data: dataStringCareersEdit,
至
data: { "careersTableSelect" : careersTableSelect },
一次調用ajax
函數,
$.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");
}
});
謝謝你們為回答這個問題所做的所有努力,我已經咨詢了一位Web開發人員,並教會了我如何在jquery中正確使用ajax。 ;)
當您從jQuery.Ajax中發布數據時,您做的根本是錯誤的。數據應該是一個對象,鍵應該是服務器端POST變量的名稱,稍后將在PHP中使用...示例:
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");
}
});
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