我有
enter code here
function ajaxFunction(){
var ajaxRequest; // The variable that makes Ajax possible!
try{
// Opera 8.0+, Firefox, Safari
ajaxRequest = new XMLHttpRequest();
} catch (e){
// Internet Explorer Browsers
try{
ajaxRequest = new ActiveXObject("Msxml2.XMLHTTP");
} catch (e) {
try{
ajaxRequest = new ActiveXObject("Microsoft.XMLHTTP");
} catch (e){
// Something went wrong
alert("Your browser is too old to run me!");
return false;
}
}
}
// Create a function that will receive data sent from the server
ajaxRequest.onreadystatechange = function(){
if(ajaxRequest.readyState == 4){
$.post('userfind.php', function(data) {
document.getElementById("myTable").style.display = "block";
var x=document.getElementById("myTable");
for (var i = 0; i < data.length; i++) {
var row=x.insertRow(-1);
var cell1=row.insertCell(-1);
var cell2=row.insertCell(-1);
...
...
和php
enter code here
<?php
session_start();
$username = "XXXXXXXX";
$password = "XXXXXXXX";
$database = "XXXXXXXX";
$link = mysql_connect("localhost", "$username", "$password");
if(!$link) {echo("Failed to establish connection to mysql server");
exit();}
$status = mysql_select_db($database);
$oId = mysql_real_escape_string($_POST["order_IDsearch"]);
if (isset($order_IDsearch)){
$result = mysql_query ("SELECT * FROM personal_info WHERE order_id= '".$oId."' ");
$myjsons = array();
while($row = mysql_fetch_assoc($result)){
$myjsons[] = $row;
}
echo json_encode($myjsons);
}
?>
如果我删除SQL条件，javascript将显示该表，标记if并标记$post
如果我离开php，它就不会显示表格，如您所见，
php页面帮助有什么问题，请
这是整个javascript ajax函数
enter code here
function ajaxFunction(){
var ajaxRequest; // The variable that makes Ajax possible!
try{
// Opera 8.0+, Firefox, Safari
ajaxRequest = new XMLHttpRequest();
} catch (e){
// Internet Explorer Browsers
try{
ajaxRequest = new ActiveXObject("Msxml2.XMLHTTP");
} catch (e) {
try{
ajaxRequest = new ActiveXObject("Microsoft.XMLHTTP");
} catch (e){
// Something went wrong
alert("Your browser is too old to run me!");
return false;
}
}
}
// Create a function that will receive data sent from the server
ajaxRequest.onreadystatechange = function(){
if(ajaxRequest.readyState == 4){
$.post('userfind.php', {orderId:"order_IDsearch"}, function(data) {
var obj = $("#myTable").show();
var x = obj.get(0);
for (var i = 0; i < data.length; i++) {
var row=x.insertRow(-1);
var cell1=row.insertCell(-1);
...
...
cell1.innerHTML = "<b><input name='edit' type='button' onClick='editRow(this)' value='Edit' /> <input name='del' type='button' onClick='delRow(this)' value='Del' /></b>";
cell2.innerHTML = data[i].user_id;
cell3.innerHTML = data[i].first_name ;
....
....
}},'json');}
}
ajaxRequest.open("POST", "userfind.php", true);
ajaxRequest.send(null);
}
这让我很困惑，请帮我修改代码，我应该重新编写代码吗？这个代码可以修改吗？
您没有从$.post向php文件传递任何内容。。。
让它像
$.post('userfind.php', {order_IDsearch: "your data"}, function(data){
// your implementation
});
这应该有效。。。
如果你的订单_IDsearch是动态的，那么就这样做吧
$.post('userfind.php', {yourData:order_IDsearch}, function(data){
// your implementation
});
在PHP方面，您必须在中访问它
$_POST['yourData'];
发送多个值
$.post('userfind.php', {key1:value1,key2:value2,...}, function(data){
// your implementation
});
Aditya Parab确实回答了您的问题，但我必须插入一些您需要理解的jQuery基础知识。
你可以（也应该）戏剧性地简化你的功能。所有ajaxRequest的内容，代码的前20行左右，都是在不使用jQueryAjax时使用的。这些东西被内置到jQuery中，你不必这么做。
此外，在jQuery中，您可以通过声明var object = $('#objectId').get(0)来获取对象的ID。您不需要document.getElementsById()。
简化功能
function ajaxFunction(){
$.post('userfind.php', {var1: "value 1"}, function(data) {
var obj = $("#myTable").show();
var x = obj.get(0); //Get the JS object from the jQuery Object
for (var i = 0; i < data.length; i++) {
var row=x.insertRow(-1);
var cell1=row.insertCell(-1);
var cell2=row.insertCell(-1);
}
...
...
});
}