请求两个 Ajax
我正在尝试进行两次 Ajax 调用来获取数据以填充网页的不同部分,正如您已经知道的那样,只会发生第二次。
所以我想我应该这样做:
callAjax1('a'); callAjax2('b');
function callAjax1(data) {
ajax(data);
}
function callAjax2(data) {
ajax(data);
}
function ajax(data) {
// calls XMLHttpRequestObject etc
}
这个想法是,现在我将拥有两个可以独立运行的独立 ajax 实例,而不是调用 ajax() 两次。
它有效..但前提是我在 ajax() 顶部添加警报以让我知道我已经到达。
所以我认为警报会在调用第二个请求之前给第一个请求时间来完成。因此,我没有设法将它们正确地分成单独的实例。那不可能吗?
我缺少什么?
一切顺利 更新
: 我在想,我还有机会吗?
tParams = new Array (2); // we intend to call ajax twice
tParams[0] = new Array('ajaxGetDataController.php', 'PROJECT', 'id');
tParams[1] = new Array('ajaxGetFileController.php', 'FILE', 'projectId');
<select name='projectSelector' onchange=\"saveData(tParams, this.value);\">\n";
// gets called, twice
function saveData(pParams, pData) // pParams are: PageToRun, Table, Field
{
if (XMLHttpRequestObject)
{
tPage = pParams[0][0]+'?table='+pParams[0][1]+'&pField='+pParams[0][2]+'&pData='+pData;
XMLHttpRequestObject.open('GET', tPage);\n
XMLHttpRequestObject.onreadystatechange = callAjax(pParams, pData);
XMLHttpRequestObject.send(null);
}
}
function callAjax(pParams, pData)
{
if (XMLHttpRequestObject.readyState == 4 && XMLHttpRequestObject.status == 200)
{
var tReceived = XMLHttpRequestObject.responseXML;
options = tReceived.getElementsByTagName('option'); // fields and their values stored in simplest XML as options
popForm(options, pParams[0][1]); // goes off to use the DOM to populate the onscreen form
pParams.shift(); // cuts off pParams[0] and moves all elements up one
if (pParams.length>0)
{
saveData(pParams, pData);
}
}
}
I'm trying to make two Ajax calls to get data to populate different bits of a web page, and as you'll already know, only the second happens.
So I thought I'd do this:
callAjax1('a'); callAjax2('b');
function callAjax1(data) {
ajax(data);
}
function callAjax2(data) {
ajax(data);
}
function ajax(data) {
// calls XMLHttpRequestObject etc
}
The idea was that instead of calling ajax() twice, now, I'd have two independent instances of ajax that would run independently.
It works .. but only if I put in an alert at the top of ajax() to let me know I've arrived.
So I'm thinking that alert gives the first request time to finish before the second is called. Therefore, I've not managed to separate them properly into separate instances. Is that not possible?
What am I missing?
All the best
J
UPDATE:
I'm thinking this, do I stand a chance?
tParams = new Array (2); // we intend to call ajax twice
tParams[0] = new Array('ajaxGetDataController.php', 'PROJECT', 'id');
tParams[1] = new Array('ajaxGetFileController.php', 'FILE', 'projectId');
<select name='projectSelector' onchange=\"saveData(tParams, this.value);\">\n";
// gets called, twice
function saveData(pParams, pData) // pParams are: PageToRun, Table, Field
{
if (XMLHttpRequestObject)
{
tPage = pParams[0][0]+'?table='+pParams[0][1]+'&pField='+pParams[0][2]+'&pData='+pData;
XMLHttpRequestObject.open('GET', tPage);\n
XMLHttpRequestObject.onreadystatechange = callAjax(pParams, pData);
XMLHttpRequestObject.send(null);
}
}
function callAjax(pParams, pData)
{
if (XMLHttpRequestObject.readyState == 4 && XMLHttpRequestObject.status == 200)
{
var tReceived = XMLHttpRequestObject.responseXML;
options = tReceived.getElementsByTagName('option'); // fields and their values stored in simplest XML as options
popForm(options, pParams[0][1]); // goes off to use the DOM to populate the onscreen form
pParams.shift(); // cuts off pParams[0] and moves all elements up one
if (pParams.length>0)
{
saveData(pParams, pData);
}
}
}
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我将为 AJAX 函数创建一个就绪状态变量:
然后在执行第二个调用之前检查就绪状态:
并确保在执行 AJAX 调用后将 ReadyState 更改回 false。这将确保第一个 AJAX 调用在第二个 AJAX 调用尝试触发之前完成执行。
I would create a ready state variable for the AJAX function:
And then check for the ready state before executing the second call:
And make sure to change the readyState back to false after the AJAX calls have executed. This will ensure the first AJAX call has finished executing before the second one tries to fire.