Javascript 使用ajax.send将数据发送到php
有人能帮我把这些数据发送到一个.php页面,在那里我可以在我的php页面上收到它吗 javascript:Javascript 使用ajax.send将数据发送到php,javascript,php,ajax,Javascript,Php,Ajax,有人能帮我把这些数据发送到一个.php页面,在那里我可以在我的php页面上收到它吗 javascript: postToSql(){ var ajax; if (window.XMLHttpRequest) { // code for IE7+, Firefox, Chrome, Opera, Safari ajax=new XMLHttpRequest(); } else { // code for IE6, IE5 ajax=new ActiveXObjec
postToSql(){
var ajax;
if (window.XMLHttpRequest)
{
// code for IE7+, Firefox, Chrome, Opera, Safari
ajax=new XMLHttpRequest();
}
else
{
// code for IE6, IE5
ajax=new ActiveXObject("Microsoft.XMLHTTP");
}
ajax.onreadystatechange=function()
{
if (ajax.readyState==4 && ajax.status==200)
{
alert(ajax.responseText); //receiving response
}
};
var name = $("#entry_1274804157").val();
//alert(name);
var company= $("#entry_1828184698").val();
var phone=$("#entry_2039177352").val();
var email=$("#entry_1545475878").val();
var comments=$("#entry_1846523632").val();
var params = {
"name":name,
"company":company,
"phone":phone,
"email":email,
"comments": comments
};
//var jsonText = JSON.stringify(params);
ajax.open("POST", "view/templates/includes/insertgoogle.php", false);
ajax.setRequestHeader("Content-type","application/x-www-form-urlencoded");
ajax.send("totalJsonStr="+params);
//alert(totalJsonStr);
// alert(params);
return true;
}
</script>
如果您想使用$\u GET 删除:
ajax.open("POST", "view/templates/includes/insertgoogle.php", true);
加:
有用链接:在哪里创建“ajax”对象
var ajax;
if (window.XMLHttpRequest)
{// code for IE7+, Firefox, Chrome, Opera, Safari
ajax=new XMLHttpRequest();
}
else
{// code for IE6, IE5
ajax=new ActiveXObject("Microsoft.XMLHTTP");
}
从这里开始:使用POST方法 完美的解决方案是,将所有值存储在一个数组中,并使用
json.stringify()
在php中,使用
json\u decode()
解码json字符串
更新
将此添加到javascript中
<script type="text/javascript">
function postToSql(){
var ajax;
if (window.XMLHttpRequest)
{// code for IE7+, Firefox, Chrome, Opera, Safari
ajax=new XMLHttpRequest();
}
else
{// code for IE6, IE5
ajax=new ActiveXObject("Microsoft.XMLHTTP");
}
ajax.onreadystatechange=function()
{
if (ajax.readyState==4 && ajax.status==200)
{
alert(ajax.responseText); //receiving response
}
}
var name = "1234";
var company= "1234";
var phone="1234";
var params = {
"name":name,
"company":company,
"phone":phone,
};
var jsonText = JSON.stringify(params);
ajax.open("POST", "view/templates/includes/insertgoogle.php", true);
ajax.setRequestHeader("Content-type","application/x-www-form-urlencoded");
ajax.send("totalJsonStr="+jsonText);
}
</script>
<form action="https://docs.google.com/asgsasdfasg/formResponse" method="POST" id="" target="_self" onsubmit="postToSql();return false;">
函数postToSql(){
var-ajax;
if(window.XMLHttpRequest)
{//IE7+、Firefox、Chrome、Opera、Safari的代码
ajax=新的XMLHttpRequest();
}
其他的
{//IE6、IE5的代码
ajax=新的ActiveXObject(“Microsoft.XMLHTTP”);
}
ajax.onreadystatechange=function()
{
if(ajax.readyState==4&&ajax.status==200)
{
警报(ajax.responseText);//接收响应
}
}
var name=“1234”;
var company=“1234”;
var phone=“1234”;
变量参数={
“姓名”:姓名,
“公司”:公司,
“电话”:电话,
};
var jsonText=JSON.stringify(params);
open(“POST”,“view/templates/includes/insertgoogle.php”,true);
setRequestHeader(“内容类型”、“应用程序/x-www-form-urlencoded”);
send(“totalJsonStr=“+jsonText”);
}
将此添加到php
<?php
if(isset($_POST["totalJsonStr"]))
{
$jsonVal = json_decode($_POST["totalJsonStr"]);
print $jsonVal->{'name'};
print $jsonVal->{'company'};
print $jsonVal->{'phone'};
}
else
{
die("No Data Found");
}
?>
由于没有任何东西对我有效,我最终使用了jquery ajax。我不一定要通过GET接收数据,我只想接收我的php页面上的数据…在帖子中给出的链接,你可以从那里获得完整的ajax代码。我在哪里向php发送数据?对不起,我错过了代码片段。现在检查更新的答案。我发现了问题,因为async设置为true,表单在onsubmit函数完成之前提交。现在请告诉我如何用php检索这些数据…json方式或ajax方式。ajax方式$name=$\u POST['name']不起作用。在回答中,我发现的问题是我在发布数据之前设置了请求头的脚本。现在检查更新后的答案。我的答案已经正确了。也许我应该告诉你,所有这些都发生在onsubmit函数调用上
var ajax;
if (window.XMLHttpRequest)
{// code for IE7+, Firefox, Chrome, Opera, Safari
ajax=new XMLHttpRequest();
}
else
{// code for IE6, IE5
ajax=new ActiveXObject("Microsoft.XMLHTTP");
}
<script type="text/javascript">
function postToSql(){
var ajax;
if (window.XMLHttpRequest)
{// code for IE7+, Firefox, Chrome, Opera, Safari
ajax=new XMLHttpRequest();
}
else
{// code for IE6, IE5
ajax=new ActiveXObject("Microsoft.XMLHTTP");
}
ajax.onreadystatechange=function()
{
if (ajax.readyState==4 && ajax.status==200)
{
alert(ajax.responseText); //receiving response
}
}
var name = "1234";
var company= "1234";
var phone="1234";
var params = {
"name":name,
"company":company,
"phone":phone,
};
var jsonText = JSON.stringify(params);
ajax.open("POST", "view/templates/includes/insertgoogle.php", true);
ajax.setRequestHeader("Content-type","application/x-www-form-urlencoded");
ajax.send("totalJsonStr="+jsonText);
}
</script>
<form action="https://docs.google.com/asgsasdfasg/formResponse" method="POST" id="" target="_self" onsubmit="postToSql();return false;">
<?php
if(isset($_POST["totalJsonStr"]))
{
$jsonVal = json_decode($_POST["totalJsonStr"]);
print $jsonVal->{'name'};
print $jsonVal->{'company'};
print $jsonVal->{'phone'};
}
else
{
die("No Data Found");
}
?>